Question

A recent study of 100 employees of city X showed that the mean of the distance they traveled to work was 35.75 km. The sample standard deviation was 5 km. Find the 95% confidence interval of the true mean.

Answer 1

Mean = 35.75

Standard deviation = 5

N = 100

Alpha = 1 - (95/100)

= 0.05

Z value for alpha = 0.05 is 1.96

Z value = 1.96

Confidence interval = mean +/- z value ( stand deviation/ square root of n)

C.I = 35.75 +/- 1.96 ( 5 / √100)

C.I = 35.75 +/- 1.96 ( 0.5)

C.I = 35.75 +/- 0.98

Lower limit = 35.75 - 0.98. Upper limit = 35.75 +0.98

Lower = 34.77

Upper = 36.73

Confidence interval = { 34.77 , 36.73 }