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 11 sales receipts for mail-order sales results in a mean sale amount of $79.00 with a standard deviation of $18.25. A random sample of 18 sales receipts for internet sales results in a mean sale amount of $96.70 with a standard deviation of $20.25. Using this data, find the 98% 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.

−35.713332 ; 0.313332

Step-by-step explanation:

Given that:

Sample size, n1 = 11

Sample mean, x1 = 79

Standard deviation, s1 = 18.25

Sample size, n2 = 18

Sample mean, x2 = 96.70

Standard deviation, s2 = 20.25

df = n1 + n2 - 2 ; 11 + 18 - 2 = 27

Tcritical = T0.01, 27 = 2.473

S = sqrt[(s1²/n1) + (s2²/n2)]

S = sqrt[(18.25^2 / 11) + (20.25^2 / 18)]

S = 7.284

(μ1 - μ2) = (x1 - x2) ± Tcritical * S

(μ1 - μ2) = (79 - 96.70) ± 2.473*7.284

(μ1 - μ2) = - 17.7 ± 18.013332

-17.7 - 18.013332 ; - 17.7 + 18.013332

−35.713332 ; 0.313332

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts

false

step-by-step explanation:

4y - 6 + 4y^2 - 9

before simplifying this contains 4 terms

after simplifying

4y -15 + 4y^2

it contains 3 terms