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 8sales receipts for mail-order sales results in a mean sale amount of $78.70 with a standard deviation of $18.75. A random sample of 17 sales receipts for internet sales results in a mean sale amount of $64.90 with a standard deviation of $26.75. 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. Construct the 90% confidence interval. Round your answers to two decimal places.

Answer 1

i am using minitab to solve the problem.

steps:-

copy the data in minitab stat basic statistics 2 sample t select summarized data in sample 1 type 8 in sample size, 78.7 in sample mean, 18.8 in standard deviation in sample 2 type 17 in sample size, 64.9 in sample mean, 26.8 in standard deviation options in confidence level type 90 in hypothesized difference type 0 select the alternative hypothesis as difference hypothesized difference ok ok.

*** SOLUTION ***

the 90% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases be:-

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