Question

K.Brew 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 10 sales receipts for mail-order sales results in a mean sale amount of $⁢87.30 with a standard deviation of $⁢26.25. A random sample of 20 sales receipts for internet sales results in a mean sale amount of $⁢98.90 with a standard deviation of $⁢21.75. Using this data, find the 90% 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 1 of 3 : Point estimate which was -11.6

Step 2 of 3 :  

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

Step 3 of 3 :

Construct the 90% confidence interval

Answer 1

Step 1

Point estimate ( 87.3 - 98.9 ) = -11.6

Step 2

Margin of Error = = 16.865224

Step 3

Confidence interval :-

DF = 15

t(α/2, DF) = t(0.1 /2, 15 ) = 1.753

Lower Limit =
Lower Limit = -28.4652
Upper Limit =
Upper Limit = 5.2652
90% Confidence interval is ( -28.4652 , 5.2652 )