Question

A gasoline station owner wanted to estimate the mean number of gallons sold per customer. He selected a random sample of 60 customers and found a mean of 8.6, with a Standard Deviation of 2.3. What would be the 95% Confidence Interval for his sampling?

8.6 +/- 1.96 (2.3/sq rt 60)

= 8.6 +/- 1.96 (2.3/7.75)

= 8.6 +/- 0.58

= 8.02 to 9.18

So there is a 95% probability that the true mean of gallons purchased would be between 8.02 and 9.18. Why is this range so wide? The 2.3-gallon Standard Deviation found in the sample is large relative to the Mean and contributes to the range. You may want to use the t Statistic if you do not have the Population Standard Deviation for the problem.

Answer 1

Solution:-

95% Confidence Interval for his sampling using t-statistics is C.I = ( 8.01, 9.19).

C.I = 8.6 + 2.001*10.29693

C.I = 8.6 + 0.5942

C.I = ( 8.006, 9.194)