The management of Discount Furniture, a chain of discount furniture stores in the Northeast, designed an incentive plan for salespeople. To evaluate this innovative plan, 12 salespeople were selected at random, and their weekly incomes before and after the plan were recorded. Was there a significant increase in the typical salesperson’s weekly income due to the innovative incentive plan? Use the .05 significance level. Estimate the p-value, and interpret it

Step-by-step explanation:

The question is incomplete. The complete question is

The management of Discount Furniture, a chain of discount furniture stores in the Northeast, designed an incentive plan for salespeople. To evaluate this innovative plan, 12 salespeople were selected at random, and their weekly incomes before and after the plan were recorded.

Salesperson Before After

Sid Mahone $320 $340

Carol Quick 290 285

Tom Jackson 421 475

Andy Jones 510 510

Jean Sloan 210 210

Jack Walker 402 500

Peg Mancuso 625 631

Anita Loma 560 560

John Cuso 360 365

Carl Utz 431 431

A. S. Kushner 506 525

Fern Lawton 505 619

Solution:

Corresponding income of salespersons before and after form matched pairs.

The data for the test are the differences between the income is salespersons.

μd = the income before minus their income after.

Bedore after diff

320 340 -20

290 285 5

421 475 - 54

510 510 0

210 210 0

402 500 - 98

625 631 -6

569 560 0

360 365 - 5

431 431 0

506 525 - 19

505 619 - 114

Sample mean, xd

= (- 20 + 5 - 54 + 0 + 0 - 98 - 6 + 0 - 5 + 0 + - 19 - 114)/12 = - 25.92

xd = - 25.92

Standard deviation = √(summation(x - mean)²/n

n = 12

Summation(x - mean)² = (- 20 + 25.92)^2 + (5 - 25.92)^2 + (- 54 + 25.92)^2+ (0 + 25.92)^2 + (0 + 25.92)^2 + ( - 98 + 25.92)^2 + ( - 6 + 25.92)^2 + (0 + 25.92)^2 + (- 5 + 25.92)^2 + (0 + 25.92)^2 + (- 19 + 25.92)^2 + (- 114 + 25.92)^2 = 17784.5168

Standard deviation = √(17784.5168/12

sd = 38.5

For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

1) The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 12 - 1 = 11

2) The formula for determining the test statistic is

t = (xd - μd)/(sd/√n)

t = ( - 25.92- 0)/(38.5/√12)

t = - 2.33

3) We would determine the probability value by using the t test calculator.

p = 0.02

4) Assume alpha = 0.05

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. We can conclude that at 5% significance level, there is a significant increase in the typical salesperson’s weekly income due to the innovative incentive plan

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