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
50
71
73
82
64
80
50
62
86
92
76
71
60
74
71
80
76
81
70
68
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)
Calculate the mean difference between posttest and pretest scores: d ˉ = 8.5 .
Calculate the standard deviation of the differences: s d ≈ 7.93 .
Find the critical t-value for a 90% confidence interval with 9 degrees of freedom: t 0.05 , 9 ≈ 1.833 .
Calculate the 90% confidence interval: ( 3.9 , 13.1 ) .
Explanation
Analyze the problem and data We are given pretest and posttest scores for 10 students and want to construct a 90% confidence interval to determine if the teaching method increases students' knowledge. We will calculate the difference between the posttest and pretest scores for each student, then find the mean and standard deviation of these differences. Using these statistics, we will calculate the margin of error and the confidence interval.
Calculate the differences First, calculate the difference between the posttest and pretest scores for each student:
d 1 = 71 − 50 = 21 d 2 = 82 − 73 = 9 d 3 = 80 − 64 = 16 d 4 = 62 − 50 = 12 d 5 = 92 − 86 = 6 d 6 = 71 − 76 = − 5 d 7 = 74 − 60 = 14 d 8 = 80 − 71 = 9 d 9 = 81 − 76 = 5 d 10 = 68 − 70 = − 2
Calculate the mean of the differences Next, calculate the sample mean of the differences: d ˉ = 10 21 + 9 + 16 + 12 + 6 − 5 + 14 + 9 + 5 − 2 = 10 85 = 8.5
Calculate the standard deviation of the differences Now, calculate the sample standard deviation of the differences:
First, calculate the squared differences from the mean ( d i − d ˉ ) 2 :
( 21 − 8.5 ) 2 = 12. 5 2 = 156.25 ( 9 − 8.5 ) 2 = 0. 5 2 = 0.25 ( 16 − 8.5 ) 2 = 7. 5 2 = 56.25 ( 12 − 8.5 ) 2 = 3. 5 2 = 12.25 ( 6 − 8.5 ) 2 = − 2. 5 2 = 6.25 ( − 5 − 8.5 ) 2 = − 13. 5 2 = 182.25 ( 14 − 8.5 ) 2 = 5. 5 2 = 30.25 ( 9 − 8.5 ) 2 = 0. 5 2 = 0.25 ( 5 − 8.5 ) 2 = − 3. 5 2 = 12.25 ( − 2 − 8.5 ) 2 = − 10. 5 2 = 110.25
Sum of squared differences = 156.25 + 0.25 + 56.25 + 12.25 + 6.25 + 182.25 + 30.25 + 0.25 + 12.25 + 110.25 = 566.5
s d = n − 1 ∑ i = 1 n ( d i − d ˉ ) 2 = 10 − 1 566.5 = 9 566.5 = 62.94 ≈ 7.93
Find the critical t-value Determine the critical t-value for a 90% confidence interval with n − 1 = 9 degrees of freedom. For a 90% confidence interval, α = 1 − 0.90 = 0.10 , so we need t α /2 , n − 1 = t 0.05 , 9 . From a t-table or calculator, t 0.05 , 9 ≈ 1.833 .
Calculate the margin of error Calculate the margin of error: E = t α /2 , n − 1 ⋅ n s d = 1.833 ⋅ 10 7.93 ≈ 1.833 ⋅ 3.162 7.93 ≈ 1.833 ⋅ 2.508 ≈ 4.60
Calculate the confidence interval Calculate the lower and upper bounds of the confidence interval:
C I l o w er = d ˉ − E = 8.5 − 4.60 = 3.9 C I u pp er = d ˉ + E = 8.5 + 4.60 = 13.1
Therefore, the 90% confidence interval for the true mean difference between the scores is ( 3.9 , 13.1 ) .
Conclusion The 90% confidence interval for the true mean difference between the posttest and pretest scores is approximately ( 3.9 , 13.1 ) . This suggests that, with 90% confidence, the true mean difference in scores lies between 3.9 and 13.1. Since both endpoints are positive, this indicates that the teaching method likely increases students' knowledge of the course material.
Examples
Confidence intervals are used in various fields to estimate population parameters. For example, a marketing team might use a confidence interval to estimate the average spending of customers on a new product. Similarly, political polls use confidence intervals to estimate the proportion of voters who support a particular candidate. In medical research, confidence intervals are used to estimate the effectiveness of a new drug or treatment. In each case, the confidence interval provides a range of plausible values for the parameter of interest, along with a level of confidence that the true value falls within that range. This helps in making informed decisions based on sample data.
