Question

How much time per week do students spend on their assignments and review at home? A random sample of 36 USC (University of Southern California) students indicates that the mean time spent on studying at home is equal to 15.3 hours per week, with a population standard deviation equals to 3.8 hours. 1) Form a 95% CI estimate for the population mean time spent on studying at home. 2) What sample size is needed to be 90 % confident of being within only one third of the margin of error in 1).

Answer 1

Solution :

Given that,

1. = 15.3

= 3.8

n = 36

A ) At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

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

= 1.960 * (3.8 / 36) = 1.241

At 95% confidence interval estimate of the population mean is,

- E < < + E

15.3 - 1.241< < 15.3 + 1.241

14.059< < 16.541

(14.059 , 16.541):

2) standard deviation = = 3.8

margin of error = E = 0.333

  At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Sample size = n = ((Z_/2 * ) / E)²

= ((1.645 * 3.8) / 0.333)²

= 352.38 = 352

Sample size = 352