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

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)