Question

7. (9 pts. We want to estimate the difference in the mean time spent standing and walking for obese people and for lean people. A random sample of 10 lean people gives a mean time spent walking and standing of 505.751 minutes with a standard devation of 107.121, and a random sample of 10 obese people gives a mean of 393.269 with a standard deviation of 67.498. The variances of the time spent walking and standing are equal for both groups, and the time spent walking and standing is approximately normal for both groups. Find a 95% confidence interval for the difference in the mean time spent standing and walking for obese people and lean people. Interpret your results in terms of the original problem.

Answer 1

Sample Mean 1 (X1) = 505.751 Sample Standard Deviation 1 (81) = 107.121 Sample Size 1 (N1) = 10 Sample Mean 2 (X2) = 393.269 Sample Standard Deviation 2 (82) = 67.498 Sample Size 2 (N2) = 10 Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df = n1 + n2 – 2 = 10 + 10 – 2 = 18. = The critical value for a = 0.05 and df 18 degrees of freedom is te = ti-a/2;n-1 = 2.101. The corresponding confidence interval is computed as shown below: Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows: (n1 – 1)sî + (n2 – 1) sa ni + n2 – 2 = (10 – 1) 107.1212 + (10 – 1) 67.4982 10 + 10 – 2 89.529 Since we assume that the population variances are equal, the standard error is computed as follows: 1 1 se = + SpV ni n2 1 1 89.529 x + 10 10 40.039 Now, we finally compute the confidence interval: CI (#1 - X2 – to x se, X1 - X2 + te x se) = (505.751 – 393.269 – 2.101 x 40.039,505.751 – 393.269 + 2.101 x 40.039) = (28.364, 196.6)

please like ??