Question

A random sample of 250 students at a university finds that these students take a mean of 15.7 credit hours per quarter with a standard deviation of 1.5 credit hours. Estimate the mean credit hours taken by a student each quarter using a 98​% confidence interval.

Answer 1

Solution :

Given that,

= 15.7

s = 1.5

n = 250

Degrees of freedom = df = n - 1 = 250 - 1 = 249

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.012 / 2 = 0.01

t _/2,df = t_0.01,249 =2.341

Margin of error = E = t_/2,df * (s /n)

= 2.341 * (1.5 / 250)

= 0.22

Margin of error = 0.22

The 98% confidence interval estimate of the population mean is,

- E <  < + E

15.7 - 0.22 < < 15.7 + 0.22

15.48 < < 15.92

(15.48, 15.92 )