Using a critical value rule:
4 Determine the critical value rule for deciding whether to reject H0. 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 H0 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 H0 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 H0: 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 H0: μ ≥ 19.5 versus Ha: μ < 19.5 by setting α equal to .01.
Step 3: In order to test H0: μ ≥ 19.5 versus Ha: μ < 19.5, we will test the modified null hypothesis H0: 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 H0: 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 H0 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 H0 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 α equals .01, the critical value −zα is −z.01 = −2.33 [see Fig. 9.4(a)].
Figure 9.4: Testing H0: 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 H0: 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 H0: 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 H0: 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 H0: 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 H0: 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 H0 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 H0: 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 H0: μ = 330 and H0: 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 H0: 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 H0 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 H0 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 H0 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 reject H0 if and only if the absolute value of the test statistic, | z | is greater than the critical value zα/2. Because α equals .05, the critical values are [see Figure 9.5(a)]
Figure 9.5: Testing H0: 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 H0: 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 H0: 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 H0: 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 H0: 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 H0: 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 H0 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 H0: μ ≤ 50 versus Ha: μ > 50 by testing H0: μ = 50 versus Ha: μ > 50. In the payment time case we have tested H0: μ ≥ 19.5 versus Ha: μ < 19.5 by testing H0: 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 μ0 be a particular number, the summary box shows how to test H0: μ = μ0 versus either Ha: μ > μ0, Ha: μ < μ0, or Ha: μ ≠ μ0:
Testing a Hypothesis about a Population Mean when σ Is Known