To determine exactly how much statistical evidence is required to reject *H*0, we consider the errors and the correct decisions that can be made in hypothesis testing. These errors and correct decisions, as well as their implications in the trash bag advertising example, are summarized in Tables 9.1 and 9.2. Across the top of each table are listed the two possible “states of nature.” Either *H*0: *μ* ≤ 50 is true, which says the manufacturer’s claim that *μ* is greater than 50 is false, or *H*0 is false, which says the claim is true. Down the left side of each table are listed the two possible decisions we can make in the hypothesis test. Using the sample data, we will either reject *H*0: *μ* ≤ 50, which implies that the claim will be advertised, or we will not reject *H*0, which implies that the claim will not be advertised.

Table 9.1: Type I and Type II Errors

Table 9.2: The Implications of Type I and Type II Errors in the Trash Bag Example

In general, the two types of errors that can be made in hypothesis testing are defined here:

**Type I and Type II Errors**

If we reject *H*0 when it is true, this is a **Type I error.**

If we do not reject *H*0 when it is false, this is a **Type II error.**

As can be seen by comparing Tables 9.1 and 9.2, if we commit a Type I error, we will advertise a false claim. If we commit a Type II error, we will fail to advertise a true claim.

We now let the symbol ** α **(pronounced

We sometimes choose *α* as high as .10, but we usually choose *α* between .05 and .01. A frequent choice for *α* is .05. In fact, our former student tells us that the network often tests advertising claims by setting the probability of a Type I error equal to .05. That is, the network will run a commercial making a claim if the sample evidence allows it to reject a null hypothesis that says the claim is not valid in favor of an alternative hypothesis that says the claim is valid with *α* set equal to .05. Since a Type I error is deciding that the claim is valid when it is not, the policy of setting *α* equal to .05 says that, in the long run, the network will advertise only 5 percent of all invalid claims made by advertisers.

One might wonder why the network does not set *α* lower—say at .01. One reason is that **it can be shown that, for a fixed sample size, the lower we set α, the higher is β, and the higher we set α, the lower is β.** Setting

Exercises for Section 9.1

CONCEPTS

**9.1** Which hypothesis (the null hypothesis, *H*0, or the alternative hypothesis, *Ha*) is the “status quo” hypothesis (that is, the hypothesis that states that things are remaining “as is”)? Which hypothesis is the hypothesis that says that a “hoped for” or “suspected” condition exists?

**9.2** Which hypothesis (*H*0 or *Ha*) is not rejected unless there is convincing sample evidence that it is false? Which hypothesis (*H*0 or *Ha*) will be accepted only if there is convincing sample evidence that it is true?

**9.3** Define each of the following:

**a** Type I error

**b** Type II error

**c** *α*

**d** *β*

**9.4** For each of the following situations, indicate whether an error has occurred and, if so, indicate what kind of error (Type I or Type II) has occurred.

**a** We do not reject *H*0 and *H*0 is true.

**b** We reject *H*0 and *H*0 is true.

**c** We do not reject *H*0 and *H*0 is false.

**d** We reject *H*0 and *H*0 is false.

**9.5** If we reject *H*0, what is the only type of error that we could be making? Explain.

**9.6** If we do not reject *H*0, what is the only type of error that we could be making? Explain.

**9.7** When testing a hypothesis, why don’t we set the probability of a Type I error to be extremely small? Explain.

METHODS AND APPLICATIONS

**9.8** **THE VIDEO GAME SATISFACTION RATING CASE** VideoGame

Recall that “very satisfied” customers give the XYZ-Box video game system a rating that is at least 42. Suppose that the manufacturer of the XYZ-Box wishes to use the random sample of 65 satisfaction ratings to provide evidence supporting the claim that the mean composite satisfaction rating for the XYZ-Box exceeds 42.

**a** Letting *μ* represent the mean composite satisfaction rating for the XYZ-Box, set up the null and alternative hypotheses needed if we wish to attempt to provide evidence supporting the claim that *μ* exceeds 42.

**b** In the context of this situation, interpret making a Type I error; interpret making a Type II error.

**9.9** **THE BANK CUSTOMER WAITING TIME CASE** WaitTime

Recall that a bank manager has developed a new system to reduce the time customers spend waiting for teller service during peak hours. The manager hopes the new system will reduce waiting times from the current 9 to 10 minutes to less than 6 minutes.

Suppose the manager wishes to use the random sample of 100 waiting times to support the claim that the mean waiting time under the new system is shorter than six minutes.

**a** Letting *μ* represent the mean waiting time under the new system, set up the null and alternative hypotheses needed if we wish to attempt to provide evidence supporting the claim that *μ* is shorter than six minutes.

**b** In the context of this situation, interpret making a Type I error; interpret making a Type II error.

**9.10** An automobile parts supplier owns a machine that produces a cylindrical engine part. This part is supposed to have an outside diameter of three inches. Parts with diameters that are too small or too large do not meet customer requirements and must be rejected. Lately, the company has experienced problems meeting customer requirements. The technical staff feels that the mean diameter produced by the machine is off target. In order to verify this, a special study will randomly sample 100 parts produced by the machine. The 100 sampled parts will be measured, and if the results obtained cast a substantial amount of doubt on the hypothesis that the mean diameter equals the target value of three inches, the company will assign a problem-solving team to intensively search for the causes of the problem.

**a** The parts supplier wishes to set up a hypothesis test so that the problem-solving team will be assigned when the null hypothesis is rejected. Set up the null and alternative hypotheses for this situation.

**b** In the context of this situation, interpret making a Type I error; interpret making a Type II error.

**c** Suppose it costs the company $3,000 a day to assign the problem-solving team to a project. Is this $3,000 figure the daily cost of a Type I error or a Type II error? Explain.

**9.11** The Crown Bottling Company has just installed a new bottling process that will fill 16-ounce bottles of the popular Crown Classic Cola soft drink. Both overfilling and underfilling bottles are undesirable: Underfilling leads to customer complaints and overfilling costs the company considerable money. In order to verify that the filler is set up correctly, the company wishes to see whether the mean bottle fill, *μ*, is close to the target fill of 16 ounces. To this end, a random sample of 36 filled bottles is selected from the output of a test filler run. If the sample results cast a substantial amount of doubt on the hypothesis that the mean bottle fill is the desired 16 ounces, then the filler’s initial setup will be readjusted.

**a** The bottling company wants to set up a hypothesis test so that the filler will be readjusted if the null hypothesis is rejected. Set up the null and alternative hypotheses for this hypothesis test.

**b** In the context of this situation, interpret making a Type I error; interpret making a Type II error.

**9.12** Consolidated Power, a large electric power utility, has just built a modern nuclear power plant. This plant discharges waste water that is allowed to flow into the Atlantic Ocean. The Environmental Protection Agency (EPA) has ordered that the waste water may not be excessively warm so that thermal pollution of the marine environment near the plant can be avoided. Because of this order, the waste water is allowed to cool in specially constructed ponds and is then released into the ocean. This cooling system works properly if the mean temperature of waste water discharged is 60°F or cooler. Consolidated Power is required to monitor the temperature of the waste water. A sample of 100 temperature readings will be obtained each day, and if the sample results cast a substantial amount of doubt on the hypothesis that the cooling system is working properly (the mean temperature of waste water discharged is 60°F or cooler), then the plant must be shut down and appropriate actions must be taken to correct the problem.

**a** Consolidated Power wishes to set up a hypothesis test so that the power plant will be shut down when the null hypothesis is rejected. Set up the null and alternative hypotheses that should be used.

**b** In the context of this situation, interpret making a Type I error; interpret making a Type II error.

**c** The EPA periodically conducts spot checks to determine whether the waste water being discharged is too warm. Suppose the EPA has the power to impose very severe penalties (for example, very heavy fines) when the waste water is excessively warm. Other things being equal, should Consolidated Power set the probability of a Type I error equal to *α* = .01 or *α* = .05? Explain.

**9.13** Consider Exercise 9.12, and suppose that Consolidated Power has been experiencing technical problems with the cooling system. Because the system has been unreliable, the company feels it must take precautions to avoid failing to shut down the plant when its waste water is too warm. Other things being equal, should Consolidated Power set the probability of a Type I error equal to *α* = .01 or *α* = .05? Explain.

9.2: *z* Tests about a Population Mean: *σ* Known

In this section we discuss hypothesis tests about a population mean that are *based on the normal distribution*. These tests are called ** z tests,** and they require that the

**Chapter 9**

Testing a “greater than” alternative hypothesis by using a critical value rule

In Section 9.1 we explained how to set up appropriate null and alternative hypotheses. We also discussed how to specify a value for *α*, the probability of a Type I error (also called the **level of significance**) of the hypothesis test, and we introduced the idea of a test statistic. We can use these concepts to begin developing a seven step hypothesis testing procedure. We will introduce these steps in the context of the trash bag case and testing a “greater than” alternative hypothesis.

**Step 1: State the null hypothesis H0 and the alternative hypothesis Ha.** In the trash bag case, we will test

**Step 2: Specify the level of significance α.** The television network will run the commercial stating that the new trash bag is stronger than the former bag if we can reject

**Step 3: Select the test statistic.** In order to test *H*0: *μ* ≤ 50 versus *Ha*: *μ* > 50, we will test the modified null hypothesis *H*0: *μ* = 50 versus *Ha*: *μ* > 50. The idea here is that if there is sufficient evidence to reject the hypothesis that *μ* equals 50 in favor of *μ* > 50, then there is certainly also sufficient evidence to reject the hypothesis that *μ* is less than or equal to 50. In order to test *H*0: *μ* = 50 versus *Ha*: *μ* > 50, we will randomly select a sample of *n* = 40 new trash bags and calculate the mean of the breaking strengths of these bags. We will then utilize the **test statistic**

A positive value of this test statistic results from an that is greater than 50 and thus provides evidence against *H*0: *μ* = 50 and in favor of *Ha*: *μ* > 50.

**Step 4: Determine the critical value rule for deciding whether to reject H0.** To decide how large the test statistic

Place the probability of a Type I error, *α*, in the right-hand tail of the standard normal curve and use the normal table (see Table A.3, page 863) to find the normal point *zα*. Here *zα*, which we call a **critical value,** is the point on the horizontal axis under the standard normal curve that gives a right-hand tail area equal to *α*.

**Reject H0: μ = 50 in favor of Ha: μ > 50 if and only if the test statistic z is greater than the critical value zα **(This is the

Figure 9.1 illustrates that since we have set *α* equal to .05, we should use the critical value *zα* = *z*.05 = 1.645 (see Table A.3). This says that we should reject *H*0 if *z* > 1.645 and we should not reject *H*0 if *z* ≤ 1.645.

Figure 9.1: The Critical Value for Testing *H*0: *μ* = 50 versus *Ha*: *μ* > 50 by Setting *α* = .05

To better understand the critical value rule, consider the standard normal curve in Figure 9.1. The area of .05 in the right-hand tail of this curve implies that values of the test statistic *z* that are greater than 1.645 are unlikely to occur if the null hypothesis *H*0: *μ* = 50 is true. There is a 5 percent chance of observing one of these values—and thus wrongly rejecting *H*0—if *H*0 is true. However, we are more likely to observe a value of *z* greater than 1.645—and thus correctly reject *H*0—if *H*0 is false. Therefore, it is intuitively reasonable to reject *H*0 if the value of the test statistic *z* is greater than 1.645.

**Step 5: Collect the sample data and compute the value of the test statistic.** When the sample of *n* = 40 new trash bags is randomly selected, the mean of the breaking strengths is calculated to be . Assuming that *σ* is known to equal 1.65, the value of the test statistic is

**Step 6: Decide whether to reject H0 by using the test statistic value and the critical value rule.** Since the test statistic value

**Step 7: Interpret the statistical results in managerial (real-world) terms and assess their practical importance.** Since we have rejected *H*0: *μ* = 50 in favor of *Ha*: *μ* > 50 by setting *α* equal to .05, we conclude (at an *α* of .05) that the mean breaking strength of the new trash bag exceeds 50 pounds. Furthermore, this conclusion has practical importance to the trash bag manufacturer because it means that the television network will approve running commercials claiming that the new trash bag is stronger than the former bag. Note, however, that the point estimate of *μ*, , indicates that *μ* is not much larger than 50. Therefore, the trash bag manufacturer can claim only that its new bag is slightly stronger than its former bag. Of course, this might be practically important to consumers who feel that, because the new bag is 25 percent less expensive and is more environmentally sound, it is definitely worth purchasing if it has any strength advantage. However, to customers who are looking only for a substantial increase in bag strength, the statistical results would not be practically important. This illustrates that, in general, a finding of statistical significance (that is, concluding that the alternative hypothesis is true) can be practically important to some people but not to others. Notice that the point estimate of the parameter involved in a hypothesis test can help us to assess practical importance. We can also use confidence intervals to help assess practical importance.

Considerations in setting *α*

We have reasoned in Section 9.1 that the television network has set *α* equal to .05 rather than .01 because doing so means that *β*, the probability of failing to advertise a true claim (a Type II error), will be smaller than it would be if *α* were set at .01. It is informative, however, to see what would have happened if the network had set *α* equal to .01. Figure 9.2 illustrates that as we decrease *α* from .05 to .01, the critical value *zα* increases from *z*.05 = 1.645 to *z*.01 = 2.33. Because the test statistic value *z* = 2.20 is less than *z*.01 = 2.33, we cannot reject *H*0: *μ* = 50 in favor of *Ha*: *μ* > 50 by setting *α* equal to .01. This illustrates the point that, the smaller we set *α*, the larger is the critical value, and thus the stronger is the statistical evidence that we are requiring to reject the null hypothesis *H*0. Some statisticians have concluded (somewhat subjectively) that (1) **if we set α equal to .05, then we are requiring strong evidence to reject H0**; and (2)

Figure 9.2: The Critical Values for Testing *H*0: *μ* = 50 versus *Ha*: *μ* > 50 by Setting *α* = .05 and .01

A *p*-value for testing a “greater than” alternative hypothesis

To decide whether to reject the null hypothesis *H*0 at level of significance *α*, steps 4, 5, and 6 of the seven-step hypoth esis testing procedure compare the test statistic value with a critical value. Another way to make this decision is to calculate a ** p-value,** which measures the likelihood of the sample results if the null hypothesis

**Step 4: Collect the sample data and compute the value of the test statistic.** In the trash bag case, we have computed the value of the test statistic to be *z* = 2.20.

**Step 5: Calculate the p-value by using the test statistic value.** The

Figure 9.3: Testing *H*0: *μ* = 50 versus *Ha*: *μ* > 50 by Using Critical Values and the *p*-Value

**Step 6: Reject H0 if the p-value is less than α.** Recall that the television network has set

© NBC, Inc. Used with permission.

Note: This logo appears on an NBC advertising standards booklet. This booklet, along with other information provided by NBC and CBS, forms the basis for much of the discussion in the paragraph to the right.

Comparing the critical value and *p*-value methods

Thus far we have considered two methods for testing *H*0: *μ* = 50 versus *Ha*: *μ* > 50 at the .05 and .01 values of *α*. Using the first method, we determine if the test statistic value *z* = 2.20 is greater than the critical values *z*.05 = 1.645 and *z*.01 = 2.33. Using the second method, we determine if the *p*-value of .0139 is less than .05 and .01. Whereas the critical value method requires that we look up a different critical value for each different *α* value, the *p*-value method requires only that we calculate a single *p*-value and compare it directly with the different *α* values. *It follows that the p-value method is the most efficient way to test a hypothesis at different α values.* This can be useful when there are different decision makers who might use different *α* values. For example, television networks do not always evaluate advertising claims by setting *α* equal to .05. The reason is that the consequences of a Type I error (advertising a false claim) are more serious for some claims than for others. For example, the consequences of a Type I error would be fairly serious for a claim about the effectiveness of a drug or for the superiority of one product over another. However, these consequences might not be as serious for a noncomparative claim about an inexpensive and safe product, such as a cosmetic. Networks sometimes use *α* values between .01 and .04 for claims having more serious Type I error consequences, and they sometimes use *α* values between .06 and .10 for claims having less serious Type I error consequences. Furthermore, one network’s policies for setting *α* can differ somewhat from those of another. As a result, reporting an advertising claim’s *p*-value to each network is the most efficient way to tell the network whether to allow the claim to be advertised. For example, most networks would evaluate the trash bag claim by choosing an *α* value between .025 and .10. Since the *p*-value of .0139 is less than all these *α* values, most networks would allow the trash bag claim to be advertised.

A summary of the seven steps of hypothesis testing

For almost every hypothesis test discussed in this book, statisticians have developed both a critical value rule and a *p*-value that can be used to perform the hypothesis test. Furthermore, it can be shown that for each hypothesis test the *p*-value has been defined so that **we can reject the null hypothesis at level of significance α if and only if the p-value is less than α **. We now summarize a seven-step procedure for performing a hypothesis test.

**The Seven Steps of Hypothesis Testing**

**1** State the null hypothesis *H*0 and the alternative hypothesis *Ha*.

**2** Specify the level of significance *α*.

**3** Select the test statistic.

**Using a critical value rule:**

**4** Determine the critical value rule for deciding whether to reject *H*0. Use the specified value of *α* to find the critical value in the critical value rule.

**5** Collect the sample data and compute the value of the test statistic.

**6** Decide whether to reject *H*0 by using the test statistic value and the critical value rule.

**Using a p-value:**

**4** Collect the sample data and compute the value of the test statistic.

**5** Calculate the *p*-value by using the test statistic value.

**6** Reject *H*0 at level of significance *α* if the *p*-value is less than *α*.

**7** Interpret your statistical results in managerial (real-world) terms and assess their practical importance.

In the real world both critical value rules and *p*-values are used to carry out hypothesis tests. For example, NBC uses critical value rules, whereas CBS uses *p*-values, to statistically verify the validity of advertising claims. Throughout this book we will continue to present both the critical value and the *p*-value approaches to hypothesis testing.

Testing a “less than” alternative hypothesis

We next consider the payment time case and testing a “less than” alternative hypothesis:

**Step 1:** In order to study whether the new electronic billing system reduces the mean bill payment time by more than 50 percent, the management consulting firm will test *H*0: *μ* ≥ 19.5 versus *Ha*: *μ* < 19.5.

**Step 2:** The management consulting firm wishes to make sure that it truthfully describes the benefits of the new system both to the Hamilton, Ohio, trucking company and to other companies that are considering installing such a system. Therefore, the firm will require very strong evidence to conclude that *μ* is less than 19.5, which implies that it will test *H*0: *μ* ≥ 19.5 versus *Ha*: *μ* < 19.5 by setting *α* equal to .01.

**Step 3:** In order to test *H*0: *μ* ≥ 19.5 versus *Ha*: *μ* < 19.5, we will test the modified null hypothesis *H*0: *μ* = 19.5 versus *Ha*: *μ* < 19.5. The idea here is that if there is sufficient evidence to reject the hypothesis that *μ* equals 19.5 in favor of *μ* < 19.5, then there is certainly also sufficient evidence to reject the hypothesis that *μ* is greater than or equal to 19.5. In order to test *H*0: *μ* = 19.5 versus *Ha*: *μ* < 19.5, we will randomly select a sample of *n* = 65 invoices paid using the billing system and calculate the mean of the payment times of these invoices. Since the sample size is large, the Central Limit Theorem applies, and we will utilize the test statistic

A value of the test statistic *z* that is less than zero results when is less than 19.5. This provides evidence to support rejecting *H*0 in favor of *Ha* because the point estimate indicates that *μ* might be less than 19.5.

**Step 4:** To decide how much less than zero the test statistic must be to reject *H*0 in favor of *Ha* by setting the probability of a Type I error equal to α, we do the following:

Place the probability of a Type I error, *α*, in the left-hand tail of the standard normal curve and use the normal table to find the critical value −*zα*. Here −*zα* is the negative of the normal point *zα*. That is, −*zα* is the point on the horizontal axis under the standard normal curve that gives a left-hand tail area equal to *α*.

**Reject H0: μ = 19.5 in favor of Ha: μ < 19.5 if and only if the test statistic z is less than the critical value −zα.** Because

Figure 9.4: Testing *H*0: *μ* = 19.5 versus *Ha*: *μ* < 19.5 by Using Critical Values and the *p*-Value

**Step 5:** When the sample of *n* = 65 invoices is randomly selected, the mean of the payment times of these invoices is calculated to be . Assuming that *σ* is known to equal 4.2, the value of the test statistic is

**Step 6:** Since the test statistic value *z* = −2.67 is less than the critical value −*z*.01 = −2.33, we can reject *H*0: *μ* = 19.5 in favor of *Ha*: *μ* < 19.5 by setting *α* equal to .01.

**Step 7:** We conclude (at an *α* of .01) that the mean payment time for the new electronic billing system is less than 19.5 days. This, along with the fact that the sample mean is slightly less than 19.5, implies that it is reasonable for the management consulting firm to conclude that the new electronic billing system has reduced the mean payment time by slightly more than 50 percent (a substantial improvement over the old system).

A *p*-value for testing a “less than” alternative hypothesis

To test *H*0: *μ* = 19.5 versus *Ha*: *μ* < 19.5 in the payment time case by using a *p*-value, we use the following steps 4, 5, and 6:

**Step 4:** We have computed the value of the test statistic in the payment time case to be *z* = −2.67.

**Step 5:** The *p*-value for testing *H*0: *μ* = 19.5 versus *Ha*: *μ* < 19.5 is the area under the standard normal curve to the left of the test statistic value *z* = −2.67. As illustrated in Figure 9.4(b), this area is .0038. The *p*-value is the probability, computed assuming that *H*0: *μ* = 19.5 is true, of observing a value of the test statistic that is less than or equal to the value *z* = −2.67 that we have actually computed from the sample data. The *p*-value of .0038 says that, if *H*0: *μ* = 19.5 is true, then only 38 in 10,000 of all possible test statistic values are at least as negative, or extreme, as the value *z* = −2.67. That is, if we are to believe that *H*0 is true, we must believe that we have observed a test statistic value that can be described as a 38 in 10,000 chance.

**Step 6:** The management consulting firm has set *α* equal to .01. **The p-value of .0038 is less than the α of .01. Therefore, we can reject H0 by setting α equal to .01.**

Testing a “not equal to” alternative hypothesis

We next consider the Valentine’s Day chocolate case and testing a “not equal to” alternative hypothesis.

**Step 1:** To assess whether this year’s sales of its valentine box of assorted chocolates will be ten percent higher than last year’s, the candy company will test *H*0: *μ* = 330 versus *Ha*: *μ* ≠ 330. Here, *μ* is the mean order quantity of this year’s valentine box by large retail stores.

**Step 2:** If the candy company does not reject *H*0: *μ* = 330 and *H*0: *μ* = 330 is false—a Type II error—the candy company will base its production of valentine boxes on a 10 percent projected sales increase that is not correct. Since the candy company wishes to have a reasonably small probability of making this Type II error, the company will set *α* equal to .05. Setting *α* equal to .05 rather than .01 makes the probability of a Type II error smaller than it would be if *α* were set at .01. Note that in optional Section 9.5 we will verify that the probability of a Type II error in this situation is reasonably small. Therefore, if the candy company ends up not rejecting *H*0: *μ* = 330 and therefore decides to base its production of valentine boxes on the ten percent projected sales increase, the company can be intuitively confident that it has made the right decision.

**Step 3:** The candy company will randomly select *n* = 100 large retail stores and will make an early mailing to these stores promoting this year’s valentine box of assorted chocolates. The candy company will then ask each sampled retail store to report its anticipated order quantity of valentine boxes and will calculate the mean of the reported order quantities. Since the sample size is large, the Central Limit Theorem applies, and we will utilize the test statistic

A value of the test statistic that is greater than 0 results when is greater than 330. This provides evidence to support rejecting *H*0 in favor of *Ha* because the point estimate indicates that *μ* might be greater than 330. Similarly, a value of the test statistic that is less than 0 results when is less than 330. This also provides evidence to support rejecting *H*0 in favor of *Ha* because the point estimate indicates that *μ* might be less than 330.

**Step 4:** To decide how different from zero (positive or negative) the test statistic must be in order to reject *H*0 in favor of *Ha* by setting the probability of a Type I error equal to *α*, we do the following:

Divide the probability of a Type I error, *α*, into two equal parts, and place the area *α*/2 in the right-hand tail of the standard normal curve and the area *α*/2 in the left-hand tail of the standard normal curve. Then use the normal table to find the critical values *zα*/2 and −*zα*/2. Here *zα*/2 is the point on the horizontal axis under the standard normal curve that gives a right-hand tail area equal to *α*/2, and −*zα*/2 is the point giving a left-hand tail area equal to *α*/2.

**Reject H0: μ = 330 in favor of Ha: μ ≠ 330 if and only if the test statistic z is greater than the critical value zα/2 or less than the critical value −zα/2.** Note that this is equivalent to saying that we should

Figure 9.5: Testing *H*0: *μ* = 330 versus *Ha*: *μ* ≠ 330 by Using Critical Values and the *p*-Value

**Step 5:** When the sample of *n* = 100 large retail stores is randomly selected, the mean of their reported order quantities is calculated to be . Assuming that *σ* is known to equal 40, the value of the test statistic is

**Step 6:** Since the test statistic value *z* = −1 is greater than − *z*.025 = −1.96 (or, equivalently, since | *z* | = 1 is less than *z*.025 = 1.96), we cannot reject *H*0: *μ* = 330 in favor of *Ha*: *μ* ≠ 330 by setting *α* equal to .05.

**Step 7:** We cannot conclude (at an *α* of .05) that the mean order quantity of this year’s valentine box by large retail stores will differ from 330 boxes. Therefore, the candy company will base its production of valentine boxes on the ten percent projected sales increase.

A *p*-value for testing a “not equal to” alternative hypothesis

To test *H*0: *μ* = 330 versus *Ha*: *μ* ≠ 330 in the Valentine’s Day chocolate case by using a *p*-value, we use the following steps 4, 5, and 6:

**Step 4:** We have computed the value of the test statistic in the Valentine’s Day chocolate case to be *z* = −1.

**Step 5:** Note from Figure 9.5(b) that the area under the standard normal curve to the right of | *z* | = 1 is .1587. Twice this area—that is, 2(.1587) = .3174—is the *p*-value for testing *H*0: *μ* = 330 versus *Ha*: *μ* ≠ 330. To interpret the *p*-value as a probability, note that the symmetry of the standard normal curve implies that twice the area under the curve to the right of | *z* | = 1 equals the area under this curve to the right of 1 plus the area under the curve to the left of −1 [see Figure 9.5(b)]. Also, note that since both positive and negative test statistic values count against *H*0: *μ* = 330, a test statistic value that is either greater than or equal to 1 or less than or equal to −1 is at least as extreme as the observed test statistic value *z* = −1. It follows that the *p*-value of .3174 says that, if *H*0: *μ* = 330 is true, then 31.74 percent of all possible test statistic values are at least as extreme as *z* = −1. That is, if we are to believe that *H*0 is true, we must believe that we have observed a test statistic value that can be described as a 31.74 percent chance.

**Step 6:** The candy company has set *α* equal to .05. **The p-value of .3174 is greater than the α of .05. Therefore, we cannot reject H0 by setting α equal to .05.**

A general procedure for testing a hypothesis about a population mean

In the trash bag case we have tested *H*0: *μ* ≤ 50 versus *Ha*: *μ* > 50 by testing *H*0: *μ* = 50 versus *Ha*: *μ* > 50. In the payment time case we have tested *H*0: *μ* ≥ 19.5 versus *Ha*: *μ* < 19.5 by testing *H*0: *μ* = 19.5 versus *Ha*: *μ* < 19.5. In general, the usual procedure for testing a “less than or equal to” null hypothesis or a “greater than or equal to” null hypothesis is to change the null hypothesis to an equality. We then test the “equal to” null hypothesis versus the alternative hypothesis. Furthermore, the critical value and *p*-value procedures for testing a null hypothesis versus an alternative hypothesis depend upon whether the alternative hypothesis is a “greater than,” a “less than,” or a “not equal to” alternative hypothesis. The following summary box gives the appropriate procedures. Specifically, letting *μ*0 be a particular number, the summary box shows how to test *H*0: *μ* = *μ*0 versus either *Ha*: *μ* > *μ*0, *Ha*: *μ* < *μ*0, or *Ha*: *μ* ≠ *μ*0:

Testing a Hypothesis about a Population Mean when *σ* Is Known

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