Question

27. In a survey of 254 students at North Campus, the mean Hours spent on homework was 11.75, with a standard deviation of 2.75. Construct a 99% confidence interval estimate for the mean Hours spent on homework for students at North Campus. Show your calculations on a separate sheet of paper for the margin of error and the construction of the confidence interval. Give your interpretation of the confidence interval estimate. SHOW WORK

Answer 1

27)

Solution :

Given that,

Point estimate = sample mean = = 11.75

Population standard deviation = = 2.75

Sample size = n = 254

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z_/2* ( /n)

= 2.576 * (2.75 / 254)

= 0.44

At 99% confidence interval estimate of the population mean is,

- E < < + E

11.75 - 0.44 < < 11.75 + 0.44

11.31 < < 12.19

(11.31 , 12.19 )