## How Much Better Is Better?

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 CHAPTER 8 Hypothesis Testing

Objectives

After completing this chapter, you should be able to

lUnderstand the definitions used in hypothesis testing.

2State the null and alternative hypotheses.

3Find critical values for the z test.

4State the five steps used in hypothesis testing.

5Test means when s is known, using the z test.

6Test means when s is unknown, using the t test.

7Test proportions, using the z test.

8Test variances or standard deviations, using the chi-square test.

9Test hypotheses, using confidence intervals.

Outline

Introduction

8–2z Test for a Mean

8–3t Test for a Mean

8–4z Test for a Proportion

8–5χ2 Test for a Variance or Standard Deviation

8–6Confidence Intervals and Hypothesis Testing

Summary

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Statistics Today

How Much Better Is Better?

Suppose a school superintendent reads an article which states that the overall mean score for the SAT is 910. Furthermore, suppose that, for a sample of students, the average of the SAT scores in the superintendent’s school district is 960. Can the superintendent conclude that the students in his school district scored higher than average? At first glance, you might be inclined to say yes, since 960 is higher than 910. But recall that the means of samples vary about the population mean when samples are selected from a specific population. So the question arises, Is there a real difference in the means, or is the difference simply due to chance (i.e., sampling error)? In this chapter, you will learn how to answer that question by using statistics that explain hypothesis testing. See Statistics Today—Revisited for the answer. In this chapter, you will learn how to answer many questions of this type by using statistics that are explained in the theory of hypothesis testing.

Introduction

Researchers are interested in answering many types of questions. For example, a scientist might want to know whether the earth is warming up. A physician might want to know whether a new medication will lower a person’s blood pressure. An educator might wish to see whether a new teaching technique is better than a traditional one. A retail merchant might want to know whether the public prefers a certain color in a new line of fashion. Automobile manufacturers are interested in determining whether seat belts will reduce the severity of injuries caused by accidents. These types of questions can be addressed through statistical hypothesis testing, which is a decision-making process for evaluating claims about a population. In hypothesis testing, the researcher must define the population under study, state the particular hypotheses that will be investigated, give the significance level, select a sample from the population, collect the data, perform the calculations required for the statistical test, and reach a conclusion.

Hypotheses concerning parameters such as means and proportions can be investigated. There are two specific statistical tests used for hypotheses concerning means: the z test and the t test. This chapter will explain in detail the hypothesis-testing procedure along with the z test and the t test. In addition, a hypothesis-testing procedure for testing a single variance or standard deviation using the chi-square distribution is explained in Section 8–5.

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The three methods used to test hypotheses are

2.The P-value method

3.The confidence interval method

The traditional method will be explained first. It has been used since the hypothesis-testing method was formulated. A newer method, called the P-value method, has become popular with the advent of modern computers and high-powered statistical calculators. It will be explained at the end of Section 8–2. The third method, the confidence interval method, is explained in Section 8–6 and illustrates the relationship between hypothesis testing and confidence intervals.

Objective 1

Understand the definitions used in hypothesis testing.

Every hypothesis-testing situation begins with the statement of a hypothesis.

A statistical hypothesis is a conjecture about a population parameter. This conjecture may or may not be true.

There are two types of statistical hypotheses for each situation: the null hypothesis and the alternative hypothesis.

The null hypothesis, symbolized by H0, is a statistical hypothesis that states that there is no difference between a parameter and a specific value, or that there is no difference between two parameters.

The alternative hypothesis, symbolized by H1, is a statistical hypothesis that states the existence of a difference between a parameter and a specific value, or states that there is a difference between two parameters.

(Note: Although the definitions of null and alternative hypotheses given here use the word parameter, these definitions can be extended to include other terms such as distributions and randomness. This is explained in later chapters.)

As an illustration of how hypotheses should be stated, three different statistical studies will be used as examples.

Objective 2

State the null and alternative hypotheses.

Situation A  A medical researcher is interested in finding out whether a new medication will have any undesirable side effects. The researcher is particularly concerned with the pulse rate of the patients who take the medication. Will the pulse rate increase, decrease, or remain unchanged after a patient takes the medication?

Since the researcher knows that the mean pulse rate for the population under study is 82 beats per minute, the hypotheses for this situation are

H0: µ = 82andH1: µ ≠ 82

The null hypothesis specifies that the mean will remain unchanged, and the alternative hypothesis states that it will be different. This test is called a two-tailed test (a term that will be formally defined later in this section), since the possible side effects of the medicine could be to raise or lower the pulse rate.

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Situation B  A chemist invents an additive to increase the life of an automobile battery. If the mean lifetime of the automobile battery without the additive is 36 months, then her hypotheses are

H0: µ = 36andH1: µ > 36

In this situation, the chemist is interested only in increasing the lifetime of the batteries, so her alternative hypothesis is that the mean is greater than 36 months. The null hypothesis is that the mean is equal to 36 months. This test is called right-tailed, since the interest is in an increase only.

Unusual Stat

Sixty-three percent of people would rather hear bad news before hearing the good news.

Situation C  A contractor wishes to lower heating bills by using a special type of insulation in houses. If the average of the monthly heating bills is \$78, her hypotheses about heating costs with the use of insulation are

H0: µ = \$78andH1: µ < \$78

This test is a left-tailed test, since the contractor is interested only in lowering heating costs.

To state hypotheses correctly, researchers must translate the conjecture or claim from words into mathematical symbols. The basic symbols used are as follows:

Equal to      =

Greater than>

Not equal to≠

Less than    <

The null and alternative hypotheses are stated together, and the null hypothesis contains the equals sign, as shown (where k represents a specified number).

 Two-tailed test Right-tailed test Left-tailed test H0: µ = k H0: µ = k H0: µ = k H1: µ ≠ k H1: µ > k H1: µ < k

The formal definitions of the different types of tests are given later in this section.

In this book, the null hypothesis is always stated using the equals sign. This is done because in most professional journals, and when we test the null hypothesis, the assumption is that the mean, proportion, or standard deviation is equal to a given specific value. Also, when a researcher conducts a study, he or she is generally looking for evidence to support a claim. Therefore, the claim should be stated as the alternative hypothesis, i. e., using < or > or . Because of this, the alternative hypothesis is sometimes called the research hypothesis.

 Table 8–1 Hypothesis-Testing Common Phrases > < Is greater than Is less than Is above Is below Is higher than Is lower than Is longer than Is shorter than Is bigger than Is smaller than Is increased Is decreased or reduced from = ≠ Is equal to Is not equal to Is the same as Is different from Has not changed from Has changed from Is the same as Is not the same as

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A claim, though, can be stated as either the null hypothesis or the alternative hypothesis; however, the statistical evidence can only support the claim if it is the alternative hypothesis. Statistical evidence can be used to reject the claim if the claim is the null hypothesis. These facts are important when you are stating the conclusion of a statistical study.

Table 8–1 shows some common phrases that are used in hypotheses and conjectures, and the corresponding symbols. This table should be helpful in translating verbal conjectures into mathematical symbols.

Example 8–1

State the null and alternative hypotheses for each conjecture.

a.A researcher thinks that if expectant mothers use vitamin pills, the birth weight of the babies will increase. The average birth weight of the population is 8.6 pounds.

b.An engineer hypothesizes that the mean number of defects can be decreased in a manufacturing process of compact disks by using robots instead of humans for certain tasks. The mean number of defective disks per 1000 is 18.

c.A psychologist feels that playing soft music during a test will change the results of the test. The psychologist is not sure whether the grades will be higher or lower. In the past, the mean of the scores was 73.

Solution

a.H0: µ = 8.6 and H1: µ > 8.6

b.H0: µ = 18 and H1: µ < 18

c.H0: µ = 73 and H1: µ ≠ 73

After stating the hypothesis, the researcher designs the study. The researcher selects the correct statistical test, chooses an appropriate level of significance, and formulates a plan for conducting the study. In situation A, for instance, the researcher will select a sample of patients who will be given the drug. After allowing a suitable time for the drug to be absorbed, the researcher will measure each person’s pulse rate.

Recall that when samples of a specific size are selected from a population, the means of these samples will vary about the population mean, and the distribution of the sample means will be approximately normal when the sample size is 30 or more. (See Section 6–3.) So even if the null hypothesis is true, the mean of the pulse rates of the sample of patients will not, in most cases, be exactly equal to the population mean of 82 beats per minute. There are two possibilities. Either the null hypothesis is true, and the difference between the sample mean and the population mean is due to chance; or the null hypothesis is false, and the sample came from a population whose mean is not 82 beats per minute but is some other value that is not known. These situations are shown in Figure 8–1.

The farther away the sample mean is from the population mean, the more evidence there would be for rejecting the null hypothesis. The probability that the sample came from a population whose mean is 82 decreases as the distance or absolute value of the difference between the means increases.

If the mean pulse rate of the sample were, say, 83, the researcher would probably conclude that this difference was due to chance and would not reject the null hypothesis. But if the sample mean were, say, 90, then in all likelihood the researcher would conclude that the medication increased the pulse rate of the users and would reject the null hypothesis. The question is, Where does the researcher draw the line? This decision is not made on feelings or intuition; it is made statistically. That is, the difference must be significant and in all likelihood not due to chance. Here is where the concepts of statistical test and level of significance are used.

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Figure 8–1

Situations in Hypothesis Testing

A statistical test uses the data obtained from a sample to make a decision about whether the null hypothesis should be rejected.

The numerical value obtained from a statistical test is called the test value.

In this type of statistical test, the mean is computed for the data obtained from the sample and is compared with the population mean. Then a decision is made to reject or not reject the null hypothesis on the basis of the value obtained from the statistical test. If the difference is significant, the null hypothesis is rejected. If it is not, then the null hypothesis is not rejected.

In the hypothesis-testing situation, there are four possible outcomes. In reality, the null hypothesis may or may not be true, and a decision is made to reject or not reject it on the basis of the data obtained from a sample. The four possible outcomes are shown in Figure 8–2. Notice that there are two possibilities for a correct decision and two possibilities for an incorrect decision.

Figure 8–2

Possible Outcomes of a Hypothesis Test

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If a null hypothesis is true and it is rejected, then a type I error is made. In situation A, for instance, the medication might not significantly change the pulse rate of all the users in the population; but it might change the rate, by chance, of the subjects in the sample. In this case, the researcher will reject the null hypothesis when it is really true, thus committing a type I error.

On the other hand, the medication might not change the pulse rate of the subjects in the sample, but when it is given to the general population, it might cause a significant increase or decrease in the pulse rate of users. The researcher, on the basis of the data obtained from the sample, will not reject the null hypothesis, thus committing a type II error.

In situation B, the additive might not significantly increase the lifetimes of automobile batteries in the population, but it might increase the lifetimes of the batteries in the sample. In this case, the null hypothesis would be rejected when it was really true. This would be a type I error. On the other hand, the additive might not work on the batteries selected for the sample, but if it were to be used in the general population of batteries, it might significantly increase their lifetimes. The researcher, on the basis of information obtained from the sample, would not reject the null hypothesis, thus committing a type II error.

A type I error occurs if you reject the null hypothesis when it is true.

A type II error occurs if you do not reject the null hypothesis when it is false.

The hypothesis-testing situation can be likened to a jury trial. In a jury trial, there are four possible outcomes. The defendant is either guilty or innocent, and he or she will be convicted or acquitted. See Figure 8–3.

Now the hypotheses are

H0: The defendant is innocent

H1: The defendant is not innocent (i.e., guilty)

Next, the evidence is presented in court by the prosecutor, and based on this evidence, the jury decides the verdict, innocent or guilty.

If the defendant is convicted but he or she did not commit the crime, then a type I error has been committed. See block 1 of Figure 8–3. On the other hand, if the defendant is convicted and he or she has committed the crime, then a correct decision has been made. See block 2.

If the defendant is acquitted and he or she did not commit the crime, a correct decision has been made by the jury. See block 3. However, if the defendant is acquitted and he or she did commit the crime, then a type II error has been made. See block 4.

Figure 8–3

Hypothesis Testing and a Jury Trial

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The decision of the jury does not prove that the defendant did or did not commit the crime. The decision is based on the evidence presented. If the evidence is strong enough, the defendant will be convicted in most cases. If the evidence is weak, the defendant will be acquitted in most cases. Nothing is proved absolutely. Likewise, the decision to reject or not reject the null hypothesis does not prove anything. The only way to prove anything statistically is to use the entire population, which, in most cases, is not possible. The decision, then, is made on the basis of probabilities. That is, when there is a large difference between the mean obtained from the sample and the hypothesized mean, the null hypothesis is probably not true. The question is, How large a difference is necessary to reject the null hypothesis? Here is where the level of significance is used.

Unusual Stat

Of workers in the United States, 64% drive to work alone and 6% of workers walk to work.

The level of significance is the maximum probability of committing a type I error. This probability is symbolized by α (Greek letter alpha). That is, P(type I error) = α.

The probability of a type II error is symbolized by β, the Greek letter beta. That is, P(type II error) = β. In most hypothesis-testing situations, β cannot be easily computed; however, α and β are related in that decreasing one increases the other.

Statisticians generally agree on using three arbitrary significance levels: the 0.10, 0.05, and 0.01 levels. That is, if the null hypothesis is rejected, the probability of a type I error will be 10%, 5%, or 1%, depending on which level of significance is used. Here is another way of putting it: When α = 0.10, there is a 10% chance of rejecting a true null hypothesis; when α = 0.05, there is a 5% chance of rejecting a true null hypothesis; and when α = 0.01, there is a 1% chance of rejecting a true null hypothesis.

In a hypothesis-testing situation, the researcher decides what level of significance to use. It does not have to be the 0.10, 0.05, or 0.01 level. It can be any level, depending on the seriousness of the type I error. After a significance level is chosen, a critical value is selected from a table for the appropriate test. If a z test is used, for example, the z table (Table E in Appendix C) is consulted to find the critical value. The critical value determines the critical and noncritical regions.

The critical value separates the critical region from the noncritical region. The symbol for critical value is C.V.

The critical or rejection region is the range of values of the test value that indicates that there is a significant difference and that the null hypothesis should be rejected.

The noncritical or nonrejection region is the range of values of the test value that indicates that the difference was probably due to chance and that the null hypothesis should not be rejected.

The critical value can be on the right side of the mean or on the left side of the mean for a one-tailed test. Its location depends on the inequality sign of the alternative hypothesis. For example, in situation B, where the chemist is interested in increasing the average lifetime of automobile batteries, the alternative hypothesis is H1: µ > 36. Since the inequality sign is >, the null hypothesis will be rejected only when the sample mean is significantly greater than 36. Hence, the critical value must be on the right side of the mean. Therefore, this test is called a right-tailed test.

A one-tailed test indicates that the null hypothesis should be rejected when the test value is in the critical region on one side of the mean. A one-tailed test is either a right-tailed test or left-tailed test, depending on the direction of the inequality of the alternative hypothesis.

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Figure 8–4

Finding the Critical Value for α = 0.01 (Right-Tailed Test)

Objective 3

Find critical values for the z test.

To obtain the critical value, the researcher must choose an alpha level. In situation B, suppose the researcher chose α = 0.01. Then the researcher must find a z value such that 1% of the area falls to the right of the z value and 99% falls to the left of the z value, as shown in Figure 8–4(a).

Next, the researcher must find the area value in Table E closest to 0.9900. The critical z value is 2.33, since that value gives the area closest to 0.9900 (that is, 0.9901), as shown in Figure 8–4(b).

The critical and noncritical regions and the critical value are shown in Figure 8–5.

Figure 8–5

Critical and Noncritical Regions for α = 0.01 (Right-Tailed Test)

Now, move on to situation C, where the contractor is interested in lowering the heating bills. The alternative hypothesis is H1: µ < \$78. Hence, the critical value falls to the left of the mean. This test is thus a left-tailed test. At α = 0.01, the critical value is –2.33, since 0.0099 is the closest value to 0.01. This is shown in Figure 8–6.

When a researcher conducts a two-tailed test, as in situation A, the null hypothesis can be rejected when there is a significant difference in either direction, above or below the mean.

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Figure 8–6

Critical and Noncritical Regions for α = 0.01 (Left-Tailed Test)

In a two-tailed test, the null hypothesis should be rejected when the test value is in either of the two critical regions.

For a two-tailed test, then, the critical region must be split into two equal parts. If α = 0.01, then one-half of the area, or 0.005, must be to the right of the mean and one-half must be to the left of the mean, as shown in Figure 8–7.

In this case, the z value on the left side is found by looking up the z value corresponding to an area of 0.0050. The z value falls about halfway between –2.57 and –2.58 corresponding to the areas 0.0049 and 0.0051. The average of –2.57 and –2.58 is [(–2.57) + (–2.58)] ÷ 2 = –2.575 so if the z value is needed to 3 decimal places, –2.575 is used; however, if the z value is rounded to 2 decimal places, –2.58 is used.

On the right side, it is necessary to find the z value corresponding to 0.99 + 0.005, or 0.9950. Again, the value falls between 0.9949 and 0.9951, so +2.575 or 2.58 can be used. See Figure 8–7.

Figure 8–7

Finding the critical values for α = 0.01 (Two-Tailed Test)

The critical values are +2.58 and –2.58, as shown in Figure 8–8.

Figure 8–8

Critical and Noncritical Regions for α = 0.01 (Two-Tailed Test)

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Similar procedures are used to find other values of α.

Figure 8–9 with rejection regions shaded shows the critical value (C.V.) for the three situations discussed in this section for values of α = 0.10, α = 0.05, and α = 0.01. The procedure for finding critical values is outlined next (where k is a specified number).

Figure 8–9

Summary of Hypothesis Testing and Critical Values

Procedure Table

Finding the Critical Values for Specific A Values, Using Table E

Step 1Draw the figure and indicate the appropriate area.

a.If the test is left-tailed, the critical region, with an area equal to a, will be on the left side of the mean.

b.If the test is right-tailed, the critical region, with an area equal to a, will be on the right side of the mean.

c.If the test is two-tailed, α must be divided by 2; one-half of the area will be to the right of the mean, and one-half will be to the left of the mean.

Step 2a.For a left-tailed test, use the z value that corresponds to the area equivalent to α in Table E.

b.For a right-tailed test, use the z value that corresponds to the area equivalent to 1 – α.

c.For a two-tailed test, use the z value that corresponds to α/2 for the left value. It will be negative. For the right value, use the z value that corresponds to the area equivalent to 1 – α/2. It will be positive.

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Example 8–2

Using Table E in Appendix C, find the critical value(s) for each situation and draw the appropriate figure, showing the critical region.

a.A left-tailed test with α = 0.10.

b.A two-tailed test with α = 0.02.

c.A right-tailed test with α = 0.005.

Solution a

Step 1Draw the figure and indicate the appropriate area. Since this is a left-tailed test, the area of 0.10 is located in the left tail, as shown in Figure 8–10.

Step 2Find the area closest to 0.1000 in Table E. In this case, it is 0.1003. Find the z value which corresponds to the area 0.1003. It is –1.28. See Figure 8–10.

Figure 8–10

Critical Value and Critical Region for part α of Example 8–2

Solution b

Step 1Draw the figure and indicate the appropriate area. In this case, there are two areas equivalent to α/2, or 0.02/2 = 0.01.

Step 2For the left z critical value, find the area closest to α/2, or 0.02/2 = 0.01. In this case, it is 0.0099.

For the right z critical value, find the area closest to 1 – α/2, or 1 – 0.02/2 = 0. 9900. In this case, it is 0.9901.

Find the z values for each of the areas. For 0.0099, z = –2.33. For the area of 0. 9901, z = 0.9901, z = +2.33. See Figure 8–11.

Figure 8–11

Critical Values and Critical Regions for part b of Example 8–2

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Solution c

Step 1Draw the figure and indicate the appropriate area. Since this is a right-tailed test, the area 0.005 is located in the right tail, as shown in Figure 8–12.

Figure 8–12

Critical Value and Critical Region for part c of Example 8–2

Step 2Find the area closest to 1 – α, or 1 – 0.005 = 0.9950. In this case, it is 0. 9949 or 0.9951.

The two z values corresponding to 0.9949 and 0.9951 are +2.57 and +2.58. Since 0. 9500 is halfway between these two values, find the average of the two values (+2.57 + 2.58) ÷ 2 = +2.575. However, 2.58 is most often used. See Figure 8–12.

Objective 4

State the five steps used in hypothesis testing.

In hypothesis testing, the following steps are recommended.

1.State the hypotheses. Be sure to state both the null and the alternative hypotheses.

2.Design the study. This step includes selecting the correct statistical test, choosing a level of significance, and formulating a plan to carry out the study. The plan should include information such as the definition of the population, the way the sample will be selected, and the methods that will be used to collect the data.

3.Conduct the study and collect the data.

4.Evaluate the data. The data should be tabulated in this step, and the statistical test should be conducted. Finally, decide whether to reject or not reject the null hypothesis.

5.Summarize the results.

For the purposes of this chapter, a simplified version of the hypothesis-testing procedure will be used, since designing the study and collecting the data will be omitted. The steps are summarized in the Procedure Table.

Procedure Table

Step 1State the hypotheses and identify the claim.

Step 2Find the critical value(s) from the appropriate table in Appendix C.

Step 3Compute the test value.

Step 4Make the decision to reject or not reject the null hypothesis.

Step 5Summarize the results.

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Applying the Concepts 8–1

The Incredible Edible Egg company recently found that eating eggs does not increase a person’s blood serum cholesterol. Five hundred subjects participated in a study that lasted for 2 years. The participants were randomly assigned to either a no-egg group or a moderate-egg group. The blood serum cholesterol levels were checked at the beginning and at the end of the study. Overall, the groups’ levels were not significantly different. The company reminds us that eating eggs is healthy if done in moderation. Many of the previous studies relating eggs and high blood serum cholesterol jumped to improper conclusions.

Using this information, answer these questions.

1.What prompted the study?

2.What is the population under study?

3.Was a sample collected?

4.What was the hypothesis?

5.Were data collected?

6.Were any statistical tests run?

7.What was the conclusion?

See page 465 for the answers.

 Exercises 8–1

1.Define null and alternative hypotheses, and give an example of each.

2.What is meant by a type I error? A type II error? How are they related?

3.What is meant by a statistical test?

4.Explain the difference between a one-tailed and a two-tailed test.

5.What is meant by the critical region? The noncritical region?

6.What symbols are used to represent the null hypothesis and the alternative hypothesis?

7.What symbols are used to represent the probabilities of type I and type II errors?

8.Explain what is meant by a significant difference.

9.When should a one-tailed test be used? A two-tailed test?

10.List the steps in hypothesis testing.

11.In hypothesis testing, why can’t the hypothesis be proved true?

12.(ans) Using the z table (Table E), find the critical value (or values) for each.

a.α = 0.05, two-tailed test

b.α = 0.01, left-tailed test

c.α = 0.005, right-tailed test

d.α = 0.01, right-tailed test

e.α = 0.05, left-tailed test

f.α = 0.02, left-tailed test

g.α = 0.05, right-tailed test

h.α = 0.01, two-tailed test

i.α = 0.04, left-tailed test

j.α = 0.02, right-tailed test

13.For each conjecture, state the null and alternative hypotheses.

a.The average age of community college students is 24.6 years.

b.The average income of accountants is \$51,497.

c.The average age of attorneys is greater than 25.4 years.

d.The average score of 50 high school basketball games is less than 88.

e.The average pulse rate of male marathon runners is less than 70 beats per minute.

f.The average cost of a DVD player is \$79.95.

g.The average weight loss for a sample of people who exercise 30 minutes per day for 6 weeks is 8.2 pounds.

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Objective 5

Test means when σ is known, using the z test.

8-2z Test for a Mean

In this chapter, two statistical tests will be explained: the z test is used when σ is known, and the t test is used when σ is unknown. This section explains the z test, and Section 8–3 explains the t test.

Many hypotheses are tested using a statistical test based on the following general formula:

The observed value is the statistic (such as the mean) that is computed from the sample data. The expected value is the parameter (such as the mean) that you would expect to obtain if the null hypothesis were true—in other words, the hypothesized value. The denominator is the standard error of the statistic being tested (in this case, the standard error of the mean).

The z test is defined formally as follows.

The z test is a statistical test for the mean of a population. It can be used when n ≥ 30, or when the population is normally distributed and σ is known.

The formula for the z test is

where

= sample mean

µ = hypothesized population mean

σ = population standard deviation

n = sample size

For the z test, the observed value is the value of the sample mean. The expected value is the value of the population mean, assuming that the null hypothesis is true. The denominator is the standard error of the mean.

The formula for the z test is the same formula shown in Chapter 6 for the situation where you are using a distribution of sample means. Recall that the central limit theorem allows you to use the standard normal distribution to approximate the distribution of sample means when n ≥ 30.

Note: Your first encounter with hypothesis testing can be somewhat challenging and confusing, since there are many new concepts being introduced at the same time. To understand all the concepts, you must carefully follow each step in the examples and try each exercise that is assigned. Only after careful study and patience will these concepts become clear.

As stated in Section 8–1, there are five steps for solving hypothesis-testing problems:

Step 1State the hypotheses and identify the claim.

Step 2Find the critical value(s).

Step 3Compute the test value.

Step 4Make the decision to reject or not reject the null hypothesis.

Step 5Summarize the results.

Example 8–3 illustrates these five steps.

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Speaking of Statistics

This study found that people who used pedometers reported having increased energy, mood improvement, and weight loss. State possible null and alternative hypotheses for the study. What would be a likely population? What is the sample size? Comment on the sample size.

Example 8–3

Professors’ Salaries

A researcher reports that the average salary of assistant professors is more than \$42,000. A sample of 30 assistant professors has a mean salary of \$43,260. At α = 0.05, test the claim that assistant professors earn more than \$42,000 per year. The standard deviation of the population is \$5230.

Solution

Step 1State the hypotheses and identify the claim.

H0: µ = \$42,000andH1: µ > \$42,000 (claim)

Step 2Find the critical value. Since α = 0.05 and the test is a right-tailed test, the critical value is z = +1.65.

Step 3Compute the test value.

Step 4Make the decision. Since the test value, +1.32, is less than the critical value, + 1.65, and is not in the critical region, the decision is to not reject the null hypothesis. This test is summarized in Figure 8–13.

Figure 8–13

Summary of the z Test of Example 8–3

Step 5Summarize the results. There is not enough evidence to support the claim that assistant professors earn more on average than \$42,000 per year.

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Comment: Even though in Example 8–3 the sample mean of \$43,260 is higher than the hypothesized population mean of \$42,000, it is not significantly higher. Hence, the difference may be due to chance. When the null hypothesis is not rejected, there is still a probability of a type II error, i.e., of not rejecting the null hypothesis when it is false.

The probability of a type II error is not easily ascertained. For now, it is only necessary to realize that the probability of type II error exists when the decision is not to reject the null hypothesis.

Also note that when the null hypothesis is not rejected, it cannot be accepted as true. There is merely not enough evidence to say that it is false. This guideline may sound a little confusing, but the situation is analogous to a jury trial. The verdict is either guilty or not guilty and is based on the evidence presented. If a person is judged not guilty, it does not mean that the person is proved innocent; it only means that there was not enough evidence to reach the guilty verdict.

Example 8–4

Costs of Men’s Athletic Shoes

A researcher claims that the average cost of men’s athletic shoes is less than \$80. He selects a random sample of 36 pairs of shoes from a catalog and finds the following costs (in dollars). (The costs have been rounded to the nearest dollar.) Is there enough evidence to support the researcher’s claim at α = 0.10? Assume σ = 19.2.

Solution

Step 1State the hypotheses and identify the claim

H0: µ = \$80andH1: µ < \$80 (claim)

Step 2Find the critical value. Since α = 0.10 and the test is a left-tailed test, the critical value is –1.28.

Step 3Compute the test value. Since the exercise gives raw data, it is necessary to find the mean of the data. Using the formulas in Chapter 3 or your calculator gives = 75.0 and σ = 19.2. Substitute in the formula

Step 4Make the decision. Since the test value, –1.56, falls in the critical region, the decision is to reject the null hypothesis. See Figure 8–14.

Figure 8–14

Critical and Test Values for Example 8–4

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Step 5Summarize the results. There is enough evidence to support the claim that the average cost of men’s athletic shoes is less than \$80.

Comment: In Example 8–4, the difference is said to be significant. However, when the null hypothesis is rejected, there is always a chance of a type I error. In this case, the probability of a type I error is at most 0.10, or 10%.

Example 8–5

Cost of Rehabilitation

The Medical Rehabilitation Education Foundation reports that the average cost of rehabilitation for stroke victims is \$24,672. To see if the average cost of rehabilitation is different at a particular hospital, a researcher selects a random sample of 35 stroke victims at the hospital and finds that the average cost of their rehabilitation is \$25,226. The standard deviation of the population is \$3251. At α = 0.01, can it be concluded that the average cost of stroke rehabilitation at a particular hospital is different from \$24,672?

Source: Snapshot, USA TODAY.

Solution

Step 1State the hypotheses and identify the claim.

H0: µ = \$24,672andH1: µ ≠ \$24,672 (claim)

Step 2Find the critical values. Since α = 0.01 and the test is a two-tailed test, the critical values are +2.58 and –2.58.

Step 3Compute the test value.

Step 4Make the decision. Do not reject the null hypothesis, since the test value falls in the noncritical region, as shown in Figure 8–15.

Figure 8–15

Critical and Test Values for Example 8–5

Step 5Summarize the results. There is not enough evidence to support the claim that the average cost of rehabilitation at the particular hospital is different from \$24,672.

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Figure 8–16

Outcomes of a Hypothesis-Testing Situation

When σ is unknown, the t test must be used. The t test will be explained in Section 8–3.

Students sometimes have difficulty summarizing the results of a hypothesis test. Figure 8–16 shows the four possible outcomes and the summary statement for each situation.

First, the claim can be either the null or alternative hypothesis, and one should identify which it is. Second, after the study is completed, the null hypothesis is either rejected or not rejected. From these two facts, the decision can be identified in the appropriate block of Figure 8–16.

For example, suppose a researcher claims that the mean weight of an adult animal of a particular species is 42 pounds. In this case, the claim would be the null hypothesis, H0: µ = 42, since the researcher is asserting that the parameter is a specific value. If the null hypothesis is rejected, the conclusion would be that there is enough evidence to reject the claim that the mean weight of the adult animal is 42 pounds. See Figure 8–17(a).

On the other hand, suppose the researcher claims that the mean weight of the adult animals is not 42 pounds. The claim would be the alternative hypothesis H1: µ ≠ 42. Furthermore, suppose that the null hypothesis is not rejected. The conclusion, then, would be that there is not enough evidence to support the claim that the mean weight of the adult animals is not 42 pounds. See Figure 8–17(b).

Figure 8–17

Outcomes of a Hypothesis-Testing Situation for Two Specific Cases

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Again, remember that nothing is being proved true or false. The statistician is only stating that there is or is not enough evidence to say that a claim is probably true or false. As noted previously, the only way to prove something would be to use the entire population under study, and usually this cannot be done, especially when the population is large.

P-Value Method for Hypothesis Testing

Statisticians usually test hypotheses at the common α levels of 0.05 or 0.01 and sometimes at 0.10. Recall that the choice of the level depends on the seriousness of the type I error. Besides listing an α value, many computer statistical packages give a P-value for hypothesis tests.

The P-value (or probability value) is the probability of getting a sample statistic (such as the mean) or a more extreme sample statistic in the direction of the alternative hypothesis when the null hypothesis is true.

In other words, the P-value is the actual area under the standard normal distribution curve (or other curve, depending on what statistical test is being used) representing the probability of a particular sample statistic or a more extreme sample statistic occurring if the null hypothesis is true.

For example, suppose that an alternative hypothesis is H1: µ > 50 and the mean of a sample is = 52. If the computer printed a P-value of 0.0356 for a statistical test, then the probability of getting a sample mean of 52 or greater is 0.0356 if the true population mean is 50 (for the given sample size and standard deviation). The relationship between the P-value and the α value can be explained in this manner. For P = 0.0356, the null hypothesis would be rejected at α = 0.05 but not at α = 0.01. See Figure 8–18.

When the hypothesis test is two-tailed, the area in one tail must be doubled. For a two-tailed test, if α is 0.05 and the area in one tail is 0.0356, the P-value will be 2(0.0356) = 0.0712. That is, the null hypothesis should not be rejected at α = 0.05, since 0. 0712 is greater than 0.05. In summary, then, if the P-value is less than a, reject the null hypothesis. If the P-value is greater than α, do not reject the null hypothesis.

The P-values for the z test can be found by using Table E in Appendix C. First find the area under the standard normal distribution curve corresponding to the z test value. For a left-tailed test, use the area given in the table; for a right-tailed test, use 1.0000 minus the area given in the table. To get the P-value for a two-tailed test, double the area you found in the tail. This procedure is shown in step 3 of Examples 8–6 and 8–7.

The P-value method for testing hypotheses differs from the traditional method somewhat. The steps for the P-value method are summarized next.

Figure 8–18

Comparison of α Values and P-Values

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Procedure Table

Solving Hypothesis-Testing Problems (P-Value Method)

Step 1State the hypotheses and identify the claim.

Step 2Compute the test value.

Step 3Find the P-value.

Step 4Make the decision.

Step 5Summarize the results.

Examples 8–6 and 8–7 show how to use the P-value method to test hypotheses.

Example 8–6

Cost of College Tuition

A researcher wishes to test the claim that the average cost of tuition and fees at a four-year public college is greater than \$5700. She selects a random sample of 36 four-year public colleges and finds the mean to be \$5950. The population standard deviation is \$659. Is there evidence to support the claim at α = 0.05? Use the P-value method.

Source: Based on information from the College Board.

Solution

Step 1State the hypotheses and identify the claim. H0: = \$5700 and H1: µ > \$5700 (claim).

Step 2Compute the test value.

Step 3Find the P-value. Using Table E in Appendix C, find the corresponding area under the normal distribution for z = 2.28. It is 0.9887. Subtract this value for the area from 1.0000 to find the area in the right tail.

1.0000 – 0.9887 = 0.0113

Hence the P-value is 0.0113.

Step 4Make the decision. Since the P-value is less than 0.05, the decision is to reject the null hypothesis. See Figure 8–19.

Figure 8–19

P-Value and α Value for Example 8–6

Step 5Summarize the results. There is enough evidence to support the claim that the tuition and fees at four-year public colleges are greater than \$5700.

Note: Had the researcher chosen α = 0.01, the null hypothesis would not have been rejected since the P-value (0.0113) is greater than 0.01.

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Example 8–7

Wind Speed

A researcher claims that the average wind speed in a certain city is 8 miles per hour. A sample of 32 days has an average wind speed of 8.2 miles per hour. The standard deviation of the population is 0.6 mile per hour. At α = 0.05, is there enough evidence to reject the claim? Use the P-value method.

Solution

Step 1State the hypotheses and identify the claim.

H0: µ = 8 (claim)andH1: µ ≠ 8

Step 2Compute the test value.

Step 3Find the P-value. Using Table E, find the corresponding area for z = 1.89. It is 0.9706. Subtract the value from 1.0000.

1.0000 – 0.9706 = 0.0294

Since this is a two-tailed test, the area of 0.0294 must be doubled to get the P-value.

2(0.0294) = 0.0588

Step 4Make the decision. The decision is to not reject the null hypothesis, since the P-value is greater than 0.05. See Figure 8–20.

Figure 8–20

P-Values and α Values for Example 8–7

Step 5Summarize the results. There is not enough evidence to reject the claim that the average wind speed is 8 miles per hour.

In Examples 8–6 and 8–7, the P-value and the α value were shown on a normal distribution curve to illustrate the relationship between the two values; however, it is not necessary to draw the normal distribution curve to make the decision whether to reject the null hypothesis. You can use the following rule:

Decision Rule When Using a P-Value

If P-value ≤ α, reject the null hypothesis.

If P-value > α, do not reject the null hypothesis.

In Example 8–6, P-value = 0.0113 and α = 0.05. Since P-value ≤ α, the null hypothesis was rejected. In Example 8–7, P-value = 0.0588 and α = 0.05. Since P-value > α, the null hypothesis was not rejected.

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The P-values given on calculators and computers are slightly different from those found with Table E. This is so because z values and the values in Table E have been rounded. Also, most calculators and computers give the exact P-value for two-tailed tests, so it should not be doubled (as it should when the area found in Table E is used).

A clear distinction between the α value and the P-value should be made. The α value is chosen by the researcher before the statistical test is conducted. The P-value is computed after the sample mean has been found.

There are two schools of thought on P-values. Some researchers do not choose an α value but report the P-value and allow the reader to decide whether the null hypothesis should be rejected.

In this case, the following guidelines can be used, but be advised that these guidelines are not written in stone, and some statisticians may have other opinions.

Guidelines for P-Values

If P-value ≤ 0.01, reject the null hypothesis. The difference is highly significant.

If P-value > 0.01 but P-value ≤ 0.05, reject the null hypothesis. The difference is significant.

If P-value > 0.05 but P-value ≤ 0.10, consider the consequences of type I error before rejecting the null hypothesis.

If P-value > 0.10, do not reject the null hypothesis. The difference is not significant.

Others decide on the α value in advance and use the P-value to make the decision, as shown in Examples 8–6 and 8–7. A note of caution is needed here: If a researcher selects α = 0.01 and the P-value is 0.03, the researcher may decide to change the α value from 0. 01 to 0.05 so that the null hypothesis will be rejected. This, of course, should not be done. If the α level is selected in advance, it should be used in making the decision.

One additional note on hypothesis testing is that the researcher should distinguish between statistical significance and practical significance. When the null hypothesis is rejected at a specific significance level, it can be concluded that the difference is probably not due to chance and thus is statistically significant. However, the results may not have any practical significance. For example, suppose that a new fuel additive increases the miles per gallon that a car can get by mile for a sample of 1000 automobiles. The results may be statistically significant at the 0.05 level, but it would hardly be worthwhile to market the product for such a small increase. Hence, there is no practical significance to the results. It is up to the researcher to use common sense when interpreting the results of a statistical test.

Applying the Concepts 8–2

Car Thefts

You recently received a job with a company that manufactures an automobile antitheft device. To conduct an advertising campaign for the product, you need to make a claim about the number of automobile thefts per year. Since the population of various cities in the United States varies, you decide to use rates per 10,000 people. (The rates are based on the number of people living in the cities.) Your boss said that last year the theft rate per 10,000 people was 44 vehicles. You want to see if it has changed. The following are rates per 10,000 people for 36 randomly selected locations in the United States.

Source: Based on information from the National Insurance Crime Bureau.

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Using this information, answer these questions.

1.What hypotheses would you use?

2.Is the sample considered small or large?

3.What assumption must be met before the hypothesis test can be conducted?

4.Which probability distribution would you use?

5.Would you select a one- or two-tailed test? Why?

6.What critical value(s) would you use?

7.Conduct a hypothesis test. Use σ = 30.3.

10.Write a brief statement summarizing your conclusion.

11.If you lived in a city whose population was about 50,000, how many automobile thefts per year would you expect to occur?

See page 465 for the answers.

 Exercises 8–2

For Exercises 1 through 13, perform each of the following steps.

a.State the hypotheses and identify the claim.

b.Find the critical value(s).

c.Compute the test value.

d.Make the decision.

e.Summarize the results.

Use diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified.

1.Walking with a Pedometer An increase in walking has been shown to contribute to a healthier life-style. A sedentary American takes an average of 5000 steps per day (and 65% of Americans are overweight). A group of health-conscious employees of a large health care system volunteered to wear pedometers for a month to record their steps. It was found that a random sample of 40 walkers took an average of 5430 steps per day, and the population standard deviation is 600 steps. At α = 0.05 can it be concluded that they walked more than the mean number of 5000 steps per day?

Source: www.msn.com/health

2.Credit Card Debt It has been reported that the average credit card debt for college seniors is \$3262. The student senate at a large university feels that their seniors have a debt much less than this, so it conducts a study of 50 randomly selected seniors and finds that the average debt is \$2995, and the population standard deviation is \$1100. With α = 0.05, is the student senate correct?

Source: USA TODAY.

1. Revenue of Large Businesses A researcher estimates that the average revenue of the largest businesses in the United States is greater than \$24 billion. A sample of 50 companies is selected, and the revenues (in billions of dollars) are shown. At α = 0.05, is there enough evidence to support the researcher’s claim? σ = 28.7.

Source: New York Times Almanac.

4.Salaries of Ph.D. Students Full-time Ph.D. students receive an average salary of \$12,837 according to the U.S. Department of Education. The dean of graduate studies at a large state university feels that Ph.D. students in his state earn more than this. He surveys 44 randomly selected students and finds their average salary is \$14,445, and the population standard deviation is \$1500. With α = 0.05, is the dean correct?

Source: U.S. Department of Education/Chronicle of Higher Education.

5.Health Care Expenses The mean annual expenditure per 25- to 34-year-old consumer for health care is \$1468. This includes health insurance, medical services, and drugs and medical supplies. Students at a large university took a survey, and it was found that for a sample of 60 students, the mean health care expense was \$1520, and the population standard deviation is \$198. Is there sufficient evidence at α = 0.01 to conclude that their health care expenditure differs from the national average of \$1468? Is the conclusion different at α = 0.05?

Source: Time Almanac.

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6.Peanut Production in Virginia The average production of peanuts in Virginia is 3000 pounds per acre. A new plant food has been developed and is tested on 60 individual plots of land. The mean yield with the new plant food is 3120 pounds of peanuts per acre, and the population standard deviation is 578 pounds. At α = 0.05, can you conclude that the average production has increased?

Source: The Old Farmer’s Almanac.

1. Heights of 1-Year-Olds The average 1-year-old (both genders) is 29 inches tall. A random sample of 30 one-year-olds in a large day care franchise resulted in the following heights. At α = 0.05, can it be concluded that the average height differs from 29 inches? Assume σ = 2.61.

Source: www.healthepic.com

8.Salaries of Government Employees The mean salary of federal government employees on the General Schedule is \$59,593. The average salary of 30 state employees who do similar work is \$58,800 with σ = \$1500. At the 0.01 level of significance, can it be concluded that state employees earn on average less than federal employees?

Source: New York Times Almanac.

9.Undergraduate School Expenses The average undergraduate cost for tuition, fees, room, and board for all institutions last year was \$26,025. A random sample of 40 institutions of higher learning this year indicated that the mean tuition, fees, room, and board for the sample was \$27,690, and the population standard deviation is \$5492. At the 0.05 level of significance, is there sufficient evidence that the cost has increased?

Source: Time Almanac.

1. Home Prices in Pennsylvania A real estate agent claims that the average price of a home sold in Beaver County, Pennsylvania, is \$60,000. A random sample of 36 homes sold in the county is selected, and the prices in dollars are shown. Is there enough evidence to reject the agent’s claim at α = 0.05? Assume σ = \$76,025.

Source: Pittsburgh Tribune-Review.

11.Use of Disposable Cups The average college student goes through 500 disposable cups in a year. To raise environmental awareness, a student group at a large university volunteered to help count how many cups were used by students on their campus. A random sample of 50 students’ results found that they used a mean of 476 cups with σ = 42 cups. At α = 0.01, is there sufficient evidence to conclude that the mean differs from 500?

Source: www.esc.mtu.edu

12.Public School Teachers’ Salaries The average salary for public school teachers for a specific year was reported to be \$39,385. A random sample of 50 public school teachers in a particular state had a mean of \$41,680, and the population standard deviation is \$5975. Is there sufficient evidence at the α = 0.05 level to conclude that the mean salary differs from \$39,385?

Source: New York Times Almanac.

13.Ages of U.S. Senators The mean age of Senators in the 109th Congress was 60.35 years. A random sample of 40 senators from various state senates had an average age of 55.4 years, and the population standard deviation is 6.5 years. At α = 0.05, is there sufficient evidence that state senators are on average younger than the Senators in Washington?

Source: CG Today.

14.What is meant by a P-value?

15.State whether the null hypothesis should be rejected on the basis of the given P-value.

a.P-value = 0.258, α = 0.05, one-tailed test

b.P-value = 0.0684, α = 0.10, two-tailed test

c.P-value = 0.0153, α = 0.01, one-tailed test

d.P-value = 0.0232, α = 0.05, two-tailed test

e.P-value = 0.002, α = 0.01, one-tailed test

16.Soft Drink Consumption A researcher claims that the yearly consumption of soft drinks per person is 52 gallons. In a sample of 50 randomly selected people, the mean of the yearly consumption was 56.3 gallons. The standard deviation of the population is 3.5 gallons. Find the P-value for the test. On the basis of the P-value, is the researcher’s claim valid?

Source: U.S. Department of Agriculture.

17.Stopping Distances A study found that the average stopping distance of a school bus traveling 50 miles per hour was 264 feet. A group of automotive engineers decided to conduct a study of its school buses and found that for 20 buses, the average stopping distance of buses traveling 50 miles per hour was 262.3 feet. The standard deviation of the population was 3 feet. Test the claim that the average stopping distance of the company’s buses is actually less than 264 feet. Find the P-value. On the basis of the P-value, should the null hypothesis be rejected at α = 0.01? Assume that the variable is normally distributed.

Source: Snapshot, USA TODAY, March 12, 1992.

1. Copy Machine Use A store manager hypothesizes that the average number of pages a person copies on the store’s copy machine is less than 40. A sample of 50 customers’ orders is selected. At α = 0.01, is there enough evidence to support the claim? Use the P-value hypothesis-testing method. Assume σ = 30.9.

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19.Burning Calories by Playing Tennis A health researcher read that a 200-pound male can burn an average of 546 calories per hour playing tennis. Thirty-six males were randomly selected and tested. The mean of the number of calories burned per hour was 544.8. Test the claim that the average number of calories burned is actually less than 546, and find the P-value. On the basis of the P-value, should the null hypothesis be rejected at α = 0.01? The standard deviation of the population is 3. Can it be concluded that the average number of calories burned is less than originally thought?

20.Breaking Strength of Cable A special cable has a breaking strength of 800 pounds. The standard deviation of the population is 12 pounds. A researcher selects a sample of 20 cables and finds that the average breaking strength is 793 pounds. Can he reject the claim that the breaking strength is 800 pounds? Find the P-value. Should the null hypothesis be rejected at α = 0.01? Assume that the variable is normally distributed.

21.Farm Sizes The average farm size in the United States is 444 acres. A random sample of 40 farms in Oregon indicated a mean size of 430 acres, and the population standard deviation is 52 acres. At α = 0.05, can it be concluded that the average farm in Oregon differs from the national mean? Use the P-value method.

Source: New York Times Almanac.

22.Farm Sizes Ten years ago, the average acreage of farms in a certain geographic region was 65 acres. The standard deviation of the population was 7 acres. A recent study consisting of 22 farms showed that the average was 63.2 acres per farm. Test the claim, at α = 0.10, that the average has not changed by finding the P-value for the test. Assume that s has not changed and the variable is normally distributed.

23.Transmission Service A car dealer recommends that transmissions be serviced at 30,000 miles. To see whether her customers are adhering to this recommendation, the dealer selects a sample of 40 customers and finds that the average mileage of the automobiles serviced is 30,456. The standard deviation of the population is 1684 miles. By finding the P-value, determine whether the owners are having their transmissions serviced at 30,000 miles. Use α = 0.10. Do you think the α value of 0.10 is an appropriate significance level?

1. Speeding Tickets A motorist claims that the South Boro Police issue an average of 60 speeding tickets per day. These data show the number of speeding tickets issued each day for a period of one month. Assume σ is 13.42. Is there enough evidence to reject the motorist’s claim at α = 0.05? Use the P-value method.
2. Sick Days A manager states that in his factory, the average number of days per year missed by the employees due to illness is less than the national average of 10. The following data show the number of days missed by 40 employees last year. Is there sufficient evidence to believe the manager’s statement at α = 0.05? σ = 3.63. Use the P-value method.

Extending the Concepts

26.Suppose a statistician chose to test a hypothesis at α = 0.01. The critical value for a right-tailed test is + 2.33. If the test value was 1.97, what would the decision be? What would happen if, after seeing the test value, she decided to choose α = 0.05? What would the decision be? Explain the contradiction, if there is one.

27.Hourly Wage The president of a company states that the average hourly wage of her employees is \$8.65. A sample of 50 employees has the distribution shown. At α = 0.05, is the president’s statement believable? Assume σ = 0.105.

 Class Frequency 8.35–8.43 2 8.44–8.52 6 8.53–8.61 12 8.62–8.70 18 8.71–8.79 10 8.80–8.88 2

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Technology Step by Step

MINITAB

Step by Step

Hypothesis Test for the Mean and the z Distribution

MINITAB can be used to calculate the test statistic and its P-value. The P-value approach does not require a critical value from the table. If the P-value is smaller than α, the null hypothesis is rejected. For Example 8–4, test the claim that the mean shoe cost is less than \$80.

1.Enter the data into a column of MINITAB. Do not try to type in the dollar signs! Name the column ShoeCost.

2.If sigma is known, skip to step 3; otherwise estimate sigma from the sample standard deviation s.

Calculate the Standard Deviation in the Sample

a)Select Calc>Column Statistics.

b)Check the button for Standard deviation.

c)Select ShoeCost for the Input variable.

d)Type s in the text box for Store the result in:.

e)Click [OK].

Calculate the Test Statistic and P-Value

3.Select Stat>Basic Statistics>1 Sample Z, then select ShoeCost in the Variable text box.

4.Click in the text box and enter the value of sigma or type s, the sample standard deviation.

5.Click in the text box for Test mean, and enter the hypothesized value of 80.

6.Click on [Options].

a)Change the Confidence level to 90.

b)Change the Alternative to less than. This setting is crucial for calculating the P-value.

7.Click [OK] twice.

One-Sample Z: ShoeCost

Since the P-value of 0.059 is less than a, reject the null hypothesis. There is enough evidence in the sample to conclude the mean cost is less than \$80.

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TI-83 Plus or TI-84 Plus

Step by Step

Hypothesis Test for the Mean and the z Distribution (Data)

1.Enter the data values into L1.

2.Press STAT and move the cursor to TESTS.

3.Press l for ZTest.

4.Move the cursor to Data and press ENTER.

5.Type in the appropriate values.

6.Move the cursor to the appropriate alternative hypothesis and press ENTER.

7.Move the cursor to Calculate and press ENTER.

Example TI8–1

This relates to Example 8–4 from the text. At the 10% significance level, test the claim that µ < 80 given the data values.

The population standard deviation σ is unknown. Since the sample size n = 36 ≥ 30, you can use the sample standard deviation s as an approximation for σ. After the data values are entered in L1 (step 1), press STAT, move the cursor to CALC, press 1 for 1-Var Stats, then press ENTER. The sample standard deviation of 19.16097224 will be one of the statistics listed. Then continue with step 2. At step 5 on the line for σ press VARS for variables, press 5 for Statistics, press 3 for Sx.

The test statistic is z = –1.565682556, and the P-value is 0.0587114841.

Hypothesis Test for the Mean and the z Distribution (Statistics)

1.Press STAT and move the cursor to TESTS.

2.Press 1 for ZTest.

3.Move the cursor to Stats and press ENTER.

4.Type in the appropriate values.

5.Move the cursor to the appropriate alternative hypothesis and press ENTER.

6.Move the cursor to Calculate and press ENTER.

Example TI8–2

This relates to Example 8–3 from the text. At the 5% significance level, test the claim that µ > 42,000 given σ = 5230, = 43,260, and n = 30.

The test statistic is z = 1.319561037, and the P-value is 0.0934908728.

Excel

Step by Step

Hypothesis Test for the Mean: z Test

Excel does not have a procedure to conduct a hypothesis test for the mean. However, you may conduct the test of the mean by using the MegaStat Add-in available on your CD. If you have not installed this add-in, do so, following the instructions from the Chapter 1 Excel Step by Step.

Example XL8–1

This example relates to Example 8–4 from the text. At the 10% significance level, test the claim that µ < 80. The MegaStat z test uses the P-value method. Therefore, it is not necessary to enter a significance level.

1.Enter the data into column A of a new worksheet.

2.From the toolbar, select Add-Ins, MegaStat>Hypothesis Tests>Mean vs. Hypothesized Value. Note: You may need to open MegaStat from the MegaStat.xls file on your computer’s hard drive.

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3.Select data input and type A1:A36 as the Input Range.

4.Type 80 for the Hypothesized mean and select the “less than” Alternative.

5.Select z test and click [OK].

The result of the procedure is shown next.

Hypothesis Test: Mean vs. Hypothesized Value

80.000  Hypothesized value

75.000  Mean data

19.161  Standard deviation

3.193  Standard error

36  n

–1.57  z

0.0587  P-value (one-tailed, lower)

Objective 6

Test means when σ is unknown, using the t test.

8-3t Test for a Mean

When the population standard deviation is unknown, the z test is not normally used for testing hypotheses involving means. A different test, called the t test, is used. The distribution of the variable should be approximately normal.

As stated in Chapter 7, the t distribution is similar to the standard normal distribution in the following ways.

1.It is bell-shaped.

2.It is symmetric about the mean.

3.The mean, median, and mode are equal to 0 and are located at the center of the distribution.

4.The curve never touches the x axis.

The t distribution differs from the standard normal distribution in the following ways.

1.The variance is greater than 1.

2.The t distribution is a family of curves based on the degrees of freedom, which is a number related to sample size. (Recall that the symbol for degrees of freedom is d.f. See Section 7–2 for an explanation of degrees of freedom.)

3.As the sample size increases, the t distribution approaches the normal distribution.

The t test is defined next.

The t test is a statistical test for the mean of a population and is used when the population is normally or approximately normally distributed, σ is unknown.

The formula for the t test is

The degrees of freedom are d.f. = n – 1.

The formula for the t test is similar to the formula for the z test. But since the population standard deviation σ is unknown, the sample standard deviation s is used instead.

The critical values for the t test are given in Table F in Appendix C. For a one-tailed test, find the α level by looking at the top row of the table and finding the appropriate column. Find the degrees of freedom by looking down the left-hand column.

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Notice that the degrees of freedom are given for values from 1 through 30, then at intervals above 30. When the degrees of freedom are above 30, some textbooks will tell you to use the nearest table value; however, in this textbook, you should always round down to the nearest table value. For example, if d.f. = 59, use d.f. = 55 to find the critical value or values. This is a conservative approach.

As the degrees of freedom get larger, the critical values approach the z values. Hence the bottom values (large sample size) are the same as the z values that were used in the last section.

Example 8–8

Find the critical t value for α = 0.05 with d.f. = 16 for a right-tailed t test.

Solution

Find the 0.05 column in the top row and 16 in the left-hand column. Where the row and column meet, the appropriate critical value is found; it is +1.746. See Figure 8–21.

Figure 8–21

Finding the Critical Value for the t Test in Table F (Example 8–8)

Example 8–9

Find the critical t value for α = 0.01 with d.f. = 22 for a left-tailed test.

Solution

Find the 0.01 column in the row labeled One tail, and find 22 in the left column. The critical value is –2.508 since the test is a one-tailed left test.

Example 8–10

Find the critical values for α = 0.10 with d.f. = 18 for a two-tailed t test.

Solution

Find the 0.10 column in the row labeled Two tails, and find 18 in the column labeled d.f. The critical values are +1.734 and –1.734.

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Example 8–11

Find the critical value for α = 0.05 with d.f. = 28 for a right-tailed t test.

Solution

Find the 0.05 column in the One-tail row and 28 in the left column. The critical value is + 1.701.

When you test hypotheses by using the t test (traditional method), follow the same procedure as for the z test, except use Table F.

Step 1State the hypotheses and identify the claim.

Step 2Find the critical value(s) from Table F.

Step 3Compute the test value.

Step 4Make the decision to reject or not reject the null hypothesis.

Step 5Summarize the results.

Remember that the t test should be used when the population is approximately normally distributed and the population standard deviation is unknown.

Examples 8–12 through 8–14 illustrate the application of the t test.

Example 8–12

Hospital Infections

A medical investigation claims that the average number of infections per week at a hospital in southwestern Pennsylvania is 16.3. A random sample of 10 weeks had a mean number of 17.7 infections. The sample standard deviation is 1.8. Is there enough evidence to reject the investigator’s claim at α = 0.05?

Source: Based on information obtained from Pennsylvania Health Care Cost Containment Council.

Solution

Step 1H0: µ = 16.3 (claim) and H1: µ ≠ 16.3.

Step 2The critical values are +2.262 and –2.262 for α = 0.05 and d.f. = 9.

Step 3The test value is

Step 4Reject the null hypothesis since 2.46 > 2.262. See Figure 8–22.

Figure 8–22

Summary of the t Test of Example 8–12

Step 5There is enough evidence to reject the claim that the average number of infections is 16.3.

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Example 8–13

Substitute Teachers’ Salaries

An educator claims that the average salary of substitute teachers in school districts in Allegheny County, Pennsylvania, is less than \$60 per day. A random sample of eight school districts is selected, and the daily salaries (in dollars) are shown. Is there enough evidence to support the educator’s claim at α = 0.10?

60   56   60   55   70   55   60   55

Source: Pittsburgh Tribune-Review.

Solution

Step 1H0: µ = \$60 and H1: µ < \$60 (claim).

Step 2At α = 0.10 and d.f. = 7, the critical value is –1.415.

Step 3To compute the test value, the mean and standard deviation must be found. Using either the formulas in Chapter 3 or your calculator, = \$58.88, and s = 5.08, you find

Step 4Do not reject the null hypothesis since –0.624 falls in the noncritical region. See Figure 8–23.

Figure 8–23

Critical Value and Test Value for Example 8–13

Step 5There is not enough evidence to support the educator’s claim that the average salary of substitute teachers in Allegheny County is less than \$60 per day.

The P-values for the t test can be found by using Table F; however, specific P-values for t tests cannot be obtained from the table since only selected values of α (for example, 0.01, 0.05) are given. To find specific P-values for t tests, you would need a table similar to Table E for each degree of freedom. Since this is not practical, only intervals can be found for P-values. Examples 8–14 to 8–16 show how to use Table F to determine intervals for P-values for the t test.

Example 8–14

Find the P-value when the t test value is 2.056, the sample size is 11, and the test is right-tailed.

Solution

To get the P-value, look across the row with 10 degrees of freedom (d.f. = n – 1) in Table F and find the two values that 2.056 falls between. They are 1.812 and 2.228. Since this is a right-tailed test, look up to the row labeled One tail, α and find the two α values corresponding to 1.812 and 2.228. They are 0.05 and 0.025, respectively. See Figure 8–24.

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Figure 8–24

Finding the P-Value for Example 8–14

Hence, the P-value would be contained in the interval 0.025 < P-value < 0.05. This means that the P-value is between 0.025 and 0.05. If α were 0.05, you would reject the null hypothesis since the P-value is less than 0.05. But if α were 0.01, you would not reject the null hypothesis since the P-value is greater than 0.01. (Actually, it is greater than 0.025.)

Example 8–15

Find the P-value when the t test value is 2.983, the sample size is 6, and the test is two-tailed.

Solution

To get the P-value, look across the row with d.f. = 5 and find the two values that 2.983 falls between. They are 2.571 and 3.365. Then look up to the row labeled Two tails, α to find the corresponding α values.

In this case, they are 0.05 and 0.02. Hence the P-value is contained in the interval 0.02 < P-value < 0.05. This means that the P-value is between 0.02 and 0.05. In this case, if α = 0.05, the null hypothesis can be rejected since P-value < 0.05; but if α = 0.01, the null hypothesis cannot be rejected since P-value > 0.01 (actually P-value > 0.02).

Note: Since many of you will be using calculators or computer programs that give the specific P-value for the t test and other tests presented later in this textbook, these specific values, in addition to the intervals, will be given for the answers to the examples and exercises.

The P-value obtained from a calculator for Example 8–14 is 0.033. The P-value obtained from a calculator for Example 8–15 is 0.031.

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To test hypotheses using the P-value method, follow the same steps as explained in Section 8–2. These steps are repeated here.

Step 1State the hypotheses and identify the claim.

Step 2Compute the test value.

Step 3Find the P-value.

Step 4Make the decision.

Step 5Summarize the results.

This method is shown in Example 8–16.

Example 8–16

Jogger’s Oxygen Uptake

A physician claims that joggers’ maximal volume oxygen uptake is greater than the average of all adults. A sample of 15 joggers has a mean of 40.6 milliliters per kilogram (ml/kg) and a standard deviation of 6 ml/kg. If the average of all adults is 36.7 ml/kg, is there enough evidence to support the physician’s claim at α = 0.05?

Solution

Step 1State the hypotheses and identify the claim.

H0: µ = 36.7andH1: µ > 36.7 (claim)

Step 2Compute the test value. The test value is

Step 3Find the P-value. Looking across the row with d.f. = 14 in Table F, you sees that 2.517 falls between 2.145 and 2.624, corresponding to α = 0.025 and α = 0.01 since this is a right-tailed test. Hence, P-value > 0.01 and P-value < 0.025 or 0.01 < P-value < 0.025. That is, the P-value is somewhere between 0.01 and 0.025. (The P-value obtained from a calculator is 0.012.)

Step 4Reject the null hypothesis since P-value < 0.05 (that is, P-value < α).

Step 5There is enough evidence to support the claim that the joggers’ maximal volume oxygen uptake is greater than 36.7 ml/kg.

Interesting Fact

The area of Alaska contains of the total area of the United States.

Students sometimes have difficulty deciding whether to use the z test or t test. The rules are the same as those pertaining to confidence intervals.

1.If σ is known, use the z test. The variable must be normally distributed if n < 30.

2.If σ is unknown but n ≥ 30, use the t test.

3.If σ is unknown and n < 30, use the t test. (The population must be approximately normally distributed.)

These rules are summarized in Figure 8–25.

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Speaking of Statistics

Can Sunshine Relieve Pain?

A study conducted at the University of Pittsburgh showed that hospital patients in rooms with lots of sunlight required less pain medication the day after surgery and during their total stay in the hospital than patients who were in darker rooms.

Patients in the sunny rooms averaged 3.2 milligrams of pain reliever per hour for their total stay as opposed to 4.1 milligrams per hour for those in darker rooms. This study compared two groups of patients. Although no statistical tests were mentioned in the article, what statistical test do you think the researchers used to compare the groups?

Figure 8–25

Using the z or t Test

Applying the Concepts 8–3

How Much Nicotine Is in Those Cigarettes?

A tobacco company claims that its best-selling cigarettes contain at most 40 mg of nicotine. This claim is tested at the 1% significance level by using the results of 15 randomly selected cigarettes. The mean was 42.6 mg and the standard deviation is 3.7 mg. Evidence suggests that nicotine is normally distributed. Information from a computer output of the hypothesis test is listed.

Sample mean = 42.6

Sample standard deviation = 3.7

Sample size = 15

Degrees of freedom = 14

P-value = 0.008

Significance level = 0.01

Test statistic t = 2.72155

Critical value t = 2.62610

1.What are the degrees of freedom?

2.Is this a z or t test?

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3.Is this a comparison of one or two samples?

4.Is this a right-tailed, left-tailed, or two-tailed test?

5.From observing the P-value, what would you conclude?

6.By comparing the test statistic to the critical value, what would you conclude?

7.Is there a conflict in this output? Explain.

8.What has been proved in this study?

See page 465 for the answers.

 Exercises 8–3

1.In what ways is the t distribution similar to the standard normal distribution? In what ways is the t distribution different from the standard normal distribution?

2.What are the degrees of freedom for the t test?

3.Find the critical value (or values) for the t test for each.

a.n = 10, a = 0.05, right-tailed

b.n = 18, a = 0.10, two-tailed

c.n = 6, a = 0.01, left-tailed

d.n = 9, a = 0.025, right-tailed

e.n = 15, a = 0.05, two-tailed

f.n = 23, a = 0.005, left-tailed

g.n = 28, a = 0.01, two-tailed

h.n = 17, a = 0.02, two-tailed

4.(ans) Using Table F, find the P-value interval for each test value.

a.t = 2.321, n = 15, right-tailed

b.t = 1.945, n = 28, two-tailed

c.t = –1.267, n = 8, left-tailed

d.t = 1.562, n = 17, two-tailed

e.t = 3.025, n = 24, right-tailed

f.t = –1.145, n = 5, left-tailed

g.t = 2.179, n = 13, two-tailed

h.t = 0.665, n = 10, right-tailed

For Exercises 5 through 18, perform each of the following steps.

a.State the hypotheses and identify the claim.

b.Find the critical value(s).

c.Find the test value.

d.Make the decision.

e.Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified.

Assume that the population is approximately normally distributed.

5.Veterinary Expenses of Cat Owners According to the American Pet Products Manufacturers Association, cat owners spend an average of \$179 annually in routine veterinary visits. A random sample of local cat owners revealed that 10 randomly selected owners spent an average of \$205 with s = \$26. Is there a significant statistical difference at α = 0.01?

Source: www.hsus.org/pets

1. Park Acreage A state executive claims that the average number of acres in western Pennsylvania state parks is less than 2000 acres. A random sample of five parks is selected, and the number of acres is shown. At α = 0.01, is there enough evidence to support the claim?

959   1187   493   6249   541

Source: Pittsburgh Tribune-Review.

7.State and Local Taxes The U.S. average for state and local taxes for a family of four is \$4172. A random sample of 20 families in a northeastern state indicates that they paid an annual amount of \$4560 with a standard deviation of \$1590. At α = 0.05, is there sufficient evidence to conclude that they pay more than the national average of \$4172?

Source: New York Times Almanac.

8.Commute Time to Work A survey of 15 large U.S. cities finds that the average commute time one way is 25.4 minutes. A chamber of commerce executive feels that the commute in his city is less and wants to publicize this. He randomly selects 25 commuters and finds the average is 22.1 minutes with a standard deviation of 5.3 minutes. At α = 0.10, is he correct?

Source: New York Times Almanac.

1. Heights of Tall Buildings A researcher estimates that the average height of the buildings of 30 or more stories in a large city is at least 700 feet. A random sample of 10 buildings is selected, and the heights in feet are shown. At α = 0.025, is there enough evidence to reject the claim?

Source: Pittsburgh Tribune-Review.

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10.Exercise and Reading Time Spent by Men Men spend an average of 29 minutes per day on weekends and holidays exercising and playing sports. They spend an average of 23 minutes per day reading. A random sample of 25 men resulted in a mean of 35 minutes exercising with a standard deviation of 6.9 minutes and an average of 20.5 minutes reading with s = 7.2 minutes. At α = 0.05 for both, is there sufficient evidence that these two results differ from the national means?

Source: Time magazine.

11.Cost of College The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was \$13,252. The following year, a random sample of 20 two-year institutions had a mean of \$15,560 and a standard deviation of \$3500. Is there sufficient evidence at the α = 0.01 level to conclude that the mean cost has increased?

Source: New York Times Almanac.

12.Income of College Students’ Parents A large university reports that the mean salary of parents of an entering class is \$91,600. To see how this compares to his university, a president surveys 28 randomly selected families and finds that their average income is \$88,500. If the standard deviation is \$10,000, can the president conclude that there is a difference? At α = 0.10, is he correct?

Source: Chronicle of Higher Education.

13.Cost of Making a Movie During a recent year the average cost of making a movie was \$54.8 million. This year, a random sample of 15 recent action movies had an average production cost of \$62.3 million with a variance of \$90.25 million. At the 0.05 level of significance, can it be concluded that it costs more than average to produce an action movie?

Source: New York Times Almanac.

1. Chocolate Chip Cookie Calories The average 1-ounce chocolate chip cookie contains 110 calories. A random sample of 15 different brands of 1-ounce chocolate chip cookies resulted in the following calorie amounts. At the α = 0.01 level, is there sufficient evidence that the average calorie content is greater than 110 calories?

Source: The Doctor’s Pocket Calorie, Fat, and Carbohydrate Counter.

1. Earnings of Financial Specialists Average weekly earnings for those working in financial activities is \$623. A group of employees working for a large national accounting firm felt that they earned more. The mean weekly earnings for a random sample of 21 employees showed an average weekly earning of \$650 with a standard deviation of \$72. At α = 0.01, can it be concluded that the employees are correct?

Source: New York Times Almanac.

1. Water Consumption The Old Farmer’s Almanac stated that the average consumption of water per person per day was 123 gallons. To test the hypothesis that this figure may no longer be true, a researcher randomly selected 16 people and found that they used on average 119 gallons per day and s = 5.3. At α = 0.05, is there enough evidence to say that the Old Farmer’s Almanac figure might no longer be correct? Use the P-value method.
2. Doctor Visits A report by the Gallup Poll stated that on average a woman visits her physician 5.8 times a year. A researcher randomly selects 20 women and obtained these data.

At α = 0.05 can it be concluded that the average is still 5.8 visits per year? Use the P-value method.

1. Number of Jobs The U.S. Bureau of Labor and Statistics reported that a person between the ages of 18 and 34 has had an average of 9.2 jobs. To see if this average is correct, a researcher selected a sample of 8 workers between the ages of 18 and 34 and asked how many different places they had worked. The results were as follows:

8    12    15    6    1    9    13    2

At α = 0.05 can it be concluded that the mean is 9.2? Use the P-value method. Give one reason why the respondents might not have given the exact number of jobs that they have worked.

1. Teaching Assistants’Stipends A random sample of stipends of teaching assistants in economics is listed. Is there sufficient evidence at the α = 0.05 level to conclude that the average stipend differs from \$15,000? The stipends listed (in dollars) are for the academic year.

Source: Chronicle of Higher Education.

1. Average Family Size The average family size was reported as 3.18. A random sample of families in a particular school district resulted in the following family sizes:

At α = 0.05, does the average family size differ from the national average?

Source: New York Times Almanac.

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Technology Step by Step

MINITAB

Step by Step

Hypothesis Test for the Mean and the t Distribution

This relates to Example 8–13. Test the claim that the average salary for substitute teachers is less than \$60 per day.

1.Enter the data into C1 of a MINITAB worksheet. Do not use the dollar sign. Name the column Salary.

2.Select Stat>Basic Statistics>1-Sample t.

3.Choose C1 Salary as the variable.

4.Click inside the text box for Test mean, and enter the hypothesized value of 60.

5.Click [Options].

6.The Alternative should be less than.

7.Click [OK] twice.

In the session window, the P-value for the test is 0.276.

One-Sample T: Salary

We cannot reject H0: There is not enough evidence in the sample to conclude the mean salary is less than \$60.

TI-83 Plus or TI-84 Plus

Step by Step

Hypothesis Test for the Mean and the t Distribution (Data)

1.Enter the data values into L1.

2.Press STAT and move the cursor to TESTS.

3.Press 2 for T-Test.

4.Move the cursor to Data and press ENTER.

5.Type in the appropriate values.

6.Move the cursor to the appropriate alternative hypothesis and press ENTER.

7.Move the cursor to Calculate and press ENTER.

Hypothesis Test for the Mean and the t Distribution (Statistics)

1.Press STAT and move the cursor to TESTS.

2.Press 2 for T-Test.

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3.Move the cursor to Stats and press ENTER.

4.Type in the appropriate values.

5.Move the cursor to the appropriate alternative hypothesis and press ENTER.

6.Move the cursor to Calculate and press ENTER.

Excel

Step by Step

Hypothesis Test for the Mean: t Test

Excel does not have a procedure to conduct a hypothesis test for the mean. However, you may conduct the test of the mean using the MegaStat Add-in available on your CD. If you have not installed this add-in, do so, following the instructions from the Chapter 1 Excel Step by Step.

Example XL8–2

This example relates to Example 8–13 from the text. At the 10% significance level, test the claim that µ < 60. The MegaStat t test uses the P-value method. Therefore, it is not necessary to enter a significance level.

1.Enter the data into column A of a new worksheet.

2.From the toolbar, select Add-Ins, MegaStat>Hypothesis Tests>Mean vs. Hypothesized Value. Note: You may need to open MegaStat from the MegaStat.xls file on your computer’s hard drive.

3.Select data input and type A1:A8 as the Input Range.

4.Type 60 for the Hypothesized mean and select the “less than” Alternative.

5.Select t test and click [OK].

The result of the procedure is shown next.

Hypothesis Test: Mean vs. Hypothesized Value

60.000 Hypothesized value

58.875 Mean data

5.083 Standard deviation

1.797 Standard error

n

7 d.f.

–0.63 t

0.2756 P-value (one-tailed, lower)

Objective 7

Test proportions, using the z test.

8-4z Test for a Proportion

Many hypothesis-testing situations involve proportions. Recall from Chapter 7 that a proportion is the same as a percentage of the population.

These data were obtained from The Book of Odds by Michael D. Shook and Robert L. Shook (New York: Penguin Putnam, Inc.):

• 59% of consumers purchase gifts for their fathers.
• 85% of people over 21 said they have entered a sweepstakes.
• 51% of Americans buy generic products.
• 35% of Americans go out for dinner once a week.

A hypothesis test involving a population proportion can be considered as a binomial experiment when there are only two outcomes and the probability of a success does not change from trial to trial. Recall from Section 5–3 that the mean is µ = np and the standard deviation is for the binomial distribution.

Since a normal distribution can be used to approximate the binomial distribution when np ≥ 5 and nq ≥ 5, the standard normal distribution can be used to test hypotheses for proportions.

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Formula for the z Test for Proportions

where

(sample proportion)

p = population proportion

n = sample size

The formula is derived from the normal approximation to the binomial and follows the general formula

We obtain from the sample (i.e., observed value), p is the expected value (i.e., hypothesized population proportion), and is the standard error.

The formula can be derived from the formula by substituting µ = nP and and then dividing both numerator and denominator by n. Some algebra is used. See Exercise 23 in this section.

The steps for hypothesis testing are the same as those shown in Section 8–3. Table E is used to find critical values and P-values.

Examples 8–17 to 8–19 show the traditional method of hypothesis testing. Example 8–20 shows the P-value method.

Sometimes it is necessary to find , as shown in Examples 8–17, 8–19, and 8–20, and sometimes is given in the exercise. See Example 8–18.

Example 8–17

People Who Are Trying to Avoid Trans Fats

A dietitian claims that 60% of people are trying to avoid trans fats in their diets. She randomly selected 200 people and found that 128 people stated that they were trying to avoid trans fats in their diets. At α = 0.05, is there enough evidence to reject the dietitian’s claim?

Source: Based on a survey by the Gallup Poll.

Solution

Step 1State the hypothesis and identify the claim.

H0: p = 0.60 (claim)andH1: p ≠ 0.60

Step 2Find the critical values. Since α = 0.05 and the test value is two-tailed, the critical values are ± 1.96.

Step 3Compute the test value. First, it is necessary to find

Substitute in the formula.

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Step 4Make the decision. Do not reject the null hypothesis since the test value falls outside the critical region, as shown in Figure 8–26.

Figure 8–26

Critical and Test Values for Example 8–17

Step 5Summarize the results. There is not enough evidence to reject the claim that 60% of people are trying to avoid trans fats in their diets.

Example 8–18

Survey on Call-Waiting Service

A telephone company representative estimates that 40% of its customers have call-waiting service. To test this hypothesis, she selected a sample of 100 customers and found that 37% had call waiting. At α = 0.01, is there enough evidence to reject the claim?

Solution

Step 1State the hypotheses and identify the claim.

H0: p = 0.40 (claim)andH1: p ≠ 0.40

Step 2Find the critical value(s). Since α = 0.01 and this test is two-tailed, the critical values are ±2.58.

Step 3Compute the test value. It is not necessary to find since it is given in the exercise; = 0.37. Substitute in the formula and solve.

p = 0.40andq = 1 – 0.40 = 0.60

Step 4Make the decision. Do not reject the null hypothesis, since the test value falls in the noncritical region, as shown in Figure 8–27.

Figure 8–27

Critical and Test Values for Example 8–18

Step 5Summarize the results. There is not enough evidence to reject the claim that 40% of the telephone company’s customers have call waiting.

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Example 8–19

Replacing \$1 Bills with \$1 Coins

A statistician read that at least 77% of the population oppose replacing \$1 bills with \$1 coins. To see if this claim is valid, the statistician selected a sample of 80 people and found that 55 were opposed to replacing the \$1 bills. At α = 0.01, test the claim that at least 77% of the population are opposed to the change.

Source: USA TODAY.

Solution

Step 1State the hypotheses and identify the claim.

H0: p = 0.77 (claim)andH1: p < 0.77

Step 2Find the critical value(s). Since α = 0.01 and the test is left-tailed, the critical value is –2.33.

Step 3Compute the test value.

Step 4Do not reject the null hypothesis, since the test value does not fall in the critical region, as shown in Figure 8–28.

Figure 8–28

Critical and Test Values for Example 8–19

Step 5There is not enough evidence to reject the claim that at least 77% of the population oppose replacing \$1 bills with \$1 coins.

Example 8–20

An attorney claims that more than 25% of all lawyers advertise. A sample of 200 lawyers in a certain city showed that 63 had used some form of advertising. At α = 0.05, is there enough evidence to support the attorney’s claim? Use the P-value method.

Solution

Step 1State the hypotheses and identify the claim.

H0: p = 0.25andH1: p > 0.25 (claim)

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Interesting Fact

Lightning is the second most common killer among storm-related hazards. On average, 73 people are killed each year by lightning. Of people who are struck by lightning, 90% do survive; however, they usually have lasting medical problems or disabilities.

Step 2Compute the test value.

Step 3Find the P-value. The area under the curve for z = 2.12 is 0.9830. Subtracting the area from 1.0000, you get 1.0000 – 0.9830 = 0.0170. The P-value is 0.0170.

Step 4Reject the null hypothesis, since 0.0170 < 0.05 (that is, P-value < 0.05). See Figure 8–29.

Figure 8–29

P-Value and α Value for Example 8–20

Step 5There is enough evidence to support the attorney’s claim that more than 25% of the lawyers use some form of advertising.

Applying the Concepts 8–4

Quitting Smoking

Assume you are part of a research team that compares products designed to help people quit smoking. Condor Consumer Products Company would like more specific details about the study to be made available to the scientific community. Review the following and then answer the questions about how you would have conducted the study.

New StopSmoke

No method has been proved more effective. StopSmoke provides significant advantages over all other methods. StopSmoke is simpler to use, and it requires no weaning. StopSmoke is also significantly less expensive than the leading brands. StopSmoke’s superiority has been proved in two independent studies.

1.What were the statistical hypotheses?

2.What were the null hypotheses?

3.What were the alternative hypotheses?

4.Were any statistical tests run?

5.Were one- or two-tailed tests run?

6.What were the levels of significance?

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7.If a type I error was committed, explain what it would have been.

8.If a type II error was committed, explain what it would have been.

9.What did the studies prove?

10.Two statements are made about significance. One states that StopSmoke provides significant advantages, and the other states that StopSmoke is significantly less expensive than other leading brands. Are they referring to statistical significance? What other type of significance is there?

See page 465 for the answers.

 Exercises 8–4

1.Give three examples of proportions.

2.Why is a proportion considered a binomial variable?

3.When you are testing hypotheses by using proportions, what are the necessary requirements?

4.What are the mean and the standard deviation of a proportion?

For Exercises 5 through 15, perform each of the following steps.

a.State the hypotheses and identify the claim.

b.Find the critical value(s).

c.Compute the test value.

d.Make the decision.

e.Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified.

5.Home Ownership A recent survey found that 68.6% of the population own their homes. In a random sample of 150 heads of households, 92 responded that they owned their homes. At the α = 0.01 level of significance, does that suggest a difference from the national proportion?

Source: World Almanac.

6.Stocks and Mutual Fund Ownership It has been found that 50.3% of U.S. households own stocks and mutual funds. A random sample of 300 heads of households indicated that 171 owned some type of stock. At what level of significance would you conclude that this was a significant difference?

Source: www.census.gov

7.Computer Hobbies It has been reported that 40% of the adult population participate in computer hobbies during their leisure time. A random sample of 180 adults found that 65 engaged in computer hobbies. At α = 0.01, is there sufficient evidence to conclude that the proportion differs from 40%?

Source: New York Times Almanac.

8.Female Physicians The percentage of physicians who are women is 27.9%. In a survey of physicians employed by a large university health system, 45 of 120 randomly selected physicians were women. Is there sufficient evidence at the 0.05 level of significance to conclude that the proportion of women physicians at the university health system exceeds 27.9%?

Source: New York Times Almanac.

9.Answering Machine Ownership It has been reported in an almanac that 78% of Americans own an answering machine. A random sample of 143 college professors at small liberal arts schools revealed that 100 owned an answering machine. At α = 0.05, test the claim that the percentage is the same as stated in the almanac.

Source: World Almanac.

10.Undergraduate Enrollment It has been found that 85.6% of all enrolled college and university students in the United States are undergraduates. A random sample of 500 enrolled college students in a particular state revealed that 420 of them were undergraduates. Is there sufficient evidence to conclude that the proportion differs from the national percentage? Use α = 0.05.

Source: Time Almanac.

11.Fatal Accidents The American Automobile Association (AAA) claims that 54% of fatal car/truck accidents are caused by driver error. A researcher studies 30 randomly selected accidents and finds that 14 were caused by driver error. Using α = 0.05, can the AAA claim be refuted?

Source: AAA/CNN.

12.Exercise to Reduce Stress A survey by Men’s Health magazine stated that 14% of men said they used exercise to reduce stress. Use α = 0.10. A random sample of 100 men was selected, and 10 said that they used exercise to relieve stress. Use the P-value method to test the claim. Could the results be generalized to all adult Americans?

13.After-School Snacks In the Journal of the American Dietetic Association, it was reported that 54% of kids said that they had a snack after school. A random sample of 60 kids was selected, and 36 said that they had a snack after school. Use α = 0.01 and the P-value method to test the claim. On the basis of the results, should parents be concerned about their children eating a healthy snack?

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14.Natural Gas Heat The Energy Information Administration reported that 51.7% of homes in the United States were heated by natural gas. A random sample of 200 homes found that 115 were heated by natural gas. Does the evidence support the claim, or has the percentage changed? Use α = 0.05 and the P-value method. What could be different if the sample were taken in a different geographic area?

15.Youth Smoking Researchers suspect that 18% of all high school students smoke at least one pack of cigarettes a day. At Wilson High School, with an enrollment of 300 students, a study found that 50 students smoked at least one pack of cigarettes a day. At α = 0.05, test the claim that 18% of all high school students smoke at least one pack of cigarettes a day. Use the P-value method.

16.Credit Card Usage For a certain year a study reports that the percentage of college students using credit cards was 83%. A college dean of student services feels that this is too high for her university, so she randomly selects 50 students and finds that 40 of them use credit cards. At α = 0.04, is she correct about her university?

Source: USA TODAY.

17.Borrowing Library Books For Americans using library services, the American Library Association (ALA) claims that 67% borrow books. A library director feels that this is not true so he randomly selects 100 borrowers and finds that 82 borrowed books. Can he show that the ALA claim is incorrect? Use α = 0.05.

Source: American Library Association; USA TODAY

18.Doctoral Students’ Salaries Nationally, at least 60% of Ph.D. students have paid assistantships. A college dean feels that this is not true in his state, so he randomly selects 50 Ph.D. students and finds that 26 have assistantships. At α = 0.05, is the dean correct?

Source: U.S. Department of Education, Chronicle of Higher Education.

19.Football Injuries A report by the NCAA states that 57.6% of football injuries occur during practices. A head trainer claims that this is too high for his conference, so he randomly selects 36 injuries and finds that 17 occurred during practices. Is his claim correct, at α = 0.05?

Source: NCAA Sports Medicine Handbook.

20.Foreign Languages Spoken in Homes Approximately 19.4% of the U.S. population 5 years old and older speaks a language other than English at home. In a large metropolitan area it was found that out of 400 randomly selected residents over 5 years of age, 94 spoke a language other than English at home. Is there sufficient evidence to conclude that the proportion is higher than the national proportion? You choose the level of significance.

Source: www.census.gov

Extending the Concepts

When np or nq is not 5 or more, the binomial table (Table B in Appendix C) must be used to find critical values in hypothesis tests involving proportions.

21.Coin Tossing A coin is tossed 9 times and 3 heads appear. Can you conclude that the coin is not balanced? Use α = 0.10. [Hint: Use the binomial table and find 2P(X ≤ 3) with p = 0.5 and n = 9.]

22.First-Class Airline Passengers In the past, 20% of all airline passengers flew first class. In a sample of 15 passengers, 5 flew first class. At α = 0.10, can you conclude that the proportions have changed?

23.Show that can be derived from by substituting µ = np and and dividing both numerator and denominator by n.

Technology Step by Step

MINITAB

Step by Step

Hypothesis Test for One Proportion and the z Distribution

MINITAB will calculate the test statistic and P-value for a test of a proportion, given the statistics from a sample or given the raw data. For Example 8–18, test the claim that 40% of all telephone customers have call-waiting service.

1.Select Stat>Basic Statistics>1 Proportion.

2.Click on the button for Summarized data. There are no data to enter in the worksheet.

3.Click in the box for Number of trials and enter 100.

4.In the Number of events box enter 37.

5.Click on [Options].

6.Type the complement of α, 99 for the confidence level.

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7.Very important! Check the box for Use test and interval based on normal distribution.

8.Click [OK] twice.

The results for the confidence interval will be displayed in the session window. Since the P-value of 0.540 is greater than α = 0.01, the null hypothesis cannot be rejected.

Test and CI for One Proportion

There is not enough evidence to conclude that the proportion is different from 40%.

TI-83 Plus or TI-84 Plus

Step by Step

Hypothesis Test for the Proportion

1.Press STAT and move the cursor to TESTS.

2.Press 5 for 1-PropZTest.

3.Type in the appropriate values.

4.Move the cursor to the appropriate alternative hypothesis and press ENTER.

5.Move the cursor to Calculate and press ENTER.

Example TI8–3

This pertains to Example 8–18 in the text. Test the claim that p = 40%, given n = 100 and = 0.37.

The test statistic is z = –0.6123724357, and the P-value is 0.5402912598.

Excel

Step by Step

Hypothesis Test for the Proportion: z Test

Excel does not have a procedure to conduct a hypothesis test for the population proportion. However, you may conduct the test of the proportion, using the MegaStat Add-in available on your CD. If you have not installed this add-in, do so, following the instructions from the Chapter 1 Excel Step by Step.

Example XL8–4

This example relates to Example 8–18 from the text. At the 1% significance level, test the claim that p = 0.40. The MegaStat test of the population proportion uses the P-value method. Therefore, it is not necessary to enter a significance level.

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1.From the toolbar, select Add-Ins, MegaStat>Hypothesis Tests>Proportion vs. Hypothesized Value. Note: You may need to open MegaStat from the MegaStat.xls file on your computer’s hard drive.

2.Type 0.37 for the Observed proportion, p.

3.Type 0.40 for the Hypothesized proportion, p.

4.Type 100 for the sample size, n.

5.Select the “not equal” Alternative.

6.Click [OK].

The result of the procedure is shown next.

Hypothesis Test for Proportion vs. Hypothesized Value

Observed   Hypothesized

0.37    0.4 p (as decimal)

37/100    40/100 p (as fraction)

1. 40. X

100    100 n

0.049   standard error

–0.61   z

0.5403   p-value (two-tailed)

Objective 8

Test variances or standard deviations, using the chi-square test.

8-5χ2 Test for a Variance or Standard Deviation

In Chapter 7, the chi-square distribution was used to construct a confidence interval for a single variance or standard deviation. This distribution is also used to test a claim about a single variance or standard deviation.

To find the area under the chi-square distribution, use Table G in Appendix C. There are three cases to consider:

1.Finding the chi-square critical value for a specific α when the hypothesis test is right-tailed.

2.Finding the chi-square critical value for a specific α when the hypothesis test is left-tailed.

3.Finding the chi-square critical values for a specific α when the hypothesis test is two-tailed.

Example 8–21

Find the critical chi-square value for 15 degrees of freedom when α = 0.05 and the test is right-tailed.

Solution

The distribution is shown in Figure 8–30.

Figure 8–30

Chi-Square Distribution for Example 8–21

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Find the α value at the top of Table G, and find the corresponding degrees of freedom in the left column. The critical value is located where the two columns meet—in this case, 24.996. See Figure 8–31.

Figure 8–31

Locating the Critical Value in Table G for Example 8–21

Example 8–22

Find the critical chi-square value for 10 degrees of freedom when α = 0.05 and the test is left-tailed.

Solution

This distribution is shown in Figure 8–32.

Figure 8–32

Chi-Square Distribution for Example 8–22

When the test is left-tailed, the α value must be subtracted from 1, that is, 1 – 0.05 = 0.95. The left side of the table is used, because the chi-square table gives the area to the right of the critical value, and the chi-square statistic cannot be negative. The table is set up so that it gives the values for the area to the right of the critical value. In this case, 95% of the area will be to the right of the value.

For 0.95 and 10 degrees of freedom, the critical value is 3.940. See Figure 8–33.

Figure 8–33

Locating the Critical Value in Table G for Example 8–22

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Example 8–23

Find the critical chi-square values for 22 degrees of freedom when α = 0.05 and a two-tailed test is conducted.

Solution

When a two-tailed test is conducted, the area must be split, as shown in Figure 8–34. Note that the area to the right of the larger value is 0.025 (0.05/2 or α/2), and the area to the right of the smaller value is 0.975 (1.00 – 0.05/2 or 1 – α/2).

Figure 8–34

Chi-Square Distribution for Example 8–23

Remember that chi-square values cannot be negative. Hence, you must use α values in the table of 0.025 and 0.975. With 22 degrees of freedom, the critical values are 36.781 and 10.982, respectively.

After the degrees of freedom reach 30, Table G gives values only for multiples of 10 (40, 50, 60, etc.). When the exact degrees of freedom sought are not specified in the table, the closest smaller value should be used. For example, if the given degrees of freedom are 36, use the table value for 30 degrees of freedom. This guideline keeps the type I error equal to or below the α value.

When you are testing a claim about a single variance using the chi-square test, there are three possible test situations: right-tailed test, left-tailed test, and two-tailed test.

If a researcher believes the variance of a population to be greater than some specific value, say, 225, then the researcher states the hypotheses as

H0: σ2 = 225andH1: σ2 > 225

and conducts a right-tailed test.

If the researcher believes the variance of a population to be less than 225, then the researcher states the hypotheses as

H0: σ2 = 225andH1: σ2 < 225

and conducts a left-tailed test.

Finally, if a researcher does not wish to specify a direction, she or he states the hypotheses as

H0: σ2 = 225andH1: σ2 ≠ 225

and conducts a two-tailed test.

Formula for the Chi-Square Test for a Single Variance

with degrees of freedom equal to n – 1 and where

n = sample size

s2 = sample variance

σ2 = population variance

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You might ask, Why is it important to test variances? There are several reasons. First, in any situation where consistency is required, such as in manufacturing, you would like to have the smallest variation possible in the products. For example, when bolts are manufactured, the variation in diameters due to the process must be kept to a minimum, or the nuts will not fit them properly. In education, consistency is required on a test. That is, if the same students take the same test several times, they should get approximately the same grades, and the variance of each of the student’s grades should be small. On the other hand, if the test is to be used to judge learning, the overall standard deviation of all the grades should be large so that you can differentiate those who have learned the subject from those who have not learned it.

Three assumptions are made for the chi-square test, as outlined here.

Unusual Stat

About 20% of cats owned in the United States are overweight.

Assumptions for the Chi-Square Test for a Single Variance

1.The sample must be randomly selected from the population.

2.The population must be normally distributed for the variable under study.

3.The observations must be independent of one another.

The traditional method for hypothesis testing follows the same five steps listed earlier. They are repeated here.

Step 1State the hypotheses and identify the claim.

Step 2Find the critical value(s).

Step 3Compute the test value.

Step 4Make the decision.

Step 5Summarize the results.

Examples 8–24 through 8–26 illustrate the traditional hypothesis-testing procedure for variances.

Example 8–24

Variation of Test Scores

An instructor wishes to see whether the variation in scores of the 23 students in her class is less than the variance of the population. The variance of the class is 198. Is there enough evidence to support the claim that the variation of the students is less than the population variance (σ2 = 225) at α = 0.05? Assume that the scores are normally distributed.

Solution

Step 1State the hypotheses and identify the claim.

H0: σ2 = 225andH1: σ2 < 225 (claim)

Step 2Find the critical value. Since this test is left-tailed and α = 0.05, use the value 1 – 0.05 = 0.95. The degrees of freedom are n – 1 = 23 – 1 = 22. Hence, the critical value is 12.338. Note that the critical region is on the left, as shown in Figure 8–35.

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Figure 8–35

Critical Value for Example 8–24

Step 3Compute the test value.

Step 4Make the decision. Since the test value 19.36 falls in the noncritical region, as shown in Figure 8–36, the decision is to not reject the null hypothesis.

Figure 8–36

Critical and Test Values for Example 8–24

Step 5Summarize the results. There is not enough evidence to support the claim that the variation in test scores of the instructor’s students is less than the variation in scores of the population.

Example 8–25

Outpatient Surgery

A hospital administrator believes that the standard deviation of the number of people using outpatient surgery per day is greater than 8. A random sample of 15 days is selected. The data are shown. At α = 0.10, is there enough evidence to support the administrator’s claim? Assume the variable is normally distributed.

Solution

Step 1State the hypotheses and identify the claim.

H0: σ2 = 64andH1: σ2 > 64 (claim)

Since the standard deviation is given, it should be squared to get the variance.

Step 2Find the critical value. Since this test is right-tailed with d.f. of 15 – 1 = 14 and α = 0.10, the critical value is 21.064.

Step 3Compute the test value. Since raw data are given, the standard deviation of the sample must be found by using the formula in Chapter 3 or your calculator. It is s = 11.2.

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Step 4Make the decision. The decision is to reject the null hypothesis since the test value, 27.44, is greater than the critical value, 21.064, and falls in the critical region. See Figure 8–37.

Figure 8–37

Critical and Test Value for Example 8–25

Step 5Summarize the results. There is enough evidence to support the claim that the standard deviation is greater than 8.

Example 8–26

Nicotine Content of Cigarettes

A cigarette manufacturer wishes to test the claim that the variance of the nicotine content of its cigarettes is 0.644. Nicotine content is measured in milligrams, and assume that it is normally distributed. A sample of 20 cigarettes has a standard deviation of 1.00 milligram. At α = 0.05, is there enough evidence to reject the manufacturer’s claim?

Solution

Step 1State the hypotheses and identify the claim.

H0: σ2 = 0.644 (claim)andH1: σ2 ≠ 0.644

Step 2Find the critical values. Since this test is a two-tailed test at α = 0.05, the critical values for 0.025 and 0.975 must be found. The degrees of freedom are 19; hence, the critical values are 32.852 and 8.907, respectively. The critical or rejection regions are shown in Figure 8–38.

Figure 8–38

Critical Values for Example 8–26

Step 3Compute the test value.

Since the standard deviation s is given in the problem, it must be squared for the formula.

Step 4Make the decision. Do not reject the null hypothesis, since the test value falls between the critical values (8.907 < 29.5 < 32.852) and in the noncritical region, as shown in Figure 8–39.

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Figure 8–39

Critical and Test Values for Example 8–26

Step 5Summarize the results. There is not enough evidence to reject the manufacturer’s claim that the variance of the nicotine content of the cigarettes is equal to 0.644.

Approximate P-values for the chi-square test can be found by using Table G in Appendix C. The procedure is somewhat more complicated than the previous procedures for finding P-values for the z and t tests since the chi-square distribution is not exactly symmetric and χ2 values cannot be negative. As we did for the t test, we will determine an interval for the P-value based on the table. Examples 8–27 through 8–29 show the procedure.

Example 8–27

Find the P-value when χ2 = 19.274, n = 8, and the test is right-tailed.

Solution

To get the P-value, look across the row with d.f. = 7 in Table G and find the two values that 19.274 falls between. They are 18.475 and 20.278. Look up to the top row and find the α values corresponding to 18.475 and 20.278. They are 0.01 and 0.005, respectively. See Figure 8–40. Hence the P-value is contained in the interval 0.005 < P-value < 0.01. (The P-value obtained from a calculator is 0.007.)

Figure 8–40

P-Value Interval for Example 8–27

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Example 8–28

Find the P-value when χ2 = 3.823, n = 13, and the test is left-tailed.

Solution

To get the P-value, look across the row with d.f. = 12 and find the two values that 3.823 falls between. They are 3.571 and 4.404. Look up to the top row and find the values corresponding to 3.571 and 4.404. They are 0.99 and 0.975, respectively. When the χ2 test value falls on the left side, each of the values must be subtracted from 1 to get the interval that P-value falls between.

1 – 0.99 = 0.01 and 1 – 0.975 = 0.025

Hence the P-value falls in the interval

0.01 < P-value < 0.025

(The P-value obtained from a calculator is 0.014.)

When the χ2 test is two-tailed, both interval values must be doubled. If a two-tailed test were being used in Example 8–28, then the interval would be 2(0.01) < P-value < 2(0.025), or 0.02 < P-value < 0.05.

The P-value method for hypothesis testing for a variance or standard deviation follows the same steps shown in the preceding sections.

Step 1State the hypotheses and identify the claim.

Step 2Compute the test value.

Step 3Find the P-value.

Step 4Make the decision.

Step 5Summarize the results.

Example 8–29 shows the P-value method for variances or standard deviations.

Example 8–29

Car Inspection Times

A researcher knows from past studies that the standard deviation of the time it takes to inspect a car is 16.8 minutes. A sample of 24 cars is selected and inspected. The standard deviation is 12.5 minutes. At α = 0.05, can it be concluded that the standard deviation has changed? Use the P-value method.

Solution

Step 1State the hypotheses and identify the claim.

H0: σ = 16.8andH1: σ ≠ 16.8 (claim)

Step 2Compute the test value.

Step 3Find the P-value. Using Table G with d.f. = 23, the value 12.733 falls between 11.689 and 13.091, corresponding to 0.975 and 0.95, respectively. Since these values are found on the left side of the distribution, each value must be subtracted from 1. Hence 1 – 0.975 = 0.025 and 1 – 0.95 = 0.05. Since this is a two-tailed test, the area must be doubled to obtain the P-value interval. Hence 0.05 < P-value < 0.10, or somewhere between 0.05 and 0.10. (The P-value obtained from a calculator is 0.085.)

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Step 4Make the decision. Since α = 0.05 and the P-value is between 0.05 and 0.10, the decision is to not reject the null hypothesis since P-value > α.

Step 5Summarize the results. There is not enough evidence to support the claim that the standard deviation has changed.

Applying the Concepts 8–5

Testing Gas Mileage Claims

Assume that you are working for the Consumer Protection Agency and have recently been getting complaints about the highway gas mileage of the new Dodge Caravans. Chrysler Corporation agrees to allow you to randomly select 40 of its new Dodge Caravans to test the highway mileage. Chrysler claims that the Caravans get 28 mpg on the highway. Your results show a mean of 26.7 and a standard deviation of 4.2. You support Chrysler’s claim.

1.Show why you support Chrysler’s claim by listing the P-value from your output. After more complaints, you decide to test the variability of the miles per gallon on the highway. From further questioning of Chrysler’s quality control engineers, you find they are claiming a standard deviation of 2.1.

2.Test the claim about the standard deviation.

3.Write a short summary of your results and any necessary action that Chrysler must take to remedy customer complaints.

4.State your position about the necessity to perform tests of variability along with tests of the means.

See pages 465 and 466 for the answers.

 Exercises 8–5

1.Using Table G, find the critical value(s) for each, show the critical and noncritical regions, and state the appropriate null and alternative hypotheses. Use σ2 = 225.

a.α = 0.05, n = 18, right-tailed

b.α = 0.10, n = 23, left-tailed

c.α = 0.05, n = 15, two-tailed

d.α = 0.10, n = 8, two-tailed

e.α = 0.01, n = 17, right-tailed

f.α = 0.025, n = 20, left-tailed

g.α = 0.01, n = 13, two-tailed

h.α = 0.025, n = 29, left-tailed

2.(ans) Using Table G, find the P-value interval for each χ2 test value.

a.χ2 = 29.321, n = 16, right-tailed

b.χ2 = 10.215, n = 25, left-tailed

c.χ2 = 24.672, n = 11, two-tailed

d.χ2 = 23.722, n = 9, right-tailed

e.χ2 = 13.974, n = 28, two-tailed

f.χ2 = 10.571, n = 19, left-tailed

g.χ2 = 12.144, n = 6, two-tailed

h.χ2 = 8.201, n = 23, two-tailed

For Exercises 3 through 9, assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified.

1. Calories in Pancake Syrup A nutritionist claims that the standard deviation of the number of calories in 1 tablespoon of the major brands of pancake syrup is 60. A sample of major brands of syrup is selected, and the number of calories is shown. At α = 0.10, can the claim be rejected?

Source: Based on information from The Complete Book of Food Counts by Corrine T. Netzer, Dell Publishers, New York.

4.High Temperatures in January Daily weather observations for southwestern Pennsylvania for the first three weeks of January show daily high temperatures as follows: 55, 44, 51, 59, 62, 60, 46, 51, 37, 30, 46, 51, 53, 57, 57, 39, 28, 37, 35, and 28 degrees Fahrenheit. The normal standard deviation in high temperatures for this time period is usually no more than 8 degrees. A meteorologist believes that with the unusual trend in temperatures the standard deviation is greater. At α = 0.05, can we conclude that the standard deviation is greater than 8 degrees?

Source: www.wunderground.com

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5.Stolen Aircraft Test the claim that the standard deviation of the number of aircraft stolen each year in the United States is less than 15 if a sample of 12 years had a standard deviation of 13.6. Use α = 0.05.

Source: Aviation Crime Prevention Institute.

1. Weights of Football Players A random sample of weights (in pounds) of football players at a local college is listed. At α = 0.05, is there sufficient evidence that the standard deviation of all football players at the college is less than 10 pounds?

Source: Football Preview, Washington Observer-Reporter.

7.Transferring Phone Calls The manager of a large company claims that the standard deviation of the time (in minutes) that it takes a telephone call to be transferred to the correct office in her company is 1.2 minutes or less. A sample of 15 calls is selected, and the calls are timed. The standard deviation of the sample is 1.8 minutes. At α = 0.01, test the claim that the standard deviation is less than or equal to 1.2 minutes. Use the P-value method.

1. Soda Bottle Content A machine fills 12-ounce bottles with soda. For the machine to function properly, the standard deviation of the sample must be less than or equal to 0.03 ounce. A sample of eight bottles is selected, and the number of ounces of soda in each bottle is given. At α = 0.05, can we reject the claim that the machine is functioning properly? Use the P-value method.
2. Calories in Doughnuts A random sample of 20 different kinds of doughnuts had the following calorie counts. At α = 0.01, is there sufficient evidence to conclude that the standard deviation is greater than 20 calories?

Source: The Doctor’s Pocket Calorie, Fat, and Carbohydrate Counter.

1. Exam Grades A statistics professor is used to having a variance in his class grades of no more than 100. He feels that his current group of students is different, and so he examines a random sample of midterm grades (listed below.) At α = 0.05, can it be concluded that the variance in grades exceeds 100?

11.Tornado Deaths A researcher claims that the standard deviation of the number of deaths annually from tornadoes in the United States is less than 35. If a sample of 11 randomly selected years had a standard deviation of 32, is the claim believable? Use α = 0.05.

Source: National Oceanic and Atmospheric Administration.

12.Interstate Speeds It has been reported that the standard deviation of the speeds of drivers on Interstate 75 near Findlay, Ohio, is 8 miles per hour for all vehicles. A driver feels from experience that this is very low. A survey is conducted, and for 50 drivers the standard deviation is 10.5 miles per hour. At α = 0.05, is the driver correct?

1. Home Run Totals A random sample of home run totals for National League Home Run Champions from 1938 to 2001 is shown. At the 0.05 level of significance, is there sufficient evidence to conclude that the variance is greater than 25?

Source: New York Times Almanac.

1. Heights of Volcanoes A sample of heights (in feet) of active volcanoes in North America, outside of Alaska, is listed below. Is there sufficient evidence that the standard deviation in heights of volcanoes outside Alaska is less than the standard deviation in heights of Alaskan volcanoes, which is 2385.9 feet? Use α = 0.05.

Source: Time Almanac.

15.Manufactured Machine Parts A manufacturing process produces machine parts with measurements the standard deviation of which must be no more than 0. 52 mm. A random sample of 20 parts in a given lot revealed a standard deviation in measurement of 0.568 mm. Is there sufficient evidence at α = 0.05 to conclude that the standard deviation of the parts is outside the required guidelines?

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Technology Step by Step

MINITAB

Step by Step

Hypothesis Test for Variance

For Example 8–25, test the administrator’s claim that the standard deviation is greater than 8. There is no menu item to calculate the test statistic and P-value directly.

Calculate the Standard Deviation and Sample Size

1.Enter the data into a column of MINITAB. Name the column OutPatients.

2.The standard deviation and sample size will be calculated and stored.

a)Select Calc>Column Statistics.

b)Check the button for Standard deviation. You can only do one of these statistics at a time.

c)Use OutPatients for the Input variable.

d)Store the result in s, then click [OK].

3.Select Edit>Edit Last Dialog Box, then do three things:

a)Change the Statistic option from Standard Deviation to N nonmissing.

b)Type n in the text box for Store the result.

c)Click [OK].

Calculate the Chi-Square Test Statistic

4.Select Calc>Calculator.

a)In the text box for Store result in variable: type in K3. The chi-square value will be stored in a constant so it can be used later.

b)In the expression, type in the formula as shown. The double asterisk is the symbol used for a power.

c)Click [OK]. The chi-square value of 27.44 will be stored in K3.

Calculate the P-Value

d)Select Calc>Probability Distributions>Chi-Square.

e)Click the button for Cumulative probability.

f)Type in 14 for Degrees of freedom.

g)Click in the text box for Input constant and type K3.

h)Type in K4 for Optional storage.

i)Click [OK]. Now K4 contains the area to the left of the chi-square test statistic.

Subtract the cumulative area from 1 to find the area on the right side of the chi-square test statistic. This is the P-value for a right-tailed test.

j)Select Calc>Calculator.

k)In the text box for Store result in variable, type in P-value.

l)The expression 1 – K4 calculates the complement of the cumulative area.

m)Click [OK].

The result will be shown in the first row of C2, 0.0168057. Since the P-value is less than α, reject the null hypothesis. The standard deviation in the sample is 11.2, the point estimate for the true standard deviation σ.

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TI-83 Plus or TI-84 Plus

Step by Step

Performing a Hypothesis Test for the Variance and Standard Deviation (Data)

1.Enter the values into L1.

2.Press PRGM, move the cursor to the program named SDHYP, and press ENTER twice.

3.Press 1 for Data.

4.Type L1 for the list and press ENTER.

5.Type the number corresponding to the type of alternative hypothesis.

6.Type the value of the hypothesized variance and press ENTER.

7.Press ENTER to clear the screen.

Example TI8–4

This pertains to Example 8–25 in the text. Test the claim that σ > 8 for these data.

25    30    5    15    18    42    16    9    10    12   12   38   8   14   27

Since P-value = 0.017 < 0.1, we reject H0 and conclude H1. Therefore, there is enough evidence to support the claim that the standard deviation of the number of people using outpatient surgery is greater than 8.

Performing a Hypothesis Test for the Variance and Standard Deviation (Statistics)

1.Press PRGM, move the cursor to the program named SDHYP, and press ENTER twice.

2.Press 2 for Stats.

3.Type the sample standard deviation and press ENTER.

4.Type the sample size and press ENTER.

5.Type the number corresponding to the type of alternative hypothesis.

6.Type the value of the hypothesized variance and press ENTER.

7.Press ENTER to clear the screen.

Example TI8–5

This pertains to Example 8–26 in the text. Test the claim that σ2 = 0.644, given n = 20 and s = 1.

Since P-value = 0.117 > 0.05, we do not reject H0 and do not conclude H1. Therefore, there is not enough evidence to reject the manufacturer’s claim that the variance of the nicotine content of the cigarettes is equal to 0.644.

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Excel

Step by Step

Hypothesis Test for the Variance: Chi-Square Test

Excel does not have a procedure to conduct a hypothesis test for the variance. However, you may conduct the test of the variance using the MegaStat Add-in available on your CD. If you have not installed this add-in, do so, following the instructions from the Chapter 1 Excel Step by Step.

Example XL8–4

This example relates to Example 8–26 from the text. At the 5% significance level, test the claim that σ2 = 0.644. The MegaStat chi-square test of the population variance uses the P-value method. Therefore, it is not necessary to enter a significance level.

1.Type a label for the variable: Nicotine in cell A1.

2.Type the observed variance: 1 in cell A2.

3.Type the sample size: 20 in cell A3.

4.From the toolbar, select Add-Ins, MegaStat>Hypothesis Tests>Chi-Square Variance Test. Note: You may need to open MegaStat from the MegaStat.xls file on your computer’s hard drive.

5.Select summary input.

6.Type A1:A3 for the Input Range.

7.Type 0.644 for the Hypothesized variance and select the “not equal” Alternative.

8.Click [OK].

The result of the procedure is shown next.

Chi-Square Variance Test

0.64  Hypothesized variance

1.00  Observed variance of nicotine

20  n

19  d.f.

29.50  Chi-square

0.1169  P-value (two-tailed)

Objective 9

Test hypotheses, using confidence intervals.

8-6Confidence Intervals and Hypothesis Testing

In hypothesis testing, another concept that might be of interest to students in elementary statistics is the relationship between hypothesis testing and confidence intervals.

There is a relationship between confidence intervals and hypothesis testing. When the null hypothesis is rejected in a hypothesis-testing situation, the confidence interval for the mean using the same level of significance will not contain the hypothesized mean. Likewise, when the null hypothesis is not rejected, the confidence interval computed using the same level of significance will contain the hypothesized mean. Examples 8–30 and 8–31 show this concept for two-tailed tests.

Example 8–30

Sugar Production

Sugar is packed in 5-pound bags. An inspector suspects the bags may not contain 5 pounds. A sample of 50 bags produces a mean of 4.6 pounds and a standard deviation of 0.7 pound. Is there enough evidence to conclude that the bags do not contain 5 pounds as stated at α = 0.05? Also, find the 95% confidence interval of the true mean.

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Solution

Now H0: µ = 5 and H1: µ ≠ 5 (claim). The critical values are +1.96 and –1.96. The test value is

Since –4.04 < –1.96, the null hypothesis is rejected. There is enough evidence to support the claim that the bags do not weigh 5 pounds.

The 95% confidence for the mean is given by

Notice that the 95% confidence interval of µ does not contain the hypothesized value µ = 5. Hence, there is agreement between the hypothesis test and the confidence interval.

Example 8–31

Hog Weights

A researcher claims that adult hogs fed a special diet will have an average weight of 200 pounds. A sample of 10 hogs has an average weight of 198.2 pounds and a standard deviation of 3.3 pounds. At α = 0.05, can the claim be rejected? Also, find the 95% confidence interval of the true mean.

Solution

Now H0: µ = 200 pounds (claim) and H1: µ 200 pounds. The t test must be used since σ is unknown. It is assumed that hog weights are normally distributed. The critical values at α = 0.05 with 9 degrees of freedom are +2.262 and –2.262. The test value is

Thus, the null hypothesis is not rejected. There is not enough evidence to reject the claim that the weight of the adult hogs is 200 pounds.

The 95% confidence interval of the mean is

The 95% confidence interval does contain the hypothesized mean µ = 200. Again there is agreement between the hypothesis test and the confidence interval.

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In summary, then, when the null hypothesis is rejected at a significance level of α, the confidence interval computed at the 1 – α level will not contain the value of the mean that is stated in the null hypothesis. On the other hand, when the null hypothesis is not rejected, the confidence interval computed at the same significance level will contain the value of the mean stated in the null hypothesis. These results are true for other hypothesis-testing situations and are not limited to means tests.

The relationship between confidence intervals and hypothesis testing presented here is valid for two-tailed tests. The relationship between one-tailed hypothesis tests and one-sided or one-tailed confidence intervals is also valid; however, this technique is beyond the scope of this textbook.

Applying the Concepts 8–6

Consumer Protection Agency Complaints

Hypothesis testing and testing claims with confidence intervals are two different approaches that lead to the same conclusion. In the following activities, you will compare and contrast those two approaches.

Assume you are working for the Consumer Protection Agency and have recently been getting complaints about the highway gas mileage of the new Dodge Caravans. Chrysler Corporation agrees to allow you to randomly select 40 of its new Dodge Caravans to test the highway mileage. Chrysler claims that the vans get 28 mpg on the highway. Your results show a mean of 26.7 and a standard deviation of 4.2. You are not certain if you should create a confidence interval or run a hypothesis test. You decide to do both at the same time.

1.Draw a normal curve, labeling the critical values, critical regions, test statistic, and population mean. List the significance level and the null and alternative hypotheses.

2.Draw a confidence interval directly below the normal distribution, labeling the sample mean, error, and boundary values.

3.Explain which parts from each approach are the same and which parts are different.

4.Draw a picture of a normal curve and confidence interval where the sample and hypothesized means are equal.

5.Draw a picture of a normal curve and confidence interval where the lower boundary of the confidence interval is equal to the hypothesized mean.

6.Draw a picture of a normal curve and confidence interval where the sample mean falls in the left critical region of the normal curve.

See page 466 for the answers.

 Exercises 8–6

1.Ski Shop Sales A ski shop manager claims that the average of the sales for her shop is \$1800 a day during the winter months. Ten winter days are selected at random, and the mean of the sales is \$1830. The standard deviation of the population is \$200. Can you reject the claim at α = 0.05? Find the 95% confidence interval of the mean. Does the confidence interval interpretation agree with the hypothesis test results? Explain. Assume that the variable is normally distributed.

2.One-Way Airfares The average one-way airfare from Pittsburgh to Washington, D.C., is \$236. A random sample of 20 one-way fares during a particular month had a mean of \$210 with a standard deviation of \$43. At α = 0.02, is there sufficient evidence to conclude a difference from the stated mean? Use the sample statistics to construct a 98% confidence interval for the true mean one-way airfare from Pittsburgh to Washington, D.C., and compare your interval to the results of the test. Do they support or contradict one another?

Source: www.fedstats.gov

3.Condominium Monthly Maintenance Fees The sales manager of a rental agency claims that the monthly maintenance fee for a condominium in the Lakewood region is \$86. Past surveys showed that the standard deviation of the population is \$6. A sample of 15 owners shows that they pay an average of \$84. Test the manager’s claim at α = 0.01. Find the 99% confidence interval of the mean. Does the confidence interval interpretation agree with the results of the hypothesis test? Explain. Assume that the variable is normally distributed.

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4.Canoe Trip Times The average time it takes a person in a one-person canoe to complete a certain river course is 47 minutes. Because of rapid currents in the spring, a group of 10 people traverse the course in 42 minutes. The standard deviation, known from previous trips, is 7 minutes. Test the claim that this group’s time was different because of the strong currents. Use α = 0.10. Find the 90% confidence level of the true mean. Does the confidence interval interpretation agree with the results of the hypothesis test? Explain. Assume that the variable is normally distributed.

5.Working at Home Workers with a formal arrangement with their employer to be paid for time worked at home worked an average of 19 hours per week. A random sample of 15 mortgage brokers indicated that they worked a mean of 21.3 hours per week with a standard deviation of 6.5 hours. At α = 0.05, is there sufficient evidence to conclude a difference? Construct a 95% confidence interval for the true mean number of paid working hours at home. Compare the results of your confidence interval to the conclusion of your hypothesis test and discuss the implications.

Source: www.bls.gov

6.Newspaper Reading Times A survey taken several years ago found that the average time a person spent reading the local daily newspaper was 10.8 minutes. The standard deviation of the population was 3 minutes. To see whether the average time had changed since the newspaper’s format was revised, the newspaper editor surveyed 36 individuals. The average time that the 36 people spent reading the paper was 12.2 minutes. At α = 0.02, is there a change in the average time an individual spends reading the newspaper? Find the 98% confidence interval of the mean. Do the results agree? Explain.

Summary

This chapter introduces the basic concepts of hypothesis testing. A statistical hypothesis is a conjecture about a population. There are two types of statistical hypotheses: the null and the alternative hypotheses. The null hypothesis states that there is no difference, and the alternative hypothesis specifies a difference. To test the null hypothesis, researchers use a statistical test. Many test values are computed by using

Two common statistical tests for hypotheses about a mean are the z test and the t test. The z test is used either when the population standard deviation is known and the variable is normally distributed or when σ is known and the sample size is greater than or equal to 30. When the population standard deviation is not known and the variable is normally distributed, the sample standard deviation is used, but a t test should be conducted instead. The z test is also used to test proportions when np ≥ 5 and nq ≥ 5.

Researchers compute a test value from the sample data in order to decide whether the null hypothesis should be rejected. Statistical tests can be one-tailed or two-tailed, depending on the hypotheses.

The null hypothesis is rejected when the difference between the population parameter and the sample statistic is said to be significant. The difference is significant when the test value falls in the critical region of the distribution. The critical region is determined by α, the level of significance of the test. The level is the probability of committing a type I error. This error occurs when the null hypothesis is rejected when it is true. Three generally agreed upon significance levels are 0.10, 0.05, and 0.01. A second kind of error, the type II error, can occur when the null hypothesis is not rejected when it is false.

Finally, you can test a single variance by using a chi-square test.

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All hypothesis-testing situations using the traditional method should include the following steps:

1.State the null and alternative hypotheses and identify the claim.

2.State an alpha level and find the critical value(s).

3.Compute the test value.

4.Make the decision to reject or not reject the null hypothesis.

5.Summarize the results.

All hypothesis-testing situations using the P-value method should include the following steps:

1.State the hypotheses and identify the claim.

2.Compute the test value.

3.Find the P-value.

4.Make the decision.

5.Summarize the results.

Important Terms

α (alpha)

alternative hypothesis

β (beta)

chi-square test

critical or rejection region

critical value

hypothesis testing

left-tailed test

level of significance

noncritical or nonrejection region

null hypothesis

one-tailed test

P-value

research hypothesis

right-tailed test

statistical hypothesis

statistical test

test value

t test

two-tailed test

type I error

type II error

z test

Important Formulas

Formula for the z test for means:

Formula for the t test for means:

Formula for the z test for proportions:

Formula for the chi-square test for variance or standard deviation:

Review Exercises

For Exercises 1 through 19, perform each of the following steps.

a.State the hypotheses and identify the claim.

b.Find the critical value(s).

c.Compute the test value.

d.Make the decision.

e.Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified.

1. High Temperatures in the United States A meteorologist claims that the average of the highest temperatures in the United States is 98°. A random sample of 50 cities is selected, and the highest temperatures are recorded. The data are shown. At a α = 0.05, can the claim be rejected? σ = 7.71.

Source: The World Almanac & Book of Facts.

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2.Salaries for Actuaries Nationwide graduates entering the actuarial field earn \$40,000. A college placement officer feels that this number is too low. She surveys 36 graduates entering the actuarial field and finds the average salary to be \$41,000. The population standard deviation is \$3000. Can her claim be supported at α = 0.05?

Source: BeAnActuary.org

3.Monthly Home Rent The average monthly rent for a one-bedroom home in San Francisco is \$1229. A random sample of 15 one-bedroom homes about 15 miles outside of San Francisco had a mean rent of \$1350. The population standard deviation is \$250. At α = 0.05, can we conclude that the monthly rent outside San Francisco differs from that in the city?

Source: New York Times Almanac.

4.Salaries for Actuaries Nationwide, the average salary of actuaries who achieve the rank of Fellow is \$150,000. An insurance executive wants to see how this compares with Fellows within his company. He checks the salaries of eight Fellows and finds the average salary to be \$155,500 with a standard deviation of \$15,000. Can he conclude that Fellows in his company make more than the national average, using α = 0.05?

Source: BeAnActuary.org

1. Debt of College Graduates A random sample of the average debt (in dollars) at graduation from 30 of the top 100 public colleges and universities is listed below. Is there sufficient evidence at α = 0.01 to conclude that the population mean debt at graduation is less than \$18,000?

Source: www.Kiplinger.com

6.Tennis Fans The Tennis Industry Association stated that the average age of a tennis fan is 32 years. To test the claim, a researcher selected a random sample of 18 tennis fans and found that the mean of their ages was 31.3 years and the standard deviation was 2.8 years. At α = 0.05 does it appear that the average age is lower than what was stated by the Tennis Industry Association? Use the P-value method, and assume the variable is approximately normally distributed.

1. Whooping Crane Eggs Once down to about 15, the world’s only wild flock of whooping cranes now numbers a record 237 birds in its Texas Coastal Bend wintering ground (www.SunHerald.com). The average whooping crane egg weighs 208 grams. A new batch of eggs was recently weighed, and their weights are listed below. At α = 0.01, is there sufficient evidence to conclude that the weight is greater than 208 grams?

Source: http:whoopers.usgs.gov

8.Union Membership Nationwide 13.7% of employed wage and salary workers are union members (down from 20.1% in 1983). A random sample of 300 local wage and salary workers showed that 50 belonged to a union. At α = 0.05, is there sufficient evidence to conclude that the proportion of union membership differs from 13.7%?

Source: Time Almanac.

9.Federal Prison Populations Nationally 60.2% of federal prisoners are serving time for drug offenses. A warden feels that in his prison the percentage is even higher. He surveys 400 inmates’ records and finds that 260 of the inmates are drug offenders. At α = 0.05, is he correct?

Source: New York Times Almanac.

10.Free School Lunches It has been reported that 59.3% of U.S. school lunches served are free or at a reduced price. A random sample of 300 children in a large metropolitan area indicated that 156 of them received lunch free or at a reduced price. At the 0.01 level of significance, is there sufficient evidence to conclude that the proportion is less than 59.3%?

Source: www.fns.usda.gov

11.Portable Radio Ownership A radio manufacturer claims that 65% of teenagers 13 to 16 years old have their own portable radios. A researcher wishes to test the claim and selects a random sample of 80 teenagers. She finds that 57 have their own portable radios. At α = 0.05, should the claim be rejected? Use the P-value method.

12.Weights of Football Players A football coach claims that the average weight of all the opposing teams’ members is 225 pounds. For a test of the claim, a sample of 50 players is taken from all the opposing teams. The mean is found to be 230 pounds. The population standard deviation is 15 pounds. At α = 0.01, test the coach’s claim. Find the P-value and make the decision.

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13.Time Until Indigestion Relief An advertisement claims that Fasto Stomach Calm will provide relief from indigestion in less than 10 minutes. For a test of the claim, 35 individuals were given the product; the average time until relief was 9.25 minutes. From past studies, the standard deviation of the population is known to be 2 minutes. Can you conclude that the claim is justified? Find the P-value and let α = 0.05.

14.Times of Videos A film editor feels that the standard deviation for the number of minutes in a video is 3.4 minutes. A sample of 24 videos has a standard deviation of 4.2 minutes. At α = 0.05, is the sample standard deviation different from what the editor hypothesized?

15.Fuel Consumption The standard deviation of the fuel consumption of a certain automobile is hypothesized to be greater than or equal to 4.3 miles per gallon. A sample of 20 automobiles produced a standard deviation of 2.6 miles per gallon. Is the standard deviation really less than previously thought? Use α = 0.05 and the P-value method.

1. Apartment Rental Rates A real estate agent claims that the standard deviation of the rental rates of apartments in a certain county is \$95. A random sample of rates in dollars is shown. At α = 0.02, can the claim be refuted?

Source: Pittsburgh Tribune-Review.

1. Games Played by NBA Scoring Leaders A random sample of the number of games played by individual NBA scoring leaders is found below. Is there sufficient evidence to conclude that the variance in games played differs from 40? Use α = 0.05.

Source: Time Almanac.

18.Tire Inflation To see whether people are keeping their car tires inflated to the correct level of 35 pounds per square inch (psi), a tire company manager selects a sample of 36 tires and checks the pressure. The mean of the sample is 33.5 psi, and the population standard deviation is 3 psi. Are the tires properly inflated? Use α = 0.10. Find the 90% confidence interval of the mean. Do the results agree? Explain.

19.Plant Leaf Lengths A biologist knows that the average length of a leaf of a certain full-grown plant is 4 inches. The standard deviation of the population is 0.6 inch. A sample of 20 leaves of that type of plant given a new type of plant food had an average length of 4.2 inches. Is there reason to believe that the new food is responsible for a change in the growth of the leaves? Use α = 0.01. Find the 99% confidence interval of the mean. Do the results concur? Explain. Assume that the variable is approximately normally distributed.

Statistics Today

How Much Better Is Better?–Revisited

Now that you have learned the techniques of hypothesis testing presented in this chapter, you realize that the difference between the sample mean and the population mean must be significant before you can conclude that the students really scored above average. The superintendent should follow the steps in the hypothesis-testing procedure and be able to reject the null hypothesis before announcing that his students scored higher than average.

Data Analysis

The Data Bank is found in Appendix D, or on the World Wide Web by following links from www.mhhe.com/bluman/

1.From the Data Bank, select a random sample of at least 30 individuals, and test one or more of the following hypotheses by using the z test. Use α = 0.05.

a.For serum cholesterol, H0: µ = 220 milligram percent (mg%).

b.For systolic pressure, H0: µ = 120 millimeters of mercury (mm Hg).

c.For IQ, H0: µ = 100.

d.For sodium level, H0: µ = 140 milliequivalents per liter (mEq/l).

2.Select a random sample of 15 individuals and test one or more of the hypotheses in Exercise 1 by using the t test. Use α = 0.05.

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3.Select a random sample of at least 30 individuals, and using the z test for proportions, test one or more of the following hypotheses. Use α = 0.05.

a.For educational level, H0: p = 0.50 for level 2.

b.For smoking status, H0: p = 0.20 for level 1.

c.For exercise level, H0: p = 0.10 for level 1.

d.For gender, H0: p = 0.50 for males.

4.Select a sample of 20 individuals and test the hypothesis H0: σ2 = 225 for IQ level. Use α = 0.05.

5.Using the data from Data Set XIII, select a sample of 10 hospitals and test H0: µ = 250 and H1: µ < 250 for the number of beds. Use α = 0.05.

6.Using the data obtained in Exercise 5, test the hypothesis H0: σ ≥ 150. Use σ = 0.05.

Chapter Quiz

Determine whether each statement is true or false. If the statement is false, explain why.

1.No error is committed when the null hypothesis is rejected when it is false.

2.When you are conducting the t test, the population must be approximately normally distributed.

3.The test value separates the critical region from the noncritical region.

4.The values of a chi-square test cannot be negative.

5.The chi-square test for variances is always one-tailed.

6.When the value of α is increased, the probability of committing a type I error is

a.Decreased

b.Increased

c.The same

d.None of the above

7.If you wish to test the claim that the mean of the population is 100, the appropriate null hypothesis is

1. a. = 100

b.µ ≥ 100

c.µ ≤ 100

d.µ = 100

8.The degrees of freedom for the chi-square test for variances or standard deviations are

a.1

b.n

c.n – 1

d.None of the above

9.For the z test, if s is unknown and n ≥ 30, one can substitute ______ for σ.

a.n

b.s

c.χ2

d.t

Complete the following statements with the best answer.

10.Rejecting the null hypothesis when it is true is called a(n) _________ error.

11.The probability of a type II error is referred to as ________.

12.A conjecture about a population parameter is called a(n) ________.

13.To test the claim that the mean is greater than 87, you would use a(n) _______-tailed test.

14.The degrees of freedom for the t test are __________.

For the following exercises where applicable:

a.State the hypotheses and identify the claim.

b.Find the critical value(s).

c.Compute the test value.

d.Make the decision.

e.Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified.

1. Ages of Professional Women A sociologist wishes to see if it is true that for a certain group of professional women, the average age at which they have their first child is 28.6 years. A random sample of 36 women is selected, and their ages at the birth of their first child are recorded. At α = 0.05, does the evidence refute the sociologist’s assertion? σ = 4.18.

16.Home Closing Costs A real estate agent believes that the average closing cost of purchasing a new home is \$6500 over the purchase price. She selects 40 new home sales at random and finds that the average closing costs are \$6600. The standard deviation of the population is \$120. Test her belief at α = 0.05.

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17.Chewing Gum Use A recent study stated that if a person chewed gum, the average number of sticks of gum he or she chewed daily was 8. To test the claim, a researcher selected a random sample of 36 gum chewers and found the mean number of sticks of gum chewed per day was 9. The standard deviation of the population is 1. At α = 0.05, is the number of sticks of gum a person chews per day actually greater than 8?

1. Hotel Rooms A travel agent claims that the average of the number of rooms in hotels in a large city is 500. At α = 0.01 is the claim realistic? The data for a sample of six hotels are shown.

713    300    292    311    598    401    618

Give a reason why the claim might be deceptive.

19.Heights of Models In a New York modeling agency, a researcher wishes to see if the average height of female models is really less than 67 inches, as the chief claims. A sample of 20 models has an average height of 65.8 inches. The standard deviation of the sample is 1.7 inches. At α = 0.05, is the average height of the models really less than 67 inches? Use the P-value method.

20.Experience of Taxi Drivers A taxi company claims that its drivers have an average of at least 12.4 years’ experience. In a study of 15 taxi drivers, the average experience was 11.2 years. The standard deviation was 2. At α = 0.10, is the number of years’ experience of the taxi drivers really less than the taxi company claimed?

21.Ages of Robbery Victims A recent study in a small city stated that the average age of robbery victims was 63.5 years. A sample of 20 recent victims had a mean of 63.7 years and a standard deviation of 1.9 years. At α = 0.05, is the average age higher than originally believed? Use the P-value method.

22.First-Time Marriages A magazine article stated that the average age of women who are getting married for the first time is 26 years. A researcher decided to test this hypothesis at α = 0.02. She selected a sample of 25 women who were recently married for the first time and found the average was 25.1 years. The standard deviation was 3 years. Should the null hypothesis be rejected on the basis of the sample?

23.Survey on Vitamin Usage A survey in Men’s Health magazine reported that 39% of cardiologists said that they took vitamin E supplements. To see if this is still true, a researcher randomly selected 100 cardiologists and found that 36 said that they took vitamin E supplements. At α = 0.05 test the claim that 39% of the cardiologists took vitamin E supplements. A recent study said that taking too much vitamin E might be harmful. How might this study make the results of the previous study invalid?

24.Breakfast Survey A dietitian read in a survey that at least 55% of adults do not eat breakfast at least 3 days a week. To verify this, she selected a random sample of 80 adults and asked them how many days a week they skipped breakfast. A total of 50% responded that they skipped breakfast at least 3 days a week. At α = 0.10, test the claim.

25.Caffeinated Beverage Survey A Harris Poll found that 35% of people said that they drink a caffeinated beverage to combat midday drowsiness. A recent survey found that 19 out of 48 people stated that they drank a caffeinated beverage to combat midday drowsiness. At α = 0.02 is the claim of the percentage found in the Harris Poll believable?

26.Radio Ownership A magazine claims that 75% of all teenage boys have their own radios. A researcher wished to test the claim and selected a random sample of 60 teenage boys. She found that 54 had their own radios. At α = 0.01, should the claim be rejected?

27.Find the P-value for the z test in Exercise 15.

28.Find the P-value for the z test in Exercise 16.

29.Pages in Romance Novels A copyeditor thinks the standard deviation for the number of pages in a romance novel is greater than 6. A sample of 25 novels has a standard deviation of 9 pages. At α = 0.05, is it higher, as the editor hypothesized?

30.Seed Germination Times It has been hypothesized that the standard deviation of the germination time of radish seeds is 8 days. The standard deviation of a sample of 60 radish plants’ germination times was 6 days. At α = 0.01, test the claim.

31.Pollution By-products The standard deviation of the pollution by-products released in the burning of 1 gallon of gas is 2.3 ounces. A sample of 20 automobiles tested produced a standard deviation of 1.9 ounces. Is the standard deviation really less than previously thought? Use α = 0.05.

32.Strength of Wrapping Cord A manufacturer claims that the standard deviation of the strength of wrapping cord is 9 pounds. A sample of 10 wrapping cords produced a standard deviation of 11 pounds. At α = 0.05, test the claim. Use the P-value method.

33.Find the 90% confidence interval of the mean in Exercise 15. Is µ contained in the interval?

34.Find the 95% confidence interval for the mean in Exercise 16. Is µ contained in the interval?

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Critical Thinking Challenges

The power of a test (1 – β) can be calculated when a specific value of the mean is hypothesized in the alternative hypothesis; for example, let H0: µ = 50 and let H1: µ = 52. To find the power of a test, it is necessary to find the value of β. This can be done by the following steps:

Step 1For a specific value of α find the corresponding value of , using , where µ is the hypothesized value given in H0: Use a right-tailed test.

Step 2Using the value of found in step 1 and the value of µ in the alternative hypothesis, find the area corresponding to z in the formula

Step 3Subtract this area from 0.5000. This is the value of β.

Step 4Subtract the value of β from 1. This will give you the power of a test. See Figure 8–41.

1.Find the power of a test, using the hypotheses given previously and α = 0.05, σ = 3, and n = 30.

2.Select several other values for µ in H1 and compute the power of the test. Generalize the results.

Figure 8–41

Relationship Among α, β, and the Power of a Test

Data Projects

Use a significance level of 0.05 for all tests below.

1.Business and Finance Use the Dow Jones Industrial stocks in data project 1 of Chapter 7 as your data set. Find the gain or loss for each stock over the last quarter. Test the claim that the mean is that the stocks broke even (no gain or loss indicates a mean of 0).

2.Sports and Leisure Use the most recent NFL season for your data. For each team, find the quarterback rating for the number one quarterback. Test the claim that the mean quarterback rating for a number one quarterback is more than 80.

3.Technology Use your last month’s itemized cell phone bill for your data. Determine the percentage of your text messages that were outgoing. Test the claim that a majority of your text messages were outgoing. Determine the mean, median, and standard deviation for the length of a call. Test the claim that the mean length of a call is longer than the value for you found for the median length.

4.Health and Wellness Use the data collected in data project 4 of Chapter 7 for this exercise. Test the claim that the mean body temperature is less than 98.6 degrees Fahrenheit.

5.Politics and Economics Use the most recent results of the Presidential primary elections for both parties. Determine what percentage of voters in your state voted for the eventual Democratic nominee for President and what percentage voted for the eventual Republican nominee. Test the claim that a majority of your state favored the candidate who won the nomination for each party.

6.Your Class Use the data collected in data project 6 of Chapter 7 for this exercise. Test the claim that the mean BMI for a student is more than 25.

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Section 8–1 Eggs and Your Health

1.The study was prompted by claims that linked foods high in cholesterol to high blood serum cholesterol.

2.The population under study is people in general.

3.A sample of 500 subjects was collected.

4.The hypothesis was that eating eggs did not increase blood serum cholesterol.

5.Blood serum cholesterol levels were collected.

6.Most likely but we are not told which test.

7.The conclusion was that eating a moderate amount of eggs will not significantly increase blood serum cholesterol level.

Section 8–2 Car Thefts

1.The hypotheses are H0: µ = 44 and H1: µ ≠ 44.

2.This sample can be considered large for our purposes.

3.The variable needs to be normally distributed.

4.We will use a z distribution.

5.Since we are interested in whether the car theft rate has changed, we use a two-tailed test.

6.Answers may vary. At the α = 0.05 significance level, the critical values are z = ±1.96.

7.The sample mean is = 55.97, and the population standard deviation is 30.30. Our test statistic is .

8.Since 2.37 > 1.96, we reject the null hypothesis.

9.There is enough evidence to conclude that the car theft rate has changed.

10.Answers will vary. Based on our sample data, it appears that the car theft rate has changed from 44 vehicles per 10,000 people. In fact, the data indicate that the car theft rate has increased.

11.Based on our sample, we would expect 55.97 car thefts per 10,000 people, so we would expect (55.97)(5) = 279.85, or about 280, car thefts in the city.

Section 8–3 How Much Nicotine Is in Those Cigarettes?

1.We have 15 – 1 = 14 degrees of freedom.

2.This is a t test.

3.We are only testing one sample.

4.This is a right-tailed test, since the hypotheses of the tobacco company are H0: µ = 40 and H1: µ > 40.

5.The P-value is 0.008, which is less than the significance level of 0.01. We reject the tobacco company’s claim.

6.Since the test statistic (2.72) is greater than the critical value (2.62), we reject the tobacco company’s claim.

7.There is no conflict in this output, since the results based on the P-value and on the critical value agree.

8.Answers will vary. It appears that the company’s claim is false and that there is more than 40 mg of nicotine in its cigarettes.

Section 8–4 Quitting Smoking

1.The statistical hypotheses were that StopSmoke helps more people quit smoking than the other leading brands.

2.The null hypotheses were that StopSmoke has the same effectiveness as or is not as effective as the other leading brands.

3.The alternative hypotheses were that StopSmoke helps more people quit smoking than the other leading brands. (The alternative hypotheses are the statistical hypotheses.)

4.No statistical tests were run that we know of.

5.Had tests been run, they would have been one-tailed tests.

6.Some possible significance levels are 0.01, 0.05, and 0.10.

7.A type I error would be to conclude that StopSmoke is better when it really is not.

8.A type II error would be to conclude that StopSmoke is not better when it really is.

9.These studies proved nothing. Had statistical tests been used, we could have tested the effectiveness of StopSmoke.

10.Answers will vary. One possible answer is that more than likely the statements are talking about practical significance and not statistical significance, since we have no indication that any statistical tests were conducted.

Section 8–5 Testing Gas Mileage Claims

1. The hypotheses are H0: µ = 28 and H1: µ < 28. The value of our test statistic is z = – 1.96, and the associated P-value is 0.02514. We would reject Chrysler’s claim that the Dodge Caravans are getting 28 mpg.
2. The hypotheses are H0: σ = 2.1 and H1: σ > 2.1. The value of our test statistic is , and the associated P-value is approximately zero. We would reject Chrysler’s claim that the standard deviation is 2.1 mpg.

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3.Answers will vary. It is recommended that Chrysler lower its claim about the highway miles per gallon of the Dodge Caravans. Chrysler should also try to reduce variability in miles per gallon and provide confidence intervals for the highway miles per gallon.

4.Answers will vary. There are cases when a mean may be fine, but if there is a lot of variability about the mean, there will be complaints (due to the lack of consistency).

Section 8–6 Consumer Protection Agency Complaints