Question

A sociologist develops a test to measure attitudes about public transportation, and 20 randomly selected subjects are given the test. Their mean score is 75 and their standard deviation is 9 Construct 98% confidence interval for the mean score of all such subjects.

Answer 1

( )

Solution :

Given that,

= 75

s =9

n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

a ) At 98% confidence level the t is

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

t /2,df = t0.01,19 = 2.539 ( using student t table)

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

= 2.539 * (9 / 20)

= 5.1

The 98% confidence interval mean is,

- E < < + E

75 - 5.1 < <75 +5.1

69.9 < < 80.1