Use the standard normal distribution or the t-distribution to construct a 90% co

Question

Use the standard normal distribution or the t-distribution to construct a 90% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results.

The gas mileages (in miles per gallon) of 45 randomly selected cars are listed below.

24 13 14 26 17 23 26 18 17 19 13 22 17 23 20 20 27 13 21 23 24 17 21 16 13 14 25 22 18 21 21 23 17 21 14 15 19 19 23 26 26 17 23 24 14

Which distribution should be used to construct the confidence interval?

A.Use a t-distribution because ngreater than or equals30 and sigmais known.

B.Use a t-distribution because ngreater than or equals30 and sigmais unknown.

C.Use a normal distribution because ngreater than or equals30 and sigmais known.

D.Use a normal distribution because ngreater than or equals30 and sigmais unknown.

E.Cannot use the standard normal distribution or the t-distribution because sigmais unknown, n less than 30, and the mileages are not normally distributed.

Select the correct choice below and, if necessary, fill in any answer boxes to complete your choice.

A.The 90% confidence interval is (____,____).

(Round to one decimal place as needed.)

B.Neither distribution can be used to construct the confidence interval.

Explanation / Answer

Mean=19.7556, SD=4.2004

n=45

n is greater than 30 and sample variance is known. Hence use a normal distribution

SD(mean)=4.2004/sqrt(45)=0.6262

Z(90% CI)=1.645

CI=[19.7556-1.645*0.6262,19.7556+1.645*0.6262]=[18.7,20.8]

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.