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 |
| 89 | 69 |
| 92 | 72 |
| 67 | 61 |
| 98 | 91 |
| 88 | 85 |
| 98 | 84 |
| 87 | 82 |
| 67 | 76 |
| 94 | 91 |
| 71 | 62 |
Step 1 of 2: Construct a $95 \%$ 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 (2)
Calculate the difference between pretest and posttest scores for each student.
Calculate the mean and standard deviation of these differences: d ˉ = 7.8 and s d ≈ 8.6769 .
Find the critical t-value for a 95% confidence interval with 9 degrees of freedom: t ≈ 2.262 .
Construct the confidence interval: ( 1.6 , 14.0 ) .
( 1.6 , 14.0 )
Explanation
Understand the problem and provided data We are given pretest and posttest scores for 10 students and want to determine if the teaching method increases students' knowledge by constructing a 95% confidence interval for the true mean difference between the scores.
Calculate the differences First, we calculate the difference between each student's pretest and posttest scores. These differences are: 89 − 69 = 20 , 92 − 72 = 20 , 67 − 61 = 6 , 98 − 91 = 7 , 88 − 85 = 3 , 98 − 84 = 14 , 87 − 82 = 5 , 67 − 76 = − 9 , 94 − 91 = 3 , 71 − 62 = 9 .
Calculate the mean of the differences Next, we calculate the mean of these differences: d ˉ = 10 20 + 20 + 6 + 7 + 3 + 14 + 5 − 9 + 3 + 9 = 10 78 = 7.8
Calculate the sample standard deviation of the differences Now, we calculate the sample standard deviation of the differences:
First, we find the squared differences from the mean: ( 20 − 7.8 ) 2 = 148.84 ( 20 − 7.8 ) 2 = 148.84 ( 6 − 7.8 ) 2 = 3.24 ( 7 − 7.8 ) 2 = 0.64 ( 3 − 7.8 ) 2 = 23.04 ( 14 − 7.8 ) 2 = 38.44 ( 5 − 7.8 ) 2 = 7.84 ( − 9 − 7.8 ) 2 = 282.24 ( 3 − 7.8 ) 2 = 23.04 ( 9 − 7.8 ) 2 = 1.44
Sum of squared differences = 148.84 + 148.84 + 3.24 + 0.64 + 23.04 + 38.44 + 7.84 + 282.24 + 23.04 + 1.44 = 677.6
Then, we divide by n − 1 = 9 and take the square root: s d = 9 677.6 = 75.2889 ≈ 8.6769
Find the critical t-value We need to find the critical t-value for a 95% confidence interval with n − 1 = 9 degrees of freedom. Using a t-table or a calculator, we find that t α /2 , n − 1 = t 0.025 , 9 ≈ 2.262 .
Calculate the margin of error Now, we calculate the margin of error: E = t α /2 , n − 1 ⋅ n s d = 2.262 ⋅ 10 8.6769 ≈ 2.262 ⋅ 3.1623 8.6769 ≈ 2.262 ⋅ 2.7439 ≈ 6.207
Construct the confidence interval Finally, we construct the 95% confidence interval: ( d ˉ − E , d ˉ + E ) = ( 7.8 − 6.207 , 7.8 + 6.207 ) = ( 1.593 , 14.007 ) Rounding to one decimal place, the 95% confidence interval is ( 1.6 , 14.0 ) .
Final Answer The 95% confidence interval for the true mean difference between the pretest and posttest scores is ( 1.6 , 14.0 ) .
Conclusion Since the confidence interval ( 1.6 , 14.0 ) lies entirely above zero, we can conclude that the teaching method likely increases students' knowledge of the course material.
Examples
Understanding confidence intervals is crucial in many real-world scenarios. For instance, a marketing team might use a confidence interval to estimate the average increase in sales after launching a new advertising campaign. If the confidence interval does not include zero, it suggests that the campaign had a statistically significant impact on sales. Similarly, in medical research, confidence intervals are used to estimate the effectiveness of a new drug or treatment. If the confidence interval for the difference in outcomes between the treatment and control groups does not include zero, it provides evidence that the treatment has a significant effect.
The 95% confidence interval for the mean difference in scores is (1.6, 14.0). Since this interval does not include zero, it indicates that the teaching method likely increases students' understanding of the course material. Thus, we conclude that the method has a positive effect on student learning.
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