Question

In: Statistics and Probability

24. A hospital would like to determine the mean length of stay for its stroke patients....

24. A hospital would like to determine the mean length of stay for its stroke patients. They observe a random sample of 6 patients, the resulting length of stay of stroke patients are: 6, 7, 5, 8, 11, 2 (in days) Suppose the true length of stay follows normal distribution, a 95% confidence interval for population mean is ____. (4 decimal places)

Use the information to answer Q25-27 A 98% confidence interval for the mean number of credit hours that Clemson students take is (12.8, 17.6) hours.

25. The length of interval is ____.

26. Then sample mean (also the point estimate) is ____.

27. The margin of error is ____.

Use the information to answer Q28

You have asked a random sample of 40 Clemson students how many credit hours they are taking. The resulting 98% confidence interval for the mean number of credit hours that Clemson students take is (12.8, 17.6) hours.

28. Which of the following statements about the confidence interval of (12.8, 17.6) hours is INCORRECT?

A. We are 98% confident that the true mean is between 12.8 and 17.6 hours

B. We are 98% confident that the population mean is between 12.8 and 17.6 hours

C. We are 98% confident that the population mean is in the interval (12.8, 17.6) hours

D There is a 98% probability that the population mean is between 12.8 and 17.6

Solutions

Expert Solution

24)

sample mean         x= 6.500
sample size             n= 6
sample std deviation s= 3.017
std error sx=s/√n= 1.2315
for 95% CI; and 5 df, critical t= 2.5710
margin of error E=t*std error                            = 3.1657
lower bound=sample mean-E = 3.3343
Upper bound=sample mean+E= 9.6657
from above 95% confidence interval for population mean =(3.3343 ,9.6657)

25)

ength of interval is =17.6-12.8 =4.8

26) sample mean=(17.6+12.8)/2 =15.2

27) margin of error is =4.8/2 =2.4

28)

D There is a 98% probability that the population mean is between 12.8 and 17.6


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