THE DISK BRAKE CASE
National Motors has equipped the ZX-900 with a new disk brake system. We define the stopping distance for a ZX-900 as the distance (in feet) required to bring the automobile to a complete stop from a speed of 35 mph under normal driving conditions using this new brake system. In addition, we define μ to be the mean stopping distance of all ZX-900s. One of the ZX-900’s major competitors is advertised to achieve a mean stopping distance of 60 ft. National Motors would like to claim in a new television commercial that the ZX-900 achieves a shorter mean stopping distance.
a Set up the null hypothesis H0 and the alternative hypothesis Ha that would be used to attempt to provide evidence supporting the claim that μ is less than 60.
b A television network will permit National Motors to claim that the ZX-900 achieves a shorter mean stopping distance than the competitor if H0 can be rejected in favor of Ha by setting α equal to .05. If the stopping distances of a random sample of n = 81 ZX-900s have a mean of , will National Motors be allowed to run the commercial? Perform the hypothesis test by using a critical value and a p-value. Assume here that σ = 6.02.
9.47 Consider part (b) of Exercise 9.46, and calculate a 95 percent confidence interval for μ. Do the point estimate of μ and confidence interval for μ indicate that μ might be far enough below 60 feet to suggest that we have a practically important result?
9.48 Recall from Exercise 8.12 that Bayus (1991) studied the mean numbers of auto dealers visited by early and late replacement buyers.
a Letting μ be the mean number of dealers visited by early replacement buyers, suppose that we wish to test H0: μ = 4 versus Ha: μ ≠ 4. A random sample of 800 early replacement buyers yields a mean number of dealers visited of . Assuming σ equals .71, calculate the p-value and test H0 versus Ha. Do we estimate that μ is less than 4 or greater than 4?
b Letting μ be the mean number of dealers visited by late replacement buyers, suppose that we wish to test H0: μ = 4 versus Ha: μ ≠ 4. A random sample of 500 late replacement buyers yields a mean number of dealers visited of . Assuming σ equals .66, calculate the p-value and test H0 versus Ha. Do we estimate that μ is less than 4 or greater than 4?
9.3: t Tests about a Population Mean: σ Unknown
If we do not know σ (which is usually the case), we can base a hypothesis test about μ on the sampling distribution of
If the sampled population is normally distributed, then this sampling distribution is a t distribution having n − 1 degrees of freedom. This leads to the following results:
A t Tests about a Population Mean: σ UnKnown
Define the test statistic
and assume that the population sampled is normally distributed. We can test H0: μ = μ0 versus a particular alternative hypothesis at level of significance α by using the appropriate critical value rule, or, equivalently, the corresponding p-value.
Here tα, tα/2, and the p-values are based on n − 1 degrees of freedom.
In the rest of this chapter and in Chapter 10 we will present most of the hypothesis testing examples by using hypothesis testing summary boxes and the seven hypothesis testing steps given in the previous section. However, to be concise, we will not formally number each hypothesis testing step. Rather, for each of the first six steps, we will set out in boldface font a key phrase that indicates that the step is being carried out. Then, we will highlight the seventh step—the business improvement conclusion—as we highlight all business improvement conclusions in this book. After Chapter 10, we will continue to use hypothesis testing summary boxes, and we will more informally use the seven steps.
As illustrated in the following example, we will often first use a critical value rule to test the hypotheses under consideration at a fixed value of α and then use a p-value to assess the weight of evidence against the null hypothesis.
EXAMPLE 9.4
In 1991 the average interest rate charged by U.S. credit card issuers was 18.8 percent. Since that time, there has been a proliferation of new credit cards affiliated with retail stores, oil companies, alumni associations, professional sports teams, and so on. A financial officer wishes to study whether the increased competition in the credit card business has reduced interest rates. To do this, the officer will test a hypothesis about the current mean interest rate, μ, charged by U.S. credit card issuers. The null hypothesis to be tested is H0: μ = 18.8%, and the alternative hypothesis is Ha: μ < 18.8%. If H0 can be rejected in favor of Ha at the .05 level of significance, the officer will conclude that the current mean interest rate is less than the 18.8% mean interest rate charged in 1991. To perform the hypothesis test, suppose that we randomly select n = 15 credit cards and determine their current interest rates. The interest rates for the 15 sampled cards are given in Table 9.3. A stem-and-leaf display and MINITAB box plot are given in Figure 9.7. The stem-and-leaf display looks reasonably mound-shaped, and both the stem-and-leaf display and the box plot look reasonably symmetrical. It follows that it is appropriate to calculate the value of the test statistic t in the summary box. Furthermore, since Ha: μ < 18.8% is of the form Ha: μ < μ0, we should reject H0: μ = 18.8% if the value of t is less than the critical value −tα = −t.05 = −1.761. Here, −t.05 = −1.761 is based on n − 1 = 15 −1 = 14 degrees of freedom and this critical value is illustrated in Figure 9.8(a). The mean and the standard deviation of the n = 15 interest rates in Table 9.3 are and s = 1.538. This implies that the value of the test statistic is
Figure 9.7: Stem-and-Leaf Display and Box Plot of the Interest Rates
Figure 9.8: Testing H0: μ = 18.8% versus Ha: μ < 18.8% by Using a Critical Value and a p-Value
Table 9.3: Interest Rates Charged by 15 Randomly Selected Credit Cards CreditCd
Since t = −4.97 is less than −t.05 = −1.761, we reject H0: μ = 18.8% in favor of Ha: μ < 18.8%. That is, we conclude (at an α of .05) that the current mean credit card interest rate is lower than 18.8%, the mean interest rate in 1991. Furthermore, the sample mean says that we estimate the mean interest rate is 18.8% − 16.827% = 1.973% lower than it was in 1991.
The p-value for testing H0: μ = 18.8% versus Ha: μ < 18.8% is the area under the curve of the t distribution having 14 degrees of freedom to the left of t = −4.97. Tables of t points (such as Table A.4, page 864) are not complete enough to give such areas for most t statistic values, so we use computer software packages to calculate p-values that are based on the t distribution. For example, the MINITAB output in Figure 9.9(a) and the MegaStat output in Figure 9.10 tell us that the p-value for testing H0: μ = 18.8% versus Ha: μ < 18.8% is .0001. Notice that both MINITAB and MegaStat round p-values to three or four decimal places. The Excel output in Figure 9.9(b) gives the slightly more accurate value of 0.000103 for the p-value. Because this p-value is less than .05, .01, and .001, we can reject H0 at the .05, .01, and .001 levels of significance. Also note that the p-value of .0001 on the MegaStat output is shaded dark yellow. This indicates that we can reject H0 at the .01 level of significance (light yellow shading would indicate significance at the .05 level, but not at the .01 level). As a probability, the p-value of .0001 says that if we are to believe that H0: μ = 18.8% is true, we must believe that we have observed a t statistic value (t = −4.97) that can be described as a 1 in 10,000 chance. In summary, we have extremely strong evidence that H0: μ = 18.8% is false and Ha: μ < 18.8% is true. That is, we have extremely strong evidence that the current mean credit card interest rate is less than 18.8%.
Figure 9.9: The MINITAB and Excel Outputs for Testing H0: μ = 18.8% versus Ha: μ < 18.8%
Figure 9.10: The MegaStat Output for Testing H0: μ = 18.8% versus Ha: μ < 18.8%
Recall that in three cases discussed in Section 9.2 we tested hypotheses by assuming that the population standard deviation σ is known and by using z tests. If σ is actually not known in these cases (which would probably be true), we should test the hypotheses under consideration by using t tests. Furthermore, recall that in each case the sample size is large (at least 30). In general, it can be shown that if the sample size is large, the t test is approximately valid even if the sampled population is not normally distributed (or mound shaped). Therefore, consider the Valentine’s Day chocolate case and testing H0: μ = 330 versus Ha: μ ≠ 330 at the .05 level of significance. To perform the hypothesis test, assume that we will randomly select n = 100 large retail stores and use their anticipated order quantities to calculate the value of the test statistic t in the summary box. Then, since the alternative hypothesis Ha: μ ≠ 330 is of the form Ha: μ ≠ μ0, we will reject H0: μ = 330 if the absolute value of t is greater than tα/2 = t.025 = 1.984 (based on n − 1 = 99 degrees of freedom). Suppose that when the sample is randomly selected, the mean and the standard deviation of the n = 100 reported order quantities are calculated to be and s = 39.1. The value of the test statistic is
Since | t | = 1.023 is less than t.025 = 1.984, we cannot reject H0: μ = 330 by setting α equal to .05. It follows that we cannot conclude (at an α of .05) that this year’s mean order quantity of the 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. The p-value for the hypothesis test is twice the area under the t distribution curve having 99 degrees of freedom to the right of | t | = 1.023. Using a computer, we find that this p-value is .3088, which provides little evidence against H0: μ = 330 and in favor of Ha: μ ≠ 330.
As another example, consider the trash bag case and note that the sample of n = 40 trash bag breaking strengths has mean and standard deviation s = 1.6438. The p-value for testing H0: μ = 50 versus Ha: μ > 50 is the area under the t distribution curve having n − 1 = 39 degrees of freedom to the right of
Using a computer, we find that this p-value is .0164, which provides strong evidence against H0: μ = 50 and in favor of Ha: μ > 50. In particular, recall that most television networks would evaluate the claim that the new trash bag has a mean breaking strength that exceeds 50 pounds by choosing on α value between .025 and .10. It follows, since the p-value of .0164 is less than all these α values, that most networks would allow the trash bag claim to be advertised.
As a third example, consider the payment time case and note that the sample of n = 65 payment times has mean and standard deviation s = 3.9612. The p-value for testing H0: μ = 19.5 versus Ha: μ < 19.5 is the area under the t distribution curve having n − 1 = 64 degrees of freedom to the left of
Using a computer, we find that this p-value is .0031, which is less than the management consulting firm’s α value of .01. It follows that the consulting firm will claim that the new electronic billing system has reduced the Hamilton, Ohio, trucking company’s mean bill payment time by more than 50 percent.
To conclude this section, note that if the sample size is small (<30) and the sampled population is not mound-shaped, or if the sampled population is highly skewed, then it might be appropriate to use a nonparametric test about the population median. Such a test is discussed in Chapter 18.
Exercises for Section 9.3
CONCEPTS
9.49 What assumptions must be met in order to carry out the test about a population mean based on the t distribution?
9.50 How do we decide whether to use a z test or a t test when testing a hypothesis about a population mean?