Question

A simple random sample of 40 colleges and universities in the United States has a mean tuition of 19,300 with a standard deviation of 10,600. Construct a 95% confidence interval for the mean tuition for all colleges and universities in the United States. Round the answers to the nearest whole numb

Answer 1

Answer:

16,015≤x≤22,585

Step-by-step explanation:

Formula for calculating the confidence interval is expressed as:

CI = xbar ± z *(s/√n) where;

xbar is the mean tuition = 19,300

s is the standard deviation = 10,600

n is the sample size = 40

z is the z-score at 95% confidence interval = 1.960

Substitute the values into the equation

CI = 19300 ± (1.960) *(10,600/√40)

CI = 19300± (1.960) *(10,600/6.325)

CI = 19300± (1.960) *(1,675.889)

CI = 19300±3,284.743

CI = [19300-3,284.743, 19300+3,284.743]

CI = [16,015.257, 22584.74]

CI = [16,015, 22585] (to nearest whole number)

Hence a 95% confidence interval for the mean tuition for all colleges and universities in the United States is 16,015≤x≤22,585