In a random sample of 8 ​people, the mean commute time to work was 35.5 minutes and the standard deviation was 7.4 minutes. A 90​% confidence interval using the​ t-distribution was calculated to be (30.5,40.5). After researching commute times to​ work, it was found that the population standard deviation is 9.3 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.

Margin of Error = 5.4088 ;

Confidence interval = (30.1 ; 40.9)

Interval estimate are almost the same

Step-by-step explanation:

Given that :

Population standard deviation, σ = 9.3

Sample size, n = 8

Xbar = 35.5

Confidence level = 90%

The confidence interval:

Xbar ± Margin of error

Margin of Error = Zcritical * σ/sqrt(n)

Zcritical at 90% = 1.645

Margin of Error = 1.645 * 9.3/sqrt(8) = 5.4088

Confidence interval :

Xbar ± Margin of error

35.5 ± 5.4088

Lower boundary = (35.5 - 5.4088) = 30.0912 = 30.1

Upper boundary = (35.5 + 5.4088) = 40.9088 = 40.9

(30.1 ; 40.9)

T distribution =. (30.5 ; 40.5)

Normal distribution = (30.1, 40.9)

the square root of what number? if you add that, then someone will probably answer the question.

part a: 1.08

part b: 27.59