Question

To determine if his teaching method increases students’ learning, a professor administers a pretest to his class at the beginning of the semester and then a posttest at the end of the semester. The results from 10

randomly chosen students are given below. Let Population 1 be the Pretest scores and Population 2 be the Posttest scores.

Test Scores
Pretest Posttest
54
56
53
53
71
71
69
96
92
90
86
93
74
57
86
87
50
67
61
54

Step 1 of 2 : Construct a 90%
confidence interval for the true mean difference between the scores to determine if the teaching method increases students’ knowledge of the course material. Round the endpoints of the interval to one decimal place, if necessary.

Answer 1

The 90% confidence interval for the true mean difference is approximately
`\(4.57\) to \(8.23\)`.

To construct a 90% confidence interval for the true mean difference between the pretest and posttest scores, you can use the following formula:

`\[ \bar{d} \pm t \left((s_d)/(√(n))\right) \]`

1. **Calculate the sample mean difference
`(\( \bar{d} \))`:**

`\[ \bar{d} = \frac{\sum{(X_{\text{posttest}} - X_{\text{pretest}})}}{n} \]\\ \bar{d} = (56 - 54 + 53 - 53 + \ldots + 61 - 54)/(10) \]\\ \bar{d} = (64)/(10) \]\\ \bar{d} = 6.4 \]`

2. **Calculate the sample standard deviation of the differences (\( s_d \)):**

`s_d = \sqrt{\frac{\sum{(X_{\text{posttest}} - X_{\text{pretest}} - \bar{d})^2}}{n-1}} \]\\ s_d = \sqrt{((56 - 54 - 6.4)^2 + (53 - 53 - 6.4)^2 + \ldots + (61 - 54 - 6.4)^2)/(9)} \]\\ s_d = \sqrt{(141.6)/(9)} \]\\s_d = √(15.7333) \]\\s_d \approx 3.967 \]`

3. **Determine the t-value for a 90% confidence interval with 9 degrees of freedom (sample size - 1). You can use a t-table or a statistical software for this. Let's assume the t-value is approximately 1.833 (for a two-tailed test).**

4. **Plug the values into the formula:**

`\text{Margin of Error} = t \left((s_d)/(√(n))\right) \]\\ \text{Margin of Error} = 1.833 \left((3.967)/(√(10))\right) \]\\ \text{Margin of Error} \approx 1.83 \]`

5. **Construct the Confidence Interval:**

`\[ \text{Confidence Interval} = \bar{d} \pm \text{Margin of Error} \]\\ \text{Confidence Interval} = 6.4 \pm 1.83 \`

Therefore, the 90% confidence interval for the true mean difference is approximately
`\(4.57\) to \(8.23\)`. This means we are 90% confident that the true mean difference between pretest and posttest scores lies between these values, suggesting an increase in students' knowledge of the course material due to the teaching method