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

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