Social Studies 201

Fall 2003

Answers to Computer Problem for Problem Set 5

This file contains only the written answers – the written answers and the tables associated with these is in the Word for Windows file p5af03.doc in the folder t:\students\public\201.

1. b. i. The first interval is a 95% interval estimate for mean household income of all undergraduate students at the University of Regina, Fall 1998. The mean income reported for the 617 students in the survey is 65.46 thousand dollars and the interval is from 61.99 to 68.93 thousand dollars. Since the sample size of n=617 is large, the sample mean has a normal distribution with mean m and standard deviation s divided by the square root of n. The sample standard deviation s=43.888 can be used as an estimate of s. For 95% confidence

level, the Z value is 1.96. The interval is

and this interval is from 62.00 to 68.92. Apart from rounding error, this is the 95% interval reported in the printout.

1.b.ii. The mean income from SLID was $53,378 or 53.378 thousand dollars. The sample of 617 undergraduates who reported household income yielded a mean of 65.46 thousand dollars. This evidence suggests that undergraduates come from households with a larger household income than the provincial average. The interval estimates support this argument. The widest of the intervals, the 99% confidence interval is from 60.90 to 70.03 thousand dollars. 99% intervals are constructed so that a researcher is 99% sure that these intervals contain the true mean. So a researcher can be fairly certain that the true household income for all undergraduates is not as low as 53.378 thousand dollars and a researcher would have reasonable certainty that the mean household income for all undergraduates exceeds that for all Saskatchewan households.

2.b. Earlier in the semester the data from the sample showed that, on average, males studied a little more than did females and spent a little more time than females at extracurricular activities, but the difference was very small. In contrast, females reported much higher average hours spent caring for dependents than did males.

The 95% interval estimates support these conclusions. For study hours, the interval for males is (15.0, 18.7) and for females is (15.6, 18.2), rounded off to the nearest tenth of an hour. These intervals are not that different and the interval for females is entirely within the interval for male study hours. Since the true mean can lie anywhere within these intervals, there is no assurance of any difference between males and females in terms of mean study hours.

For mean extracurricular hours, the 95% interval estimate for males is (1.2, 2.4) and for females is (1.2, 2.1). Again, there is no assurance of any difference in overall mean extracurricular hours for the two groups – the interval estimates for the means overlap too much.

For mean hours spent caring for dependents, the male interval estimate is (1.3, 4.6) and the female interval is (4.3, 8.3). While there is some chance that the true mean for each is in the overlapping region for the two intervals (between 4.3 and 4.6 hours for each), the intervals overlap very little. The male interval generally lies to the left of the female interval, at lower number of hours.

In summary, the conclusions are the same as earlier, although these are only samples so if all students were surveyed, the true values would be somewhat different from these sample values.

3.b. There are two tests of hypotheses here. For (i), the null hypothesis is that the true mean grade for all undergraduates is 75%. The question asks whether the true mean equals 75% and since there is no indication of whether it might be less than or greater than 75%, this is a two tailed hypothesis test that the mean differs from 75%. Since the sample size is n=572, the sample mean has a normal distribution and the t-value in the printout is really a Z-value. If the 0.05 significance level is adopted, the region of rejection is all Z-values of less than –1.96 or greater than +1.96. The Z-value from the table is –3.521, less than 1.96, so the null hypothesis can be rejected. At the 0.05 level of significance, the alternative hypothesis that the mean is 75% is accepted.

For the second test, where the null hypothesis is that the gpa for all undergraduates is 74%, the test is the same. This is again a two-tailed test with large sample size. In this case the Z-value is –0.142, a small Z-value, greater than –1.96 and less than +1.96. As a result, the sample mean does not lie in the region of rejection of the null hypothesis. At the 0.05 level of significance, the null hypothesis that the mean grade point average for all undergraduate students is 74% cannot be rejected.

4. b. The null hypothesis here is that the mean household income of all undergraduate students equals that mean income of all Saskatchewan households. In this case we have a suspicion that undergraduates come from higher income households, so a one tailed test could be used here. In this case, the alternative hypothesis is that the mean household income of all undergraduates exceeds the Saskatchewan mean of 53.378 thousand dollars. The sample size of n=616 is large so the sample mean is normally distributed and the t-value is really a Z-value in this case. Since there is a strong suspicion that the undergraduate mean exceeds the Saskatchewan mean, select a small significance level, say 0.01. This makes it difficult to reject the null hypothesis and, if the null hypothesis is rejected, a researcher is fairly certain that this is the correct conclusion. For a one-tailed test at the 0.01 level of significance, the critical Z-value is 2.33. Since the Z-value from the table is 6.839, this is a lot greater than 2.33 and the null hypothesis is rejected. At the 0.01 level of significance, there is strong evidence that undergraduates come from higher income households than the mean for the population as a whole.

This result is consistent with the findings in question 1, where the same conclusion resulted. In 1, the interval estimate came nowhere close to crossing the provincial mean of 53.378. The hypothesis test showed that there is strong evidence that undergraduates come from households with a higher mean household income than the Saskatchewan mean.