You wish to test the following claim (
H
a
) at a significance level of
α
=
0.001
. For the context of this problem,
μ
d
=
μ
2
−
μ
1
where the first data set represents a pre-test and the second data set represents a post-test.

H
o
:
μ
d
=
0

H
a
:
μ
d
>
0

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:
pre-test	post-test
53.6	66.4
51.8	58.9
57.7	60.4
56.2	58.6
55.4	57.5
59.1	60.8
56.4	52.1
49.6	50.1
51.8	55
62	71.9
45.7	50
60.7	72.5
72.6	63.3
46.4	53.5
46	51.5
50.3	53.7
72.6	66.4
49.3	50.3

What is the test statistic for this sample?
test statistic =
-1.2204
Incorrect (Report answer accurate to 4 decimal places.)

What is the p-value for this sample?
p-value =
(Report answer accurate to 4 decimal places.)

This problem is asking for a paired-sample \( t \)-test, as we are comparing two related sets of data (pre-test and post-test). Here's how to go about solving the problem step by step:

### Step 1: State the Hypotheses

- Null Hypothesis \( H_0 \): \( \mu_d = 0 \) (there is no difference between the pre-test and post-test scores).
- Alternative Hypothesis \( H_a \): \( \mu_d > 0 \) (the post-test scores are higher than the pre-test scores).

### Step 2: Calculate the Differences
For each pair, compute the difference between the post-test and pre-test scores:

\[
d_i = \text{post-test} - \text{pre-test}
\]

Let's compute these differences:

| Pre-test | Post-test | Difference \( d_i \) |
|----------|------------|----------------------|
| 53.6     | 66.4       | 12.8                 |
| 51.8     | 58.9       | 7.1                  |
| 57.7     | 60.4       | 2.7                  |
| 56.2     | 58.6       | 2.4                  |
| 55.4     | 57.5       | 2.1                  |
| 59.1     | 60.8       | 1.7                  |
| 56.4     | 52.1       | -4.3                 |
| 49.6     | 50.1       | 0.5                  |
| 51.8     | 55.0       | 3.2                  |
| 62.0     | 71.9       | 9.9                  |
| 45.7     | 50.0       | 4.3                  |
| 60.7     | 72.5       | 11.8                 |
| 72.6     | 63.3       | -9.3                 |
| 46.4     | 53.5       | 7.1                  |
| 46.0     | 51.5       | 5.5                  |
| 50.3     | 53.7       | 3.4                  |
| 72.6     | 66.4       | -6.2                 |
| 49.3     | 50.3       | 1.0                  |

### Step 3: Calculate the Mean and Standard Deviation of the Differences

Let’s denote the differences by \( d_i \). Now we calculate the sample mean of the differences \( \bar{d} \) and the sample standard deviation \( s_d \).

\[
\bar{d} = \frac{1}{n} \sum_{i=1}^{n} d_i
\]

\[
s_d = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (d_i - \bar{d})^2}
\]

Where:
- \( n \) is the number of pairs.
- \( d_i \) is the difference for each pair.

### Step 4: Calculate the Test Statistic

The test statistic for a paired \( t \)-test is given by:

\[
t = \frac{\bar{d}}{s_d / \sqrt{n}}
\]

Where:
- \( \bar{d} \) is the mean of the differences.
- \( s_d \) is the standard deviation of the differences.
- \( n \) is the number of data pairs.

### Step 5: Find the p-value

After calculating the test statistic, you can find the p-value using the t-distribution with \( n-1 \) degrees of freedom. This is a one-tailed test because the alternative hypothesis is \( \mu_d > 0 \).

---

I'll calculate these steps for you now.The test statistic for the sample is \( t = 2.2771 \).

The corresponding p-value is \( p = 0.0180 \) (accurate to four decimal places).

Since the p-value \( 0.0180 \) is greater than the significance level \( \alpha = 0.001 \), we do not reject the null hypothesis \( H_0 \). There isn't enough evidence at the 0.001 significance level to conclude that the post-test scores are significantly greater than the pre-test scores.

Would you like to ask any questions or need further clarification on the process?

Here are 5 related questions to expand the topic:

1. How does the sample size affect the outcome of a t-test?
2. What is the difference between one-tailed and two-tailed hypothesis tests?
3. Why do we use the t-distribution instead of the normal distribution in this case?
4. How can we interpret the p-value in the context of hypothesis testing?
5. What are the assumptions behind using a paired t-test?

**Tip:** Always check the assumptions of normality and paired structure before applying a paired-sample t-test to ensure valid results.

You wish to test the following claim (H_a) at a significance level of α = 0.001. For the context of this problem, μ_d = μ_2 − μ_1 where the first data set represents a pre-test and the second data set represents a post-test.

H_o: μ_d = 0
H_a: μ_d > 0

You believe the population of difference scores is normally distributed, but you do not know the standard deviation. You obtain the following sample of 
pre-test: 53.6, 51.8, 57.7, 56.2, 55.4, 59.1, 56.4, 49.6, 51.8, 62, 45.7, 60.7, 72.6, 46.4, 46, 50.3, 72.6, 49.3
post-test: 66.4, 58.9, 60.4, 58.6, 57.5, 60.8, 52.1, 50.1, 55, 71.9, 50, 72.5, 63.3, 53.5, 51.5, 53.7, 66.4, 50.3

What is the test statistic for this sample?
test statistic = -1.2204
Incorrect (Report answer accurate to 4 decimal places.)

What is the p-value for this sample?
p-value = (Report answer accurate to 4 decimal places.)

Perform a paired t-test to analyze the pre-test and post-test data sets, testing the hypothesis H_a: μ_d > 0 at a significance level of α = 0.001. Calculate the test statistic and p-value, and understand how to apply hypothesis testing in normally distributed populations.

Math Problem Statement

You wish to test the following claim ( H a ) at a significance level of α

0.001 . For the context of this problem, μ d

H o : μ d

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