Question

# You are interested in finding a 90% confidence interval for the average commute that non-residential students...

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 10 randomly selected non-residential college students. Round answers to 3 decimal places where possible.

10 13 24 14 11 10 22 9 21 22

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

c. If many groups of 10 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.

Level of Significance , α = 0.10

sample std dev , s = √(Σ(X- x̅ )²/(n-1) ) =5.9479

Sample Size ,n = 10

Sample Mean,  x̅ = ΣX/n = 15.6000

b)

Level of Significance , α = 0.1

degree of freedom=DF=n-1=9

't value='tα/2=1.8331          [Excel formula =t.inv(α/2,df) ]

Standard Error , SE =    s/√n =5.9479/ √10=1.880898

margin of error , E=t*SE =1.8331*1.88090=3.447899

confidence interval is

Interval Lower Limit = x̅ - E = 15.60-3.447899=12.152101

Interval Upper Limit = x̅ + E = 15.60-3.447899=19.047899

90%confidence interval is (12.15< µ <19.05)

c)

If many groups of 10 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About    (90) percent of these confidence intervals will contain the true population mean number of commute miles and about (10) percent will not contain the true population mean number of commute miles.

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