Confidence intervals for the difference between two means: Paired samples Section 10.3

Objectives Construct confidence intervals with paired samples

Construct confidence intervals with paired samples Objective 1 Construct confidence intervals with paired samples

Paired Samples Suppose we select sixteen volunteers and they are given a test in which they had to push a button in response to the appearance of an image on a screen. Their reaction times are measured. Then the subjects consumed enough alcohol to raise their blood alcohol level to 0.05%. They then took the reaction test again. Now, we have gathered two samples of data, a sample of reaction times before alcohol consumption, and a sample after alcohol consumption. These are paired samples, because each value in one sample can be paired with the value from the same person in the other sample. The pairs of data are called matched pairs.

Notation We can compute the means of the two original samples as well as the mean of the sample of differences between each matched pair. The data for our experiment, along with the means, is presented in the table: 0.05% 0% Difference 1 102 103 -1 2 100 99 3 77 69 8 4 61 50 11 5 85 96 -11 6 26 24 7 95 71 115 109 9 64 53 10 98 89 107 12 44 27 17 13 47 -3 14 92 -8 15 70 66 16 94 86 Sample Mean 81.3 74.8 6.5 Means of the two original samples: 𝑥 1 and 𝑥 2 Mean of the sample differences between each matched pair: 𝑑

Relationship Between 𝑥 1 , 𝑥 2 , and 𝑑 The values of 𝑥 1 , 𝑥 2 , and 𝑑 are 𝑥 1 = 81.3 , 𝑥 2 = 74.8, and 𝑑 = 6.5 Simple arithmetic shows that the mean of the differences, 𝑑 , is the same as the difference between the sample means. In other words, 𝑑 = 𝑥 1 − 𝑥 2 . The same relationship holds for the populations. If we let 𝜇 1 and 𝜇 2 represent the population means and 𝜇 𝑑 represent the population mean of the difference, then 𝜇 𝑑 = 𝜇 1 − 𝜇 2 .

Confidence Interval Using Matched Pairs Since 𝜇 𝑑 = 𝜇 1 − 𝜇 2 , a confidence interval for the mean 𝜇 𝑑 is also a confidence interval for the difference 𝜇 1 − 𝜇 2 . The paired data reduce the two-sample problem to a one-sample problem. Suppose we want to construct a confidence interval for the population mean increase 𝜇 𝑑 of our experiment. The method for computing a confidence interval for 𝜇 𝑑 is the usual method for computing a confidence interval for a population mean.

Notation and Assumptions Notation: 𝑑 is the sample mean of the differences between the values in the matched pairs. 𝑠 𝑑 is the sample standard deviation of the differences between the values in the matched pairs. 𝜇 𝑑 is the population mean difference for the matched pairs. Assumptions: We have two paired random samples. Either the sample size is large (𝑛 > 30), or the differences between the matched pairs come from a population that is approximately normal.

Confidence Interval for the Mean Difference Let 𝑑 be the sample mean of the differences between matched pairs, and let 𝑠 𝑑 be the sample standard deviation. Let 𝜇 𝑑 be the population mean difference between matched pairs. A level 100(1 − 𝛼)% confidence interval for 𝜇 𝑑 is 𝑑 – 𝑡 𝛼 2 𝑠 𝑑 𝑛 < 𝜇 𝑑 < 𝑑 + 𝑡 𝛼 2 𝑠 𝑑 𝑛

Example Suppose we select sixteen volunteers and they are given a test in which they had to push a button in response to the appearance of an image on a screen. Their reaction times are measured. Then the subjects consumed enough alcohol to raise their blood alcohol level to 0.05%. They then took the reaction test again. Construct a 95% confidence for 𝜇 𝑑 , the mean difference in reaction times. 0.05% 0% Difference 1 102 103 -1 2 100 99 3 77 69 8 4 61 50 11 5 85 96 -11 6 26 24 7 95 71 115 109 9 64 53 10 98 89 107 12 44 27 17 13 47 -3 14 92 -8 15 70 66 16 94 86 Sample Mean 81.3 74.8 6.5

Solution First, we check the assumptions. Since the sample size is small (𝑛 = 16), we construct a boxplot for the differences to check for outliers or strong skewness. There are no outliers and no evidence of strong skewness, so we may proceed. We compute the sample mean difference 𝑑 , and the sample standard deviation of the differences 𝑠 𝑑 . The sample mean and standard deviation are 𝑑 = 6.500 𝑠 𝑑 = 9.93311 The sample size is 𝑛 = 16, so the degrees of freedom is 16 − 1 = 15. The confidence level is 95%. From Table A.3, we find the critical value to be 𝑡 𝛼 2 = 2.131

Solution The standard error is 𝑠 𝑑 𝑛 = 9.93311 16 =2.48328 and the margin of error is 𝑡 𝛼 2 𝑠 𝑑 𝑛 = 2.131(2.48328) = 5.292. The 95% confidence interval is: Point estimate ± margin of error 6.5 – 5.292 < 𝜇 𝑑 < 6.5 + 5.292 1.2 < 𝜇 𝑑 < 11.8 We are 95% confident that the mean difference is between 1.2 and 11.8. In particular, the confidence interval does not contain 0, and all the values in the confidence interval are positive. We can be fairly certain that the mean reaction time is greater when the blood alcohol level is 0.05%.

Matched Pairs and Margin of Error Matched pairs usually have a smaller margin of error than the margin of error for two independent samples. To see this, we will compute the sample standard deviations for the reaction time data in the previous example. Let 𝑠 1 denote the sample standard deviation for the blood level 0.05% sample and let 𝑠 2 denote the sample standard deviation for the blood level 0 sample. We have 𝑠 1 = 22.71 𝑠 2 = 27.39 If the samples had been independent, the standard error would have been 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 = 22.71 2 16 + 27.39 2 16 = 8.90 Because we were able to use the sample of differences, the standard error was only 2.48. This smaller value results in a smaller margin of error.

You Should Know… The assumptions for confidence interval using matched pairs How to construct confidence intervals with paired samples