A study is conducted to asses the extend to which patients who had coronary arte

Question

A study is conducted to asses the extend to which patients who had coronary artery bypass surgery were maintaining their prescribed exercise program The following data reflect the number of times patients reported exercising over the previous month: 14, 11, 8, 6, 5, 3, 6, 13, 12, 8, 1, 4 (i) Compute the mean and standard deviation. (ii) Assuming that the data are approximately normal distributed, compute a 95% confidence interval for the population mean. (iii) Assuming that the data are approximately normal distributed, we can use the data as a pilot study to decide on a sample size n for a future study to obtain a margin of error of 2 for a 95% confidence interval. Select the appropriate n. (iv) What would be the appropriate n for a 99% confidence interval? (v) Without additional computations conclude which confidence interval is wider: 1) [Circle one:] 90% Confidence Interval 95% Confidence Interval 2) [Circle one:] 80% Confidence Interval, n = 50 80% Confidence Interval, n = 10

Explanation / Answer

(1)

2) for 11 degree of freedom and 95% CI, t=2.201

hence confidence interval =mean +/- t*std error =4.9363 ; 10.2303

3)margin of error E=2

for 95% CI, z=1.96

hence sample size n =(std deviation*z/E)2 =~17

4) for 99% CI, z=2.5758

hence from above n=~29

5)95% confidence interval

80% confidence interval, n=10

X 14.000 11.000 8.000 6.000 5.000 3.000 6.000 13.000 12.000 8.000 1.000 4.000 mean(X) 7.583 std deviation(S) 4.166 std error =S/(n)1/2 1.203

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