K.Brew sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet. R
andom 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 $81.90 with a standard deviation of $22.25. A random sample of 8 sales receipts for internet sales results in a mean sale amount of $88.30 with a standard deviation of $23.25. 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. Construct the 98% confidence interval.
I see it like this: we mupliply almost 4 by a little more then 5, so the result will be around 20 (4*5=20) : so the decimal point must be after the 20 (and it is, the result is 20.67