Question

A sociologist develops a test to measure attitudes towards public transportation, and n = 79 randomly selected subjects are given the test. Their mean score is x ¯ = 74.3 and their standard deviation is s = 18.7. Find the margin of error E for the 95% confidence interval for the mean score of all such subjects, and round to the nearest tenth.

Answer 1

Solution :

Given that,

= 74.3

s = 18.7

n = 79

Degrees of freedom = df = n - 1 = 79 - 1 = 78

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,78 =1.991

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

= 1.991 * (18.7/ 79)

= 4.19

Margin of error = 4.19