A state legislator wants to determine whether his voters' performance rating $(0-100)$ has changed from last year to this year. The following table shows the legislator's performance from the same ten randomly selected voters for last year and this year. Use this data to find the $95 \%$ confidence interval for the true difference between the population means. Assume that the populations of voters' performance ratings are normally distributed for both this year and last year.
| Rating (last year) | 74 | 69 | 70 | 89 | 44 | 58 | 45 | 73 | 82 | 83 |
| :----------------- | :- | :- | :- | :- | :- | :- | :- | :- | :- | :- |
| Rating (this year) | 86 | 57 | 72 | 70 | 63 | 83 | 71 | 92 | 52 | 55 |
Calculate the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.
Answer (2)
To find the margin of error for the 95% confidence interval of the true difference in voters' performance ratings, we calculated the differences, their mean, the sample standard deviation, and the t-critical value. Finally, we applied these values to find the margin of error to be approximately 15.785500. This value represents the uncertainty in estimating the true average performance rating change from last year to this year.
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Calculate the differences between this year's and last year's ratings.
Find the mean of the differences: d ˉ = 1.4 .
Compute the standard deviation of the differences: s d ≈ 22.067069 .
Determine the margin of error: E ≈ 15.7855 .
15.785500
Explanation
Understand the problem and provided data We are given paired data of voter performance ratings from last year and this year for 10 voters. We want to find the margin of error for a 95% confidence interval for the true difference between the population means. We assume that the populations of voters' performance ratings are normally distributed for both this year and last year.
Calculate the differences First, we need to calculate the difference between the ratings for each voter (this year - last year). The differences are: d 1 = 86 − 74 = 12 d 2 = 57 − 69 = − 12 d 3 = 72 − 70 = 2 d 4 = 70 − 89 = − 19 d 5 = 63 − 44 = 19 d 6 = 83 − 58 = 25 d 7 = 71 − 45 = 26 d 8 = 92 − 73 = 19 d 9 = 52 − 82 = − 30 d 10 = 55 − 83 = − 28
Calculate the mean of the differences Next, we calculate the sample mean of the differences: d ˉ = 10 ∑ i = 1 10 d i = 10 12 − 12 + 2 − 19 + 19 + 25 + 26 + 19 − 30 − 28 = 10 14 = 1.4
Calculate the standard deviation of the differences Now, we calculate the sample standard deviation of the differences: s d = 10 − 1 ∑ i = 1 10 ( d i − d ˉ ) 2 First, we calculate the squared differences: ( 12 − 1.4 ) 2 = 111.36 ( − 12 − 1.4 ) 2 = 179.56 ( 2 − 1.4 ) 2 = 0.36 ( − 19 − 1.4 ) 2 = 416.16 ( 19 − 1.4 ) 2 = 309.76 ( 25 − 1.4 ) 2 = 556.96 ( 26 − 1.4 ) 2 = 608.36 ( 19 − 1.4 ) 2 = 309.76 ( − 30 − 1.4 ) 2 = 985.96 ( − 28 − 1.4 ) 2 = 864.36
Sum of squared differences = 111.36 + 179.56 + 0.36 + 416.16 + 309.76 + 556.96 + 608.36 + 309.76 + 985.96 + 864.36 = 4382.6
s d = 9 4382.6 = 486.955556 ≈ 22.067069
Find the t-critical value We are given a 95% confidence interval, so α = 1 − 0.95 = 0.05 , and α /2 = 0.025 . We have n − 1 = 10 − 1 = 9 degrees of freedom. Looking up the t-critical value t 0.025 with 9 degrees of freedom in a t-table, we find t 0.025 = 2.262 .
Calculate the margin of error Now we calculate the margin of error: E = t α /2 ⋅ n s d = 2.262 ⋅ 10 22.067069 = 2.262 ⋅ 3.16227766 22.067069 ≈ 2.262 ⋅ 6.978585 ≈ 15.7855
Round the margin of error Rounding the margin of error to six decimal places, we get 15.785500 .
Examples
Understanding confidence intervals is very useful in political science. For example, a legislator wants to know how the public views their performance. By sampling a group of voters, the legislator can estimate the true average performance rating with a certain level of confidence. The margin of error tells the legislator how much uncertainty there is in the estimate. A smaller margin of error means the legislator can be more confident that the true average performance rating is close to the estimated value. This helps in making informed decisions about policies and strategies.
