Question

You are interested in finding a 90% confidence interval for the average commute that non-residential students have to their college. The data below show the number of commute miles for 14 randomly selected non-residential college students. Round answers to 3 decimal places where possible.

10

21

26

20

18

12

11

16

17

6

14

25

25

6

a. To compute the confidence interval use a ? z t  distribution.

b. With 90% confidence the population mean commute for non-residential college students is between  and   miles.

c. If many groups of 14 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About  percent of these confidence intervals will contain the true population mean number of commute miles and about  percent will not contain the true population mean number of commute miles.

Answer 1

a)

use a t distribution

b)

sample mean, xbar = 27.84
sample standard deviation, s = 17.5317
sample size, n = 13
degrees of freedom, df = n - 1 = 12

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.782

ME = tc * s/sqrt(n)
ME = 1.782 * 17.5317/sqrt(13)
ME = 5.7289

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (27.84 - 1.782 * 17.5317/sqrt(13) , 27.84 +1.782 * 17.5317/sqrt(13)))
CI = (19.175 , 36.504)

between 19.175 and 36.504

c)

About 90% contain population mean and about 10% not contain population mean

Let me know in the comment section if anything is not clear. I will reply ASAP!
If you liked the answer, please give an upvote. This will be quite encouraging for me.Thank-you!