Question

stion 3 plete answer and out of Tag question A group of 78 people enrolled in a weight-loss program that involved adhering to a special diet and to a daily exercise program. After six months, their mean weight loss was 25 pounds, with a sample standard deviation of 9 pounds. A second group of 43 went on the special diet, but did not exercise. After six months, their mean weight loss was 14 pounds, with a sample standard deviation of 7 pounds. a. What function would be used to construct a confidence interval for the difference in weight loss between the exercisers and the non-exercisers? 2SampTint b. Construct a 97.5% confidence interval for the difference in weight loss between the exercisers and the non- exercisers. Round your answers to two decimal places. ( 7.64 14.36 C. Based on your confidence interval, on average, which program has a higher weight loss? ✓ Diet and exercise. Diet and no exercise They provide the same results on average All of the above are possible. Nevtrane

Answer 1

We need to construct the 97.5% confidence interval for the difference between the population means ji - M2, for the case that the population standard deviations are not known. The following information has been provided about each of the samples: Sample Mean 1 (X1) = 25 Sample Standard Deviation 1 (81) = 9 Sample Size 1 (N1) = 78 Sample Mean 2 (X2) = 14 Sample Standard Deviation 2 (S2) = 7 Sample Size 2 (N2) = 43 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 = 78 + 43 – 2 = 119. = The critical value for a = 0.025 and df 119 degrees of freedom is te = ti-a/2;n-1 2.27. 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: = (1 - 1)x+(02 - 1)s ni + n2 - 2 = (78 – 1) × 92 + (43 – 1) × 72 78 + 43 – 2 8.349 Since we assume that the population variances are equal, the standard error is computed as follows: 1 1 se = SpV ni n2 1 1 8.349 x + 78 43 1.586 Now, we finally compute the confidence interval: CI = (X1 - X2 – tę x se, X1 - X2 + te x se) = (25 – 14 – 2.27 x 1.586, 25 – 14 +2.27 x 1.586) = (7.4, 14.6) Therefore, based on the data provided, the 97.5% confidence interval for the difference between the population means Mi – M2 is 7.4 < Mi – M2 < 14.6, which indicates that we are 97.5% confident that the true difference between population means is contained by the interval (7.4, 14.6).

a) Two sample t- test.

b) The 97.5 confidence interval is (7.4, 14.6).

c) Diet and no exercise.

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