Question

J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet. Random samples of sales receipts were studied for mail-order sales and internet sales, with the total purchase being recorded for each sale. A random sample of 1616 sales receipts for mail-order sales results in a mean sale amount of $74.50$⁢74.50 with a standard deviation of $17.25$⁢17.25. A random sample of 99 sales receipts for internet sales results in a mean sale amount of $84.40$⁢84.40 with a standard deviation of $21.25$⁢21.25. Using this data, find the 99%99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases. Assume that the population variances are not equal and that the two populations are normally distributed.

Step 2 of 3 :  

Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.

Answer 1

Step 2

Margin of Error =

Confidence interval :-

t(α/2, DF) = t(0.01 /2, 14 ) = 2.977

Lower Limit = -34.5878
Upper Limit = 14.7878
99% Confidence interval is ( -34.5878 , 14.7878 )