Math 207 Solutions (Significance-Testing Problems—for exam practice)

9.11

Note that , so it is appropriate to use the large-sample significance test for a population mean (and we can estimate the population standard deviation with the sample standard deviation).

a. The null hypothesis is the “status quo,” so (where is the average meat weight for the population of meat trays created by the supermarket). The quality control manger is interested in either under-filling the tray (potential customer dissatisfaction) or over-filling the tray (potential loss of money). Hence, the alternative hypothesis is .

b. The test statistic is . That is, our particular sample average is just 0.33 standard deviations above the null-hypothesized value. Because the alternative hypothesis is two-sided, the p-value 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.)

c. If the average amount of ground beef for all trays is actually 1 pound, then there is a 74% chance of getting our particular sample mean or a sample mean more extreme. This is not at all surprising and provides no evidence against the null hypothesis that the mean is 1 pound (the results are not statistically significant at any reasonable significance level). Note the confidence interval you found in exercise 8.33 includes the value 1, further indicating that 1 pound is a likely value for the mean.

9.15

Note that , so it is appropriate to use the large-sample significance test for a population mean (and we can estimate the population standard deviation with the sample standard deviation). Let be the average number of hours worked by the population of non-college-graduates. Then we want to test the hypotheses

(The alternative hypothesis is one-sided—this comes from the research question asked in the problem.)

The test statistic is . That is, our particular sample average is 2.63 standard deviations above the null-hypothesized average.

(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.)

a. Because the alternative hypothesis is on-sided the p-value is . At any significance level greater than or equal to 0.0043, we can reject the null hypothesis and conclude that the average hours worked by non-college-graduates is greater than 7.4 hours/day.

b. If the average number of hours worked by non-college-graduates is really 7.4 hours/day, then there is only a 0.0043 chance of getting our particular sample mean or a smaller one. This is very unlikely. Hence, we have very strong evidence that non-college-graduates work, on average, more hours than college-graduates. (These results are significant at even the more stringent 0.01 level.)

c. It’s possible to “spin” these results to claim no significance at the (really stringent) 0.001 level. But that’s not what statisticians do (we don’t “spin”). Yet there is the issue of practical significance. These results are definitely statistically significant, but are they of practical importance? The observed difference in average hours worked is 0.5 hours/day. Is an extra 30 minutes a day really of practical importance? (Personally, I’d answer “yes” to this question, but perhaps non-college-graduates would answer “no.”)