Two sample hypothesis of dependent samples using the t dist

Help very confused. Here is the problem. I have been able to complete some just not sure of the rest.

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

Was there a significant increase in the typical salesperson's weekly income due to the innovative incentive plan? Use the 0.05 significance level. Estimate the p-value, and interpret it.

Click to expand...

Step 1: State the null and alternate hypotheses (ok)

H0: µd ≤ 0
The mean differnce is less than or qual to zero, there was no significant increase.
H1: µd > 0
The mean differnce is greater than zero, there was a significant increase.

Step 2: Select a level of significance (ok)

α = 0.05

Step 3: Identify the test statistic.

Sample size is to small we us T distribution & because we have a paired sample were intersted in the distribution of the differences between before and after. The test statistic will be the paired test.

Step 4: Formulate the decision rule (DONT THINK ITS RIGHT)

This is a one tail test so the criticaal value of t is that for which the probailty of getting a larger t value

tc = 2.073873058
df = 22

Reject the null hypothesis if the observed value t is greater than 1.796 (SHOULD BE THIS ANSWER)

Step 5: Make the decision (ALL WRONG HELP)

. . . . . . . . .before. . . . . . .after
-----------------------------------------
dbar: . . ..428.3333. . . . .454.2500
stdev:. . .120.060086. . .131.4486592
n: . . . . . .12. . . . . . .. . . .12
t:

Because the statistic exceeds the critical value we reject the null hypothesis
We conclude that there was a significant increase in the typical salesmans income

Step 6: Compute the p value (HELP)

We're testing the claim that there is a raise in weekly incomes.

I get \(\displaystyle t_{s}=-0.5042\)

Critical value=1.7171

The p-value is the area to the right of your test statistic. The area that corresponds to your z-score of -0.5042 is 0.6904. Therefore, the p-value is 0.6904.

You should fail to reject the null hypothesis because the test statistic is not in the rejection region. Also, p-vlaue > 0.05. Which means you fail to reject.

Interpretation: There is not enough evidence at the 5% level to support the claim that there is a significant increase in weekly incomes due to the incentive plan.