pre-test: 51,59,92,87,78,44,92,68,76,76
post-test: 60,59,50,67,62,56,64,56,56,55

Can we say that the review section significantly harms test performance with a significance level of  = 0.002. Assume a Normal distribution.
For this study, we should use

The null and alternative hypotheses would be:
:
:

The test statistic =  (please show your answer to 3 decimal places.)

The p-value =  (Please show your answer to 4 decimal places.)

The p-value is

Based on this, we should  the null hypothesis.

As such, the final conclusion is that ...
There is not sufficient evidence to conclude that the review section significantly harms test performance with a significance level of  = 0.002
There is sufficient evidence to conclude that the review section significantly harms test performance with a significance level of  = 0.002

To determine whether the review section significantly harms test performance at a significance level of \( \alpha = 0.002 \), we can perform a **paired t-test** because the data comes in pairs (pre-test and post-test scores for each student).

### Step 1: Define the hypotheses

The null and alternative hypotheses are as follows:

- \( H_0 \): The review section does **not** significantly harm test performance, i.e., there is no difference between pre-test and post-test scores. Mathematically, this can be written as \( \mu_d = 0 \), where \( \mu_d \) is the mean of the differences between paired scores.
- \( H_1 \): The review section significantly harms test performance, i.e., the post-test scores are lower than the pre-test scores. Mathematically, this can be written as \( \mu_d > 0 \) (because we are testing for harm, we expect pre-test scores to be higher than post-test scores).

### Step 2: Calculate the differences

We subtract the post-test scores from the pre-test scores to get the differences:

\[
\text{Differences} = \text{Pre-test} - \text{Post-test} = [51-60, 59-59, 92-50, 87-67, 78-62, 44-56, 92-64, 68-56, 76-56, 76-55]
\]
\[
\text{Differences} = [-9, 0, 42, 20, 16, -12, 28, 12, 20, 21]
\]

### Step 3: Calculate the test statistic

To compute the test statistic, we first need the **mean** and **standard deviation** of the differences.

- Mean of the differences (\( \bar{d} \)):
  \[
  \bar{d} = \frac{\sum \text{Differences}}{n} = \frac{-9 + 0 + 42 + 20 + 16 - 12 + 28 + 12 + 20 + 21}{10} = \frac{138}{10} = 13.8
  \]

- Standard deviation of the differences (\( s_d \)):
  \[
  s_d = \sqrt{\frac{\sum (\text{Difference} - \bar{d})^2}{n-1}}
  \]
  First, we calculate the squared differences:
  \[
  (-9 - 13.8)^2, (0 - 13.8)^2, (42 - 13.8)^2, \ldots
  \]
  Summing them up and then dividing by \( n-1 \) gives the standard deviation.

Let's compute these values:

\( s_d = \sqrt{\frac{(-9 - 13.8)^2 + (0 - 13.8)^2 + \dots}{9}} = \sqrt{366.93} \approx 19.152 \)

Now, the test statistic is given by:
\[
t = \frac{\bar{d}}{s_d / \sqrt{n}} = \frac{13.8}{19.152 / \sqrt{10}} = \frac{13.8}{6.056} \approx 2.278
\]

### Step 4: Find the p-value

Using a t-distribution table (with \( n-1 = 9 \) degrees of freedom), we can find the p-value for \( t = 2.278 \).

Looking up the value or using a calculator for the t-distribution:
\[
\text{p-value} \approx 0.0255
\]

### Step 5: Compare the p-value with \( \alpha = 0.002 \)

The p-value \( 0.0255 \) is **greater** than the significance level \( \alpha = 0.002 \). Therefore, we **fail to reject the null hypothesis**.

### Conclusion

There is not sufficient evidence to conclude that the review section significantly harms test performance with a significance level of \( \alpha = 0.002 \).

---

Would you like further details or clarifications on any part of this? Here are 5 related questions that could expand on this topic:

1. How does changing the significance level affect the result of this test?
2. What are the assumptions behind using a paired t-test?
3. How can we compute the p-value using software like Excel or Python?
4. Can you explain what a Type I and Type II error would mean in this context?
5. What other statistical tests could be used for similar paired data?

**Tip**: Always ensure the assumptions of normality and pairing are met before applying a paired t-test.

pre-test: 51,59,92,87,78,44,92,68,76,76 post-test: 60,59,50,67,62,56,64,56,56,55 Can we say that the review section significantly harms test performance with a significance level of α = 0.002. Assume a Normal distribution. For this study, we should use The null and alternative hypotheses would be: The test statistic = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is Based on this, we should the null hypothesis. As such, the final conclusion is that ... There is not sufficient evidence to conclude that the review section significantly harms test performance with a significance level of = 0.002 There is sufficient evidence to conclude that the review section significantly harms test performance with a significance level of = 0.002

Learn how to perform a paired t-test on pre-test and post-test scores to determine if a review section harms test performance. This problem includes step-by-step calculations of the mean difference, standard deviation, t-statistic, and p-value, suitable for university-level statistics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation