Question

Land's Bend 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 12 sales receipts for mail-order sales results in a mean sale amount of $89.60 with a standard deviation of $17.75. A random sample of 18 sales receipts for internet sales results in a mean sale amount of $96.60 with a standard deviation of $25.75. Using this data, find the 80% 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 critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Step 2 of 3: Find the standard error of the sampling distribution to be used in constructing the confidence interval. Round your answer to the nearest two decimal places.

Step 3 of 3: Construct the 80% confidence interval. Round your answers to two decimal places.

Answer 1

	pop 1	pop 2
sample mean x =	89.600	96.600
std deviation s=	17.750	25.750
sample size n=	12	18
std error se=s/√n=	5.124	6.069
degree freedom=(se12+se22)2/(se12/(n1-1)+se22/(n2-1))=			27

step 1 of 3:

for 80 % CI & 27 df value of t=

1.314

from excel: t.inv(0.9,27)

step 2:

standard error of difference Se=√(S21/n1+S22/n2)=

7.94

Step 3 of 3:

margin of error E=t*std error   =		10.4372
lower bound=mean difference-E=		-17.44
Upper bound=mean differnce +E=		3.44