Lecture 47 Sections 11.1 – 11.3 Fri, Apr 16, 2004 Paired Samples Lecture 47 Sections 11.1 – 11.3 Fri, Apr 16, 2004

Paired vs. Independent Samples Paired samples – Each observation in one sample is related to a specific observation in the other sample. Independent samples – There is no direct relationship between the observations in one sample and the observations in the other sample.

Example of Paired Samples Typically, paired data is obtained from “before and after” studies. For each subject, the value before treatment and the value after treatment are observed. A company offers an SAT preparation course. Each students score is measured both before and after taking the course.

Example of Independent Samples Typically, these are two samples that are selected independently from two different populations. A company offers an SAT preparation course. It administers the SAT test to two groups. Group 1 has had the SAT prep course. Group 2 has not had the SAT prep course.

Let’s Do It! Let’s do it! 11.2, p. 625 – Paired Samples versus Independent Samples.

Paired Samples When the samples are paired, we are interested in the differences between the paired values. Let d be the difference between a randomly selected pair in a sample. Thend is the average difference for the sample. What is the sampling distribution ofd?

The Sampling Distribution ofd Let D be the mean of the population of differences. That is the same as the difference of the population means: D = A – B. Let D be the standard deviation of the population of differences. That is the standard deviation of the differences for all pairs.

The Sampling Distribution ofd The point estimate of D isd. If the sample size is at least 30, then d has a normal distribution with Mean D Standard deviation D/n. In that case, Z = (d – D)/(D/n).

The Sampling Distribution ofd For small sample sizes, it will be necessary that the population of differences have a normal distribution. In that case, we can use Z if D is known and t if D is not known. t = (d – D)/(sD/n).

Testing Hypotheses Concerning Paired Samples See Example 11.2, p. 630 – Comparing Test Scores. Is there sufficient evidence that the video improved the scores? The differences between the before and after scores is 6, 2, 2, 9, -4, 14, 2, 4, 9, 1. We computed = 4.5 and sD = 5.126.

Testing Hypotheses Concerning Paired Samples The hypotheses are H0: D = 0. H1: D > 0. The level of significance is 1%. The test statistic is t = (d – D)/(sD/n) = (4.5 – 0)/(5.126/10) = 2.776.

Testing Hypotheses Concerning Paired Samples The p-value is P(t9 > 2.776) = tcdf(2.776, 99, 9) = 0.01077. Or, using the t table, 0.01 < p-value < 0.025.

Testing Hypotheses Concerning Paired Samples We conclude that the evidence is not sufficient at the 1% level of significance to conclude that watching the video improved the scores. This example can be done on the TI-83 using the T-Test.

Confidence Intervals Concerning Paired Samples In the same example, use a 95% confidence interval to estimate the mean difference in scores. The formula is d  t  (sD/n). It evaluates to 4.5  3.667.

Let’s Do It! Let’s do it! 11.4, p. 636 – Pulse Rates. Everyone together take his pulse for 15 seconds. Stand up and then sit down and immediately take pulse for 15 seconds. Stand up and run in place for 30 seconds and immediately take pulse.

Let’s Do It! Pulse 1 2 3 4 5 6 7 8 9 10 Resting (x1) Standing (x2) Running (x3) d1 = x2 – x1 d2 = x3 – x1 d3 = x3 – x2 Pulse 11 12 13 14 15 16 17 18 19 20 Resting (x1) Standing (x2) Running (x3) d1 = x2 – x1 d2 = x3 – x1 d3 = x3 – x2

Assignment Page 626: Exercises 1, 2, 3, 6.