Question

comprete: chapter iu rroblem set Attempts: Do No Harm: 9 8. Confidence intervals for estimating the difference in population means Aa Aa E Elissa Epel, a professor of health psychology at the University of California-San Francisco, studied women in high- and low-stress situations. She found that women with higher cortisol responses to stress ate significantly more sweet food and consumed more calories on the stress day compared with those with low cortisol responses, and compared with themselves on lower stress days. Increases in negative mood in response to the stressors were also significantly related to greater food consumption. These results suggest that psychophysiological responses to stress may influence subsequent eating behavior. Over time, these alterations could impact both weight and health. You are interested in studying whether students living in the dorms or students living off campus are more stressed. You ask a sample of n = 35 students living in the dorms and n2 - 50 students living off campus to record their daily stress level for a week. The average stress level for students living in the dorms was M = 5.7 (on a 10-point stress scale), with a standard deviation of S1 - 1.4. The average stress level for students living off campus was M26.3 (on a 10-point stress scale), with a standard deviation of S2 - 1.3. To develop a confidence interval for the population mean difference wi - Hz, you need to calculate the estimated standard error of the difference of sample means, S(M - M2). The estimated standard error is SML-M2) - 0.296 Use the Distributions tool to develop a 99% confidence interval for the difference in the mean stress level of students living in the dorms and students living off campus. Select a Distribution

Answer 1

For the testing of equality of means of two independent normal distributions, i.e., testing the null hypothesis H.:41-42 = 0 against the alternative hypothesis H:4-470, the 100(1-a)% confidence limits to the difference of means 4-4, are given by (5-3) Flaka+m-2*S.E.(57 –) Where ta is the critical value of the t-distribution with df = n +n, -2, at level of significance, a corresponding to the both-tailed test. For the given problem, we have x = 5.7.x2 = 6.3 n = 35.n,= 50, and S.E.(-7)=0.296 Now for a =0.01, the critical value of the t-distribution with df=83 is, to 005.83 = 2.636. Therefore, the confidence limits to 4-4, are calculated as (5.7-6.3)+(2.636x0.296)=(-1.38.0.18) Hence, the 99% confidence interval for 44-4, is -1.38 to 0.18. This means that you are 99% confident that the unknown difference between the mean stress level of the population of students living in the dorms and the population of students living off campus is located within this interval.

Now for a =0.10, the critical value of the t-distribution with df=83 is, to 05.83 = 1.663. Therefore, the confidence limits to th - My are calculated as (5.7-6.3) 7(1.663x0.296)=(-1.09.-0.11) Hence, the 90% confidence interval for 44-4, is -1.09 to -0.11. This means that you are 90% confident that the unknown difference between the mean stress level of the population of students living in the dorms and the population of students living off campus is located within this interval. The new confidence interval is smaller than the original one, because the new level of confidence is greater than the original one.

Answer 2

it should be smaller and less'