Question

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

copy data

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

answer

( -2.721 )

Explanation:

Step 1: Calculate the differences \( d \)

First, we find the differences \( d = \text{After} - \text{Before} \) for each participant:
- For the first pair: \( 3 - 4 = -1 \)
- Second: \( 4 - 4 = 0 \)
- Third: \( 4 - 3 = 1 \)
- Fourth: \( 4 - 8 = -4 \)
- Fifth: \( 4 - 4 = 0 \)
- Sixth: \( 8 - 11 = -3 \)
- Seventh: \( 10 - 11 = -1 \)
- Eighth: \( 5 - 6 = -1 \)
- Ninth: \( 8 - 10 = -2 \)
- Tenth: \( 7 - 7 = 0 \)
- Eleventh: \( 4 - 5 = -1 \)
- Twelfth: \( 5 - 6 = -1 \)

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 \( d \) (\( s_d \))

First, find the squared differences from the mean:
- \( (-1 - (-1.0833))^2 \approx 0.0069 \)
- \( (0 - (-1.0833))^2 \approx 1.1736 \)
- \( (1 - (-1.0833))^2 \approx 4.3403 \)
- \( (-4 - (-1.0833))^2 \approx 8.4931 \)
- \( (0 - (-1.0833))^2 \approx 1.1736 \)
- \( (-3 - (-1.0833))^2 \approx 3.6531 \)
- \( (-1 - (-1.0833))^2 \approx 0.0069 \)
- \( (-1 - (-1.0833))^2 \approx 0.0069 \)
- \( (-2 - (-1.0833))^2 \approx 0.8331 \)
- \( (0 - (-1.0833))^2 \approx 1.1736 \)
- \( (-1 - (-1.0833))^2 \approx 0.0069 \)
- \( (-1 - (-1.0833))^2 \approx 0.0069 \)

Sum these squared differences:
\( 0.0069 + 1.1736 + 4.3403 + 8.4931 + 1.1736 + 3.6531 + 0.0069 + 0.0069 + 0.8331 + 1.1736 + 0.0069 + 0.0069 \approx 20.973 \)

The sample variance \( s_d^2 = \frac{\sum (d - \bar{d})^2}{n - 1} = \frac{20.973}{11} \approx 1.9066 \)
The sample standard deviation \( s_d = \sqrt{1.9066} \approx 1.3808 \)

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

For a paired t - test, the test statistic is given by:
\( t=\frac{\bar{d}- \mu_d}{s_d/\sqrt{n}} \)
We are testing the claim that \( \mu_d < 0 \) (since we expect after to be less than before, so \( d=\text{After}-\text{Before}<0 \) on average), so \( \mu_d = 0 \) (under the null hypothesis \( H_0:\mu_d = 0 \), \( H_1:\mu_d<0 \))

Substitute \( \bar{d}=-1.0833 \), \( s_d = 1.3808 \), \( n = 12 \) into the formula:
\( t=\frac{-1.0833 - 0}{1.3808/\sqrt{12}}=\frac{-1.0833}{1.3808/3.4641}\)
First, calculate \( 1.3808/3.4641\approx0.40 \) (more precisely, \( 1.3808\div3.4641\approx0.3986 \))
Then \( t=\frac{-1.0833}{0.3986}\approx - 2.717 \) (Wait, but the given answer is - 2.721. Maybe due to more precise calculations)

Let's recalculate with more precision:

\( \sum d=-13 \), \( \bar{d}=\frac{-13}{12}\approx - 1.0833333 \)

\( \sum (d - \bar{d})^2 \):
- For \( d=-1 \): \( (-1 + 1.0833333)^2=(0.0833333)^2 = 0.0069444 \)
- \( d = 0 \): \( (0 + 1.0833333)^2 = 1.1736111 \)
- \( d = 1 \): \( (1 + 1.0833333)^2=(2.0833333)^2 = 4.3402778 \)
- \( d=-4 \): \( (-4 + 1.0833333)^2=(-2.9166667)^2 = 8.5076389 \)
- \( d = 0 \): \( 1.1736111 \)
- \( d=-3 \): \( (-3 + 1.0833333)^2=(-1.9166667)^2 = 3.6736111 \)
- \( d=-1 \): \( 0.0069444 \)
- \( d=-1 \): \( 0.0069444 \)
- \( d=-2 \): \( (-2 + 1.0833333)^2=(-0.9166667)^2 = 0.8402778 \)
- \( d = 0 \): \( 1.1736111 \)
- \( d=-1 \): \( 0.0069444 \)
- \( d=-1 \): \( 0.0069444 \)

Sum these values:
\( 0.0069444+1.1736111 + 4.3402778+8.5076389+1.1736111+3.6736111+0.0069444+0.0069444+0.8402778+1.1736111+0.0069444+0.0069444 \)

Let's add step by step:

\( 0.0069444+1.1736111 = 1.1805555 \)

\( 1.1805555+4.3402778 = 5.5208333 \)

\( 5.5208333+8.5076389 = 14.0284722 \)

\( 14.0284722+1.1736111 = 15.2020833 \)

\( 15.2020833+3.6736111 = 18.8756944 \)

\( 18.8756944+0.0069444 = 18.8826388 \)

\( 18.8826388+0.0069444 = 18.8895832 \)

\( 18.8895832+0.8402778 = 19.729861 \)

\( 19.729861+1.1736111 = 20.9034721 \)

\( 20.9034721+0.0069444 = 20.9104165 \)

\( 20.9104165+0.0069444 = 20.9173609 \)

So \( \sum (d - \bar{d})^2=20.9173609 \)

Then \( s_d^2=\frac{20.9173609}{11}\approx1.901578 \)

\( s_d=\sqrt{1.901578}\approx1.379 \)

Now, \( t=\frac{\bar{d}}{s_d/\sqrt{n}}=\frac{-1.0833333}{1.379/\sqrt{12}} \)

\( \sqrt{12}\approx3.4641016 \)

\( s_d/\sqrt{n}=1.379/3.4641016\approx0.3981 \)

\( t=\frac{-1.0833333}{0.3981}\approx - 2.721 \)

Answer:

\(-2.721\)