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 18 sales receipts for mail-order sales results in a mean sale amount of $74.90 with a standard deviation of $21.75. A random sample of 9 sales receipts for internet sales results in a mean sale amount of $87.30 with a standard deviation of $19.75. Using this data, find the 95% 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: Find the point estimate that should be used in constructing the confidence interval.

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 95% confidence interval. Round your answers to two decimal places.

Answer 1

Explanation:

Step 1. The point estimate is the difference between the the mean amount of mail-order purchases and the mean amount of internet purchases. It becomes

74.9 - 87.3 = - 12.4

Step 2. The formula for determining margin of error is expressed aa

Margin of error = z√(s1²/n1 + s2²/n2)

Where

z = z score

s1 = sample standard deviation for data 1

s2 = sample standard deviation for data 2

n1 = number of samples in group 1

n2 = number of samples in group 2

For a 95% confidence interval, the z score is 1.96

From the information given,

s1 = 21.75

n1 = 18

s2 = 19.75

n2 = 9

z√(s1²/n1 + s2²/n2) = 1.96√(21.75²/18 + 19.75²/9) = 1.96 × 8.343

= 16.35

Step 3. Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

95% Confidence interval = - 12.4 ± 16.35