High School

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You wish to test the following claim [tex]\left( H _a\right)[/tex] at a significance level of [tex]\alpha = 0.02[/tex]. For the context of this problem, [tex]\mu_d = \text{PostTest} - \text{PreTest}[/tex]. Each row gives the scores of a single individual.

[tex]\[

\begin{array}{l}

H_0: \mu_d = 0 \\

H_a: \mu_d \ \textless \ 0

\end{array}

\][/tex]

You believe the population of difference scores is normally distributed, but you do not know the standard deviation. You obtain the following sample of data:

[tex]\[

\begin{array}{|r|r|}

\hline

\text{pre-test} & \text{post-test} \\

\hline

64.7 & 44.8 \\

31.2 & 30.9 \\

15.2 & -29.5 \\

53.9 & 39.3 \\

28.2 & 42.9 \\

49.8 & 48.2 \\

47.2 & 56.1 \\

45.8 & 21.2 \\

21.0 & -9.4 \\

41.9 & 28.3 \\

11.4 & -11.0 \\

33.7 & 7.4 \\

29.5 & 40.3 \\

47.5 & 65.2 \\

48.6 & 29.3 \\

60.4 & 66.9 \\

\hline

\end{array}

\][/tex]

What is the test statistic for this sample?
[tex]\text{test statistic} = \square[/tex] (Report answer accurate to 4 decimal places.)

What is the [tex]p[/tex]-value for this sample?
[tex]p\text{-value} = \square[/tex] (Report answer accurate to 4 decimal places.)

Answer :

To solve this problem, we're testing the claim that the average difference in scores, [tex]\(\mu_d\)[/tex], is less than zero. This is a paired t-test problem since we have paired data (pre-test and post-test scores for the same individuals).

Here's how you would approach it:

1. Calculate the Difference Scores:
For each individual, subtract the pre-test score from the post-test score to find the difference:
[tex]\[
\text{difference score} = \text{post-test score} - \text{pre-test score}
\][/tex]

2. Conduct the Paired t-Test:
- Null Hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu_d = 0\)[/tex]
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu_d < 0\)[/tex]

Since we do not know the standard deviation of the population, we'll use a t-test for the sample of differences.

3. Calculate the Test Statistic:
The test statistic for a paired t-test is calculated using the formula:
[tex]\[
t = \frac{\bar{d} - \mu_d}{s_d / \sqrt{n}}
\][/tex]
where:
- [tex]\(\bar{d}\)[/tex] = average of the difference scores
- [tex]\(s_d\)[/tex] = standard deviation of the difference scores
- [tex]\(n\)[/tex] = number of pairs

4. Find the Test Statistic:
The calculated test statistic is approximately [tex]\(-2.1623\)[/tex].

5. Calculate the p-value:
The p-value associated with this test statistic and the alternative hypothesis ([tex]\(H_a: \mu_d < 0\)[/tex]) is approximately [tex]\(0.0236\)[/tex].

6. Make a Decision:
Compare the p-value to the significance level [tex]\(\alpha = 0.02\)[/tex]:
- Since [tex]\(0.0236 > 0.02\)[/tex], we do not reject the null hypothesis [tex]\(H_0\)[/tex].

Therefore, there is not enough evidence at the [tex]\(0.02\)[/tex] significance level to support the claim that the mean of the difference scores is less than zero.

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