Question

in a random sample of 8 people the mean commute time to work was 34.5 minutes and the standard deviation was 7.2 minutes. A 90% confidence interval using the t distribution was calculated to be (29.7, 39.3). After researching commute times to work, it was found that the population standard deviation is 9.2 minutes. Find the margin of error and construct a 90% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results.

Answer 1

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.1 /2) = 1.645
34.5 ± Z (0.1/2 ) * 9.2/√(8)
Lower Limit = 34.5 - Z(0.1/2) 9.2/√(8)
Lower Limit = 29.1493
Upper Limit = 34.5 + Z(0.1/2) 9.2/√(8)
Upper Limit = 39.8507
90% Confidence interval is ( 29.1493 , 39.8507 )

Width =  39.8507 - 29.1493 = 10.7014

Margin of Error = Z (0.1/2 ) * 9.2/√(8) = 5.3507

Calculate confidence interval from t distribution is (29.7, 39.3).

Width = 39.3 - 29.7 = 9.6

Since two confidence interval overlap each othe i.e they are almost equal.

Confidence interval from Z test is more wider than t test, hence Z test CI is more reliable than t test CI.