In a random sample of 28 people, the mean commute time to work was 31.7 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a t-distribution to construct a 80% confidence interval for the population mean mu. What is the margin of error of mu? Interpret the results.

a)The 80% of confidence interval for the Population 'μ' is

(29.9121 ,33.4879)

b) Margin of error of mean '  'μ' is  =  1.7879

Step-by-step explanation:

step(i):-

Given random sample 'n' =28

The mean of the sample x⁻ =31.7 min

The standard deviation of the sample's' =7.2

Step(ii):-

80% of confidence intervals:

The 80% of confidence interval for the Population 'μ' is determined by

Degrees of freedom

γ = n-1 = 28-1 =27

The 80% of confidence interval for the Population 'μ' is determined by

(31.7 -1.7879 , 31.7 +1.7879)

(29.9121 ,33.4879)

b)

Margin of error of mean is determined by

Margin of error =1.7879

javeir earned 56 dollors more than kyle so kyle and javier had 14w but javier made 56 dollors more so 14w plus 56 equals 70w so kyle has 14w

step-by-step explanation: hope i plz give me brainliest with the points so this is not constant cause it says last week so its clearly not every week

answer: the answer is e

step-by-step explanation: