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 17 sales receipts for mail-order sales results in a mean sale amount of $82.70 with a standard deviation of $16.25. A random sample of 10 sales receipts for internet sales results in a mean sale amount of $66.90 with a standard deviation of $20.25. Using this data, find the 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

Explanation:

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

Where

x1 = mean sale amount for mail order sales = 82.70

x2 = mean sale amount for internet sales = 66.9

s1 = sample standard deviation for mail order sales = 16.25

s2 = sample standard deviation for internet sales = 20.25

n1 = number of mail order sales = 17

n2 = number of internet sales = 10

For a 99% confidence interval, we would determine the z score from the t distribution table because the number of samples are small

Degree of freedom =

(n1 - 1) + (n2 - 1) = (17 - 1) + (10 - 1) = 25

z = 2.787

Margin of error =

z√(s²/n1 + s2²/n2) = 2.787√(16.25²/17 + 20.25²/10) = 20.956190