Question

This study investigates the hours per month spent on the Internet by U.S. residents, age 18 to 24. The sample results are: n= 75, x-bar=28.5, and s=23.1 . We are to construct a 95% confidence interval based on this information.

a) What is critical value of t (that is, t*) for this confidence interval?

b) the standard error of the mean (SEM) is:

c) Rounded to one decimal place, as x-bar was, the margin of error (MOE) for your confidence interval will be:

d) Which of the following conclusions is CORRECT?

We are 95% sure the mean hours spent per week on the Internet by U.S. residents, age 18 to 24, is between 23.5 and 34.1.

We are 95% sure the mean hours spent per month on the Internet by college students is between 23.2 and 33.8.

We are 95% sure the mean hours spent per month on the Internet by U.S. residents, age 18 to 24, is between 23.2 and 33.8.

We are 95% sure the mean hours spent per month on the Internet by U.S. residents, age 18 to 24, is between 23.5 and 34.1.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 28.5

sample standard deviation = s = 23.1

sample size = n = 75

Degrees of freedom = df = n - 1 = 75 - 1 = 74

a) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,74 = 1.993

b) standard error = SE = (s /n) = ( 23.1/ 75) = 2.67

c) Margin of error = E = t/2,df * SE

= 1.993 * 2.67

Margin of error = E = 5.3

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

  ± E  

= 28.5 ± 5.3

= ( 23.2, 33.8 )

d) We are 95% sure the mean hours spent per month on the Internet by U.S. residents, age 18 to 24, is between 23.2 and 33.8.