Math Problem Statement

An anger-management course claims that, after completing its seminar, participants will lose their tempers less often. Always a skeptic, you decide to test this claim. A random sample of 12 seminar participants is chosen, and these participants are asked to record the number of times that they lost their tempers in the two weeks prior to the course. After the course is over, the same participants are asked to record the number of times that they lost their tempers in the next two weeks. The following table lists the results of the survey. Using these data, test the claim at the 0.05 level of significance assuming that the population distribution of the paired differences is approximately normal. Let participants before completing the anger-management course be Population 1 and let participants after completing the anger-management course be Population 2.

Number of Times Temper Was Lost during a Two-Week Period Before 8 9 7 3 11 5 10 9 10 3 4 5 After 6 5 5 3 8 6 9 10 6 4 5 4

Step 2 of 3 : Compute the value of the test statistic. Round your answer to three decimal places.

Solution

To compute the value of the test statistic for the paired tt-test, follow these steps:

Step 1: Compute the differences

For each participant, calculate the difference between the "Before" and "After" counts: di=BeforeiAfterid_i = \text{Before}_i - \text{After}_i

ParticipantBeforeAfterDifference (did_i)
1862
2954
3752
4330
51183
656-1
71091
8910-1
91064
1034-1
1145-1
12541

Step 2: Calculate the mean (dˉ\bar{d}) and standard deviation (sds_d) of the differences

dˉ=din,sd=(didˉ)2n1\bar{d} = \frac{\sum d_i}{n}, \quad s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}}

  1. Compute di\sum d_i: di=2+4+2+0+31+11+411+1=13\sum d_i = 2 + 4 + 2 + 0 + 3 - 1 + 1 - 1 + 4 - 1 - 1 + 1 = 13

  2. Compute dˉ\bar{d}: dˉ=din=1312=1.083\bar{d} = \frac{\sum d_i}{n} = \frac{13}{12} = 1.083

  3. Compute (didˉ)2(d_i - \bar{d})^2 for each participant, and then sum: (didˉ)2=[(21.083)2,(41.083)2,,(11.083)2](d_i - \bar{d})^2 = [(2 - 1.083)^2, (4 - 1.083)^2, \dots, (1 - 1.083)^2]

Participantdid_i(didˉ)2(d_i - \bar{d})^2
12(21.083)2=0.841(2 - 1.083)^2 = 0.841
24(41.083)2=8.477(4 - 1.083)^2 = 8.477
32(21.083)2=0.841(2 - 1.083)^2 = 0.841
40(01.083)2=1.173(0 - 1.083)^2 = 1.173
53(31.083)2=3.673(3 - 1.083)^2 = 3.673
6-1(11.083)2=4.473(-1 - 1.083)^2 = 4.473
71(11.083)2=0.007(1 - 1.083)^2 = 0.007
8-1(11.083)2=4.473(-1 - 1.083)^2 = 4.473
94(41.083)2=8.477(4 - 1.083)^2 = 8.477
10-1(11.083)2=4.473(-1 - 1.083)^2 = 4.473
11-1(11.083)2=4.473(-1 - 1.083)^2 = 4.473
121(11.083)2=0.007(1 - 1.083)^2 = 0.007

(didˉ)2=41.381\sum (d_i - \bar{d})^2 = 41.381

  1. Compute sds_d: sd=(didˉ)2n1=41.38111=3.762=1.940s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}} = \sqrt{\frac{41.381}{11}} = \sqrt{3.762} = 1.940

Step 3: Compute the test statistic

The test statistic for the paired tt-test is: t=dˉsd/nt = \frac{\bar{d}}{s_d / \sqrt{n}}

Substitute the values: t=1.0831.940/12=1.0831.940/3.464=1.0830.560=1.933t = \frac{1.083}{1.940 / \sqrt{12}} = \frac{1.083}{1.940 / 3.464} = \frac{1.083}{0.560} = 1.933

Final Answer:

The value of the test statistic is: t=1.933t = 1.933

Would you like me to continue with further steps or interpret the result?


Related Questions:

  1. How do you determine the critical value for this test?
  2. What does the tt-value indicate in the context of this problem?
  3. How does the level of significance affect the decision to reject the null hypothesis?
  4. What are the assumptions of a paired tt-test?
  5. How could this test be applied to other areas of study?

Tip:

Always check that the differences are approximately normally distributed before performing a paired tt-test.

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Math Problem Analysis

Mathematical Concepts

Statistics
Hypothesis Testing
Paired t-Test

Formulas

Mean of differences: \( \bar{d} = \frac{\sum d_i}{n} \)
Standard deviation of differences: \( s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}} \)
Test statistic for paired t-test: \( t = \frac{\bar{d}}{s_d / \sqrt{n}} \)

Theorems

Central Limit Theorem (for assuming approximate normality)
t-Distribution for small sample sizes

Suitable Grade Level

College-level Statistics