Having trouble with this one, please show all work. Thank you! Dr. J.V. is inter

Question

Having trouble with this one, please show all work. Thank you! Dr. J.V. is interested in finding the average number of hours that students spend doing Mathematics homework per week. He conducts a survey of 81 students. The average is 8 hours and the standard deviation is 2 hours. (a) Find a 90% confidence interval for the real number of hours that students study per week? (b) Juan claims that he studies on average 13 hours per week. Is this reasonable? Explain (c) What is the sample size need in order to have a margin of error of 1.4 hours?

Explanation / Answer

n = 81
Mean = 8
Stdev = 2

a. 90% interval is Mean +/- Z*Sigma/sqrt(n) = 8 +/- 1.645*2/sqrt(81) = 7.63 to 8.37

b. Z = (13-8)/(2/sqrt(9)) = 7.5, which is very high. Indicating that we reject null and conclude that his claim is unreasonable, given such a high Z value.

c. 1.4 = Z*Sigma/sqrt(n)
Here the Z value or the confidence interval is not known , so we will assume it be 1.645 or 90% confidence interval

So,
1.4 = 1.645*2/sqrt(n)
n = (1.645*2/1.4)^2
n = 5.5225 or 6 people

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