Question

Question 3 4 pts In a test of weight loss programs, 44 adults used the Atkins weight loss program. After 12 months, their mean weight loss was determined to be 2.4 lbs., with a variance of 23.6 1b2. Assuming the sample is a simple random sample obtained from a population with a normal distribution, construct a 95% confidence interval estimate of the standard deviation of the weight loss for all such subjects. Keep your numerical answers rounded to 3 decimal places. 99% CI estimate of - A Lower Limit: B. Upper Limit: C. Do the results of the study indicate whether the program is helpful in losing weight? i. Yes, as the mean average weight loss is greater than zero. ii. Yes, as the estimate for the dispersion is substantial. iii. Maybe, but we need more information to say definitively. iv. No, as the standard deviation estimate implies that the mean is no different than zero. v. None of the above. HTML Editor

Answer 1

We need to construct the 95% confidence interval for the population variance. We have been provided with the following information about the sample variance and sample size: $2 = 23.6 n = 44 = The critical values for a = 0.05 and df = 43 degrees of freedom are XŹ xi-a/2,1-1 = 26.7854 and xi = xâ/2,n-1 = 62.9904. The corresponding confidence interval is computed as shown below: (n − 1)s (n − 1)s CI(Variance) = . Xa/2,n-1 Xi a/2,n-1 = (44 – 1) × 23.6 (44 – 1) x 23.6 62.9904 26.7854 = (16.1104, 37.8863)

Now that we have the limits for the confidence interval, the limits for the 95% confidence interval for the population standard deviation are obtained by simply taking the squared root of the limits of the confidence interval for the variance, so then: CI(Standard Deviation) = (v16.1104, V 37.8863 = (4.0138, 6.1552)

We need to construct the 99% confidence interval for the population variance. We have been provided with the following information about the sample variance and sample size: $2 II 23.6 n = 44 = XŹ -a/2,n–1 The critical values for a = 0.01 and df = 43 degrees of freedom are xź 22.8595 and x = x 2,n-1 = 70.6159. The corresponding confidence interval is computed as shown below: (n − 1)s (n − 1)s CI(Variance) = x../2,1-1 Xi-a/2,n-1 (44 – 1) x 23.6 (44 – 1) x 23.6 70.6159 22.8595 (14.3707, 44.393)

Now that we have the limits for the confidence interval, the limits for the 99% confidence interval for the population standard deviation are obtained by simply taking the squared root of the limits of the confidence interval for the variance, so then: CI(Standard Deviation) = (V14.3707, V 44.393 = (3.7909, 6.6628)

CCorrect opton "V"

Non of the above