Paired Samples Lecture 39 Section 11.3 Tue, Nov 15, 2005

Independent Samples In a paired study, two observations are made on each subject, producing one sample of bivariate data. In a paired study, two observations are made on each subject, producing one sample of bivariate data. Or we could think of it as two samples of paired data. Or we could think of it as two samples of paired data. Often these are “before” and “after” observations. Often these are “before” and “after” observations. By comparing the “before” mean to the “after” mean, we can determine whether the intervening treatment had an effect. By comparing the “before” mean to the “after” mean, we can determine whether the intervening treatment had an effect.

Independent Samples On the other hand, with independent samples, there is no logical way to “pair” the data. On the other hand, with independent samples, there is no logical way to “pair” the data. One sample might be from a population of males and the other from a population of females. One sample might be from a population of males and the other from a population of females. Or one might be the treatment group and the other the control group. Or one might be the treatment group and the other the control group. The samples could be of different sizes. The samples could be of different sizes.

Paired Samples We have two samples of the same size n. We have two samples of the same size n. For each value x in the first sample, there is a corresponding value y in the second sample. For each value x in the first sample, there is a corresponding value y in the second sample. We are interested in the differences D = Y – X. We are interested in the differences D = Y – X. See Example 11.1, p. 679 – Weight Change. See Example 11.1, p. 679 – Weight Change. The two lists must be in the same order! The two lists must be in the same order!

Example Example 11.1 on the TI-83: Example 11.1 on the TI-83: Enter {148, 176, 153, …, 132} into list L 1 (before) Enter {148, 176, 153, …, 132} into list L 1 (before) Enter {154, 181, 151, …, 128} into list L 2 (after) Enter {154, 181, 151, …, 128} into list L 2 (after) Compute the differences: Compute the differences: L 2 – L 1  L 3 (after – before) L 2 – L 1  L 3 (after – before) L 3 = {6, 5, -2, …, -4}. L 3 = {6, 5, -2, …, -4}. Call this variable D. Call this variable D. Use 1-Var Stats to find the mean  d and standard deviation s D of D. Use 1-Var Stats to find the mean  d and standard deviation s D of D.  d = 1.75, s D =  d = 1.75, s D =

The Distribution of D Let n be the sample size. Let n be the sample size. If the samples are large, then D has (approx.) a normal distribution. If the samples are large, then D has (approx.) a normal distribution. However, if the samples are small, then we will need an additional assumption. However, if the samples are small, then we will need an additional assumption. The populations are normal. The populations are normal. In that case, D will have a normal distribution, too. In that case, D will have a normal distribution, too.

The Distribution of D To keep this section simple and to follow the book’s example, we will use only the t distribution. To keep this section simple and to follow the book’s example, we will use only the t distribution.

The Paired t-Test The hypotheses. The hypotheses. H 0 :  D = 0 H 0 :  D = 0 H 1 :  D  0 or  D 0 H 1 :  D  0 or  D 0 The test statistic. The test statistic. The other steps are the same. The other steps are the same.

Example Conduct the paired t-test for the data in Example Conduct the paired t-test for the data in Example H 0 :  D = 0 H 0 :  D = 0 H 1 :  D > 0 H 1 :  D > 0  = 0.05  = 0.05

Example Compute t: Compute t: The differences are 6, 5, -2, 4, 2, 2, 1, -4. The differences are 6, 5, -2, 4, 2, 2, 1, -4.  d =  d = s D = s D = t = 1.75/(3.412/  8) = t = 1.75/(3.412/  8) = Compute the p-value: Compute the p-value: p-value = P(t > 1.451) = (df = 7) p-value = P(t > 1.451) = (df = 7)

Example Since p-value > , do not reject H 0. Since p-value > , do not reject H 0. At the 5% level of significance, those who stop smoking do not gain weight. At the 5% level of significance, those who stop smoking do not gain weight.

Practice Work Example 11.2 on the TI-83 as a hypothesis-testing problem. Work Example 11.2 on the TI-83 as a hypothesis-testing problem. Enter the data into L 1 and L 2. Enter the data into L 1 and L 2. Store the differences in L 3. Store the differences in L 3. Use T-Test with the data in L 3. Use T-Test with the data in L 3.

Confidence Intervals A confidence for  D is A confidence for  D is Use the t table or the TI-83. Use the t table or the TI-83.

Example Find a 90% confidence interval for  D based on the data in Example Find a 90% confidence interval for  D based on the data in Example The 90% confidence interval is 1.75  The 90% confidence interval is 1.75  On the TI-83, use TInterval. On the TI-83, use TInterval.

Practice Find a 99% confidence interval for  D based on the data in Example Find a 99% confidence interval for  D based on the data in Example 11.2.