Question

As part of his class project, a Statistics student took a random sample of 50 college students and recorded how many hours a week they spent on the Internet. The sample reveals an average of 6.9 hrs. Calculate the 90% confidence Interval for average Internet usage among college students. Assume that the standard deviation of Internet usage for college students is 4.5 hrs/week. Answer 1&2

1) A) E =1.58 B) E = 2.02 C) E =1.05 D) E = 2.58

2) A) 5.85 < < μ 7.95 B) 5.32 < < μ 8.48 C) 4.32 < μ < 9.48 D) 4.88 < < μ 8.92

Answer 1

At 90% confidence interval the critical value is z_0.05 = 1.645

1) Margin of error(E) = z_0.05 *

                             = 1.645 * 4.5/

                             = 1.05

Option - C is correct

2) The 90% confidence interval is

+/- E

= 6.9 +/- 1.05

= 5.85, 7.95

Option - A is correct.