Math 207 Solutions – Assignment 7

9.8

Note that , so it is appropriate to use the large-sample significance test for a population mean, since the distribution of the sample average is approximately normal. (Also note this example is completely free of context—a contrived problem with “floating numbers in space.” Not very exciting, but at least you get practice calculating the power.)

a. Recall the probability of a Type II error is . Then, for a 0.05 significance level and a one-sided (greater than) alternative hypothesis, what does it mean to “accept ”? At the 0.05 significance level, the “acceptance” (that is, “do not reject”) region for the test statistic, z, is (since it is a one-sided alternative). In terms of the acceptance region is

(Although I do not include a picture with this solution—only because of the limitations of Word—I strongly encourage you to draw one, and I might require an appropriate normal-curve graph on the exam.)

So the acceptance region is and the rejection region is .

b. If then

c. If then (this makes sense, given that 2.3 is the null-hypothesized value of the mean, and we defined our test to have a 0.05 Type I error rate)

If then

If then (technically, this value greater than zero, but it’s not included on Table 3, and practically it’s very close to zero)

d. Since power = we have the following power values:

	power
2.3	1 – 0.9484 = 0.0516
2.4	1 – 0.3409 = 0.6591
2.5	1 – 0.0071 = 0.9929
2.6	1 – 0 = 1

The power curve is shown below. (As we discussed in class, the farther you move into the alternative hypothesis, the higher the power of detecting that particular value.) Note: It’s fine if you connect the dots with a smooth curve.

9.23

Note that both sample sizes are larger than 30, so it is appropriate to use the large-sample inference procedures (the difference in sample averages will follow an approximate normal distribution).

a. Let be the mean lead level in all of Section 1, and let be the mean lead level in all of Section 2. Then, the hypotheses are and . The test statistic is . (That is, our difference in sample proportions is 2.26 standard deviations below the null-hypothesized value.) Then the p-value is . (The p-value is doubled because the alternative hypothesis is two-sided.) Although a normal curve picture is not included with these solutions, it is an important part of the solution process—I haven’t included it simply because I can’t draw them in Word.

If the population lead level means are really the same, there is only a 2.38% chance that we would obtain our sample data or data more extreme. This is very strong evidence against the null hypothesis (and strong evidence that there is a difference in average lead levels for these different sections of the city). The difference is significant at the 0.05 level (since 0.0238 < 0.05).

b. The 95% confidence interval is = (-3.55 parts per million, -0.25 parts per million). Aside: note the null hypothesized value of 0 parts/million is not included in this confidence interval. Therefore, the difference is statistically significant at the 0.05 level. This agrees with our answer in part a.

c. In practice, there is only concern if the mean difference is more than 5 parts per million. Although we found statistically significant evidence that the mean lead levels are different, our confidence interval does not include the value –5 parts/million. Hence, the results are not practically significant (that is, not of practical importance to the environmental engineers).

9.46

a. Let be the proportion of adults with children that are frequent moviegoers and let be the proportion adults without children that are frequent moviegoers. We want to test the hypotheses and .

Since , , , and , we can use the large-sample significance test for two proportions (the difference in sample proportions will follow an approximate normal distribution).

For the sample of adults with children, . For the sample of adults without children, . Furthermore, the pooled estimate of the common moviegoer proportion (under the null hypothesis) is .

Then the test statistic is . (That is, our difference in sample proportions is 0.74 standard deviations away from the null-hypothesized difference.)

And the p-value is . (The p-value is doubled because the alternative hypothesis is two-sided.) Although a normal curve picture is not included with these solutions, it is an important part of the solution process—I haven’t included it simply because I can’t draw them in Word.

If the moviegoer proportions are really the same, then the probability of obtaining our data or data more extreme is 0.4592. This is not at all unusual and provides no evidence against the null hypothesis. The results are not statistically significant at any reasonable significance level. That is, we have no evidence that the proportion of moviegoer parents is different from the proportion of moviegoer non-parents.

a. If a significant difference were found (say, that the population of adults without children has a higher proportion of moviegoers), then appropriate advertising strategies could be employed (e.g., no advertising of children’s movies or toys). We’d need to make sure, though, that the significant difference was actually practically significant (this would be a question for the marketing team).

Extra Problem

Recall the definition of the probability of a Type II error: . In this problem, we must determine the sample size, n. Let be the mean systolic blood pressure (in mm Hg) for the company’s young male executives. Then we want to test the hypotheses

(Although I do not include pictures with this solution—only because of the limitations of Word—I strongly encourage you to draw them, and I might require appropriate normal-curve graphs on the exam.)

Step 1

For 0.05 significance level and a one-sided (greater than) alternative hypothesis, what does it mean to “accept ”? At the 0.05 significance level, the “acceptance” (that is, “do not reject”) region for the test statistic, z, is (since it is a one-sided alternative).

Step 2

In terms of the acceptance region is

So we “accept ” if .

Step 3

Then, because we’re interested in the specific average value of 133 mm Hg in the alternative,

Now we standardize with the 133 mm Hg value of the mean (keeping the standard deviation the same):

.

Step 4

We want a power of 0.8, which means Hence, we need to find n such that

. Since is the 20^th percentile of the standard normal distribution (reverse look-up in Table 3), we have:

. Since we always round up on sample-size-determination problems, the director should take a sample of size 56. (Note that this sample is “large”, at least 30, so the method we used—the large-sample test—is appropriate.)

More explanation of power:

The medical director decided (based on his expert opinion) the value in the alternative that he wants the test to detect with a high probability. Based on his hypotheses and a 0.05 significance test, if samples of size n=56 are repeatedly taken, then 80% of the tests will reject when the true mean is really 133 mm Hg (that is, the testing method makes the correct decision—when —80% of the time).

Solutions to Additional Problems (not to be turned in, but necessary for exam preparation)

10.9 (only part a)

Let be the mean carbon monoxide diffusing capacity for all smokers.

We want to test

Note that the sample size is small (n = 20 < 30), so we need to use the t-procedures (rather than z-procedures). The distribution of the sample diffusing capacities is mound-shaped, so the condition of a normal population is met.

The test statistic is .

The p-value is , where T has a t-distribution with 19 degrees of freedom. Because of symmetry of the t-distribution, from Table 4 we can say that the p-value is less than 0.005 (if you say the p-value is essentially 0, then that’s fine, too). (Although a picture is not included with these solutions, it is an important part of the solution process—I haven’t included it simply because I can’t draw them in Word.)

If the mean carbon monoxide diffusing capacity for all smokers is really 100 DL, then there is less than a 0.5% chance that we would obtain our particular sample mean or a smaller mean. This is very surprising; therefore, we have strong evidence against the mean equaling 100 DL. Furthermore, since p-value < 0.005 < 0.01, the results are statistically significant at the a = 0.01 level.

10.26

a. Let be the mean compartment pressure for the population of runners at rest and let be the mean compartment pressure for the population of cyclists at rest. Then the hypotheses are:

Note that the sample sizes are small (n = 10 < 30), so we should use the t-procedures (rather than the z-procedures). We are not given graphical information about the samples, so we cannot be sure that they look mound-shaped. Once we complete our analysis, we should state that it is contingent on the original populations being close to normal. Furthermore, we need to assume that the data are from random samples of people.

Since by our rule of thumb we can reasonably assume the population variances are equal. Then the pooled estimate of the population variance is:

The test statistic is

The p-value is , where T has a t distribution with (10 + 10 – 2) = 18 degrees of freedom. From Table 4, we can say that the p-value is between 0.05 and 0.10 (Minitab gives the exact value: 0.0708). Hence, the results are not statistically significant at the 0.05 level. (Although a picture is not included with these solutions, it is an important part of the solution process—I haven’t included it simply because I can’t draw them in Word.)

b. Let be the mean compartment pressure for the population of runners at 80% maximal and let be the mean compartment pressure for the population of cyclists at 80% maximal.

Note that the sample sizes are small, so we should use the t-procedures (rather than the z-procedures). We are not given graphical information about the samples, so we cannot be sure that they look mound-shaped. Once we complete our analysis, we should state that it is contingent on the original populations being close to normal. Furthermore, we need to assume that the data is from a random sample of people.

Since by our rule of thumb we can reasonably assume the population variances are equal. Then the pooled estimate of the population variance is:

There are 18 degrees of freedom and we want 95% confidence, so Then a 95% confidence interval for (-) is: (-3.32, 4.72)

We are 95% confident that the difference in mean compartment pressures at 80% maximal between runners and cyclists is between –3.32 mm Hg and 4.72 mm Hg. (Per usual, our confidence is in the method we use to create the interval—the method is correct 95% of the time.) Note that 0 is included in the interval, so the difference is not statistically significant at the 0.05 level.

c. At maximal oxygen consumption, so by our rule of thumb, we should not use the pooled t-test. (We should use the two-sample t-test that does not assume equal variances. This has a grungy formula for calculating the degrees of freedom, but Minitab can easily do this calculation for us.)

10.41

a. Each subject was presented with both signs in random order. If the subject’s reaction time in general is high, both responses will be high, and vice versa. If a matched pairs design was not used, the large variability from subject to subject might mask the variability due to the difference in sign types. The matched pairs design eliminates the subject-to-subject variability.

b. Let be the mean for the population of (prohibitive sign – permissive sign) reaction differences. Then, the hypotheses are: .

Since the sample size is small (n = 10 < 30), the t-procedures should be used (rather than the z-procedures). Note that the sample distribution of differences looks fairly skewed (this is easy to see in a histogram from Minitab), therefore we should be leery of using the t-procedures (I will carry out the test for the homework, but in practice, the t-procedures perhaps should not be used—but it turns out the data are very, very significant, so a non-parametric test will almost surely agree with the t-test results, so we needn’t worry).

The test statistic is . The p-value is where T has a t-distribution with 9 degrees of freedom. If the population mean difference in reaction times is really 0 milliseconds, there is essentially no chance that we would obtain our particular sample average or a more extreme average. This is very strong evidence that the mean difference is different from 0 milliseconds (and, in fact, greater than 0). The difference is significant at the 0.05 level. (Although a picture is not included with these solutions, it is an important part of the solution process—I haven’t included it simply because I can’t draw them in Word.)

c. There are 9 degrees of freedom and we want 95% confidence, so Then the 95% confidence interval is: = (80.51 milliseconds, 133.29 milliseconds). Note that 0 is not included in the interval, showing the results are statistically significant at the 0.05 level (as we already mentioned in part b). Furthermore, we can see that this result is probably practically significant as well as statistically significant.

10.88

Let be the mean weight loss (initial weight – final weight) for the population of obese people on the diet. Since the sample size is small (n = 8 <30), the t-procedures should be used (rather than the z-procedures). Furthermore, a paired t-test should be used. Note that the sample distribution of weight losses is slightly skewed right (although it is difficult to determine with only 8 observations), but the t-procedures should be robust enough to handle this (since it’s only a slight violation of the normality condition). Also, we must assume that the obese people were randomly selected.

There are 7 degrees of freedom and if we want 95% confidence, Then the 95% confidence interval for is: = (35.845 pounds, 48.405 pounds). We are 95% confident that the mean weight loss for all obese people on the diet is between 35.845 pounds and 48.405 pounds. Our confidence is in the method we use—if the sampling were (hypothetically) done repeatedly, then 95% of the intervals created would contain the true mean weight loss. Note that 0 pounds is not included in the confidence interval (i.e., it is not among the likely values at the 95% confidence level), so the results are statistically significant at the 0.05 level. Furthermore, based on the confidence interval endpoints, it seems these results are not only statistically significant, but also of practical importance (that’s a lot of pounds lost!).

10.100

Let be the mean weight for all cans. Since the sample size is small (n = 9 < 30), t-procedures should be used (rather than z-procedures). We are not given graphical information about the sample, so we cannot be sure that it looks mound-shaped. Once we complete our analysis, we should state that it is contingent on the original population being close to normal.

We want to test

The test statistic is .

The p-value is , where T has a t-distribution with 8 degrees of freedom. From Table 4, we can say that the p-value is between 0.05 and 0.10 (Minitab gives the exact value: 0.055). Hence, the results are not statistically significant at the 0.05 level (that is, we do not have sufficient evidence to indicate the mean weight is less than that claimed on the label).