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 1717 sales receipts for mail-order sales results in a mean sale amount of $78.60$⁢78.60 with a standard deviation of $24.75$⁢24.75. A random sample of 99 sales receipts for internet sales results in a mean sale amount of $72.10$⁢72.10 with a standard deviation of $25.25$⁢25.25. 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

The data is given for the company sells through mail-order purchase and via Internet purchase .

	Mail-order purchase	Internet purchase
sample mean
sample standard deviation
sample size

Let, true mean amount of mail-order purchase

   the mean amount of internet purchase

Calculation for 99% confidence interval for the true mean difference

For non-equal variance, the confidence interval is calculated as-

Degrees of freedom (df):

Critical value:

for 99% confidence interval,

So, the 99% confidence interval for the true mean difference for mail-order purchase and internet purchase is , i.e.,

Hence we are 99% confident that the confidence interval (-28.1878, 41.1878) contains the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases.

Margin of error:

The margin of error use in calculating the confidence interval is-

Margin of error for the 99% confidence interval is calculated to be as Margin of error=34.687848