Question

question 2 of 6, step 2 of 3
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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 0.10 level of significance assuming that the population distribution of the paired differences is approximately normal. let d = (participants after completing the anger-management course) − (participants before completing the anger-management course).

number of times temper was lost during a two-week period
before 4 4 3 8 4 11 11 6 10 7 5 6
after 3 4 4 4 4 8 10 5 8 7 4 5

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step 2 of 3: compute the value of the test statistic. round your answer to three decimal places.

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Explanation:

Step 1: Calculate the paired differences \( d \)

For each participant, calculate \( d = \text{After} - \text{Before} \):

• \( 3 - 4 = -1 \)
• \( 4 - 4 = 0 \)
• \( 4 - 3 = 1 \)
• \( 4 - 8 = -4 \)
• \( 4 - 4 = 0 \)
• \( 8 - 11 = -3 \)
• \( 10 - 11 = -1 \)
• \( 5 - 6 = -1 \)
• \( 8 - 10 = -2 \)
• \( 7 - 7 = 0 \)
• \( 4 - 5 = -1 \)
• \( 5 - 6 = -1 \)

So the differences \( d \) are: \(-1, 0, 1, -4, 0, -3, -1, -1, -2, 0, -1, -1\)

Step 2: Calculate the mean of \( d \) (\( \bar{d} \))

The formula for the mean is \( \bar{d} = \frac{\sum d}{n} \), where \( n = 12 \).

First, sum the differences:
\( \sum d = -1 + 0 + 1 + (-4) + 0 + (-3) + (-1) + (-1) + (-2) + 0 + (-1) + (-1) = -13 \)

Then, \( \bar{d} = \frac{-13}{12} \approx -1.0833 \)

Step 3: Calculate the standard deviation of \( d \) (\( s_d \))

The formula for the sample standard deviation is \( s_d = \sqrt{\frac{\sum (d - \bar{d})^2}{n - 1}} \)

First, calculate \( (d - \bar{d})^2 \) for each \( d \):

• For \( d = -1 \): \( (-1 - (-1.0833))^2 = (0.0833)^2 \approx 0.0069 \) (and there are 5 such values: -1, -1, -1, -1, -1)
• For \( d = 0 \): \( (0 - (-1.0833))^2 = (1.0833)^2 \approx 1.1736 \) (and there are 3 such values: 0, 0, 0)
• For \( d = 1 \): \( (1 - (-1.0833))^2 = (2.0833)^2 \approx 4.3403 \) (1 value)
• For \( d = -4 \): \( (-4 - (-1.0833))^2 = (-2.9167)^2 \approx 8.5076 \) (1 value)
• For \( d = -3 \): \( (-3 - (-1.0833))^2 = (-1.9167)^2 \approx 3.6736 \) (1 value)
• For \( d = -2 \): \( (-2 - (-1.0833))^2 = (-0.9167)^2 \approx 0.8403 \) (1 value)

Now sum these squared differences:
\( 5 \times 0.0069 + 3 \times 1.1736 + 4.3403 + 8.5076 + 3.6736 + 0.8403 \)
\( = 0.0345 + 3.5208 + 4.3403 + 8.5076 + 3.6736 + 0.8403 \)
\( = 20.9171 \)

Then, \( s_d = \sqrt{\frac{20.9171}{12 - 1}} = \sqrt{\frac{20.9171}{11}} \approx \sqrt{1.90155} \approx 1.379 \)

Step 4: Calculate the test statistic \( t \)

The formula for the paired - t test statistic is \( t=\frac{\bar{d}- \mu_d}{s_d/\sqrt{n}} \). Here, the null hypothesis is \( H_0:\mu_d = 0 \) (since we are testing if there is a change, and the claim is that participants lose their tempers less often, so \( \mu_d=\text{After}-\text{Before}\leq0 \) and alternative \( H_1:\mu_d < 0 \), but for the test statistic calculation, \( \mu_d = 0 \) under \( H_0 \))

So \( t=\frac{\bar{d}-0}{s_d/\sqrt{n}}=\frac{\bar{d}\sqrt{n}}{s_d} \)

Substitute \( \bar{d}=- 1.0833 \), \( n = 12 \), and \( s_d\approx1.379 \):

\( t=\frac{-1.0833\times\sqrt{12}}{1.379}=\frac{-1.0833\times3.4641}{1.379}=\frac{-3.752}{1.379}\approx - 2.721 \)

Answer:

\(-2.721\)