Hypothesis Testing: Two Population Means and Two Population Proportions: Matched or Paired Samples 1.10 2008/06/17 16:32:48 GMT-5 2008/10/27 17:35:12.013 GMT-5 Susan Dean deansusan@deanza.edu Barbara Illowsky illowskybarbara@deanza.edu Susan Dean deansusan@deanza.edu Barbara Illowsky illowskybarbara@deanza.edu Connexions cnx@cnx.org elementary statistics This module provides an overview of Hypothesis Testing: Matched or Paired Samples as a part of Collaborative Statistics collection (col10522) by Barbara Illowsky and Susan Dean. Simple random sampling is used. Differences are calculated from the matched or paired samples. The matched pairs have differences that either come from a population that is normal or the number of differences is greater than 30 or both.In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, μ d , is then tested using a Student-t test for a single population mean with n-1 degrees of freedom where n is the number of differences. The test statistic (t-score) is: t = x d ¯ − μ d ( s d n ) Matched or paired samples A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in the table. The "before" value is matched to an "after" value. Subject: A B C D E F G H Before 6.6 6.5 9.0 10.3 11.3 8.1 6.3 11.6 After 6.8 2.5 7.4 8.5 8.1 6.1 3.4 2.0 Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level. Corresponding "before" and "after" values form matched pairs. After Data Before Data Difference 6.8 6.6 0.2 2.4 6.5 -4.1 7.4 9 -1.6 8.5 10.3 -1.8 8.1 11.3 -3.2 6.1 8.1 -2 3.4 6.3 -2.9 2 11.6 -9.6 The data for the test are the differences: {0.2, -4.1, -1.6, -1.8, -3.2, -2, -2.9, -9.6}The sample mean and sample standard deviation of the differences are: x d = -3.13 and s d = 2.91 Verify these values.Let μ d be the population mean for the differences. We use the subscript d to denote "differences."Random Variable: X d = the average difference of the sensory measurements H o : μ d ≥ 0 There is no improvement. ( μ d is the population mean of the differences.) H a : μ d < 0 There is improvement. The score should be lower after hypnotism so the difference ought to be negative to indicate improvement.Distribution for the test: The distribution is a student-t with df = n - 1 = 8 - 1 = 7 . Use t 7 . (Notice that the test is for a single population mean.)Calculate the p-value using the Student-t distribution: p-value = 0.0095Graph:

X d is the random variable for the differences.The sample mean and sample standard deviation of the differences are: x d = -3.13 s d = 2.91 Compare α and the p-value: α = 0.05 and p-value = 0.0095 . α > p-value.Make a decision: Since α> p-value, reject H o .This means that μ d < 0 and there is improvement.Conclusion: At a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after hypnotism. Hypnotism appears to be effective in reducing pain.For the TI-83+ and TI-84 calculators, you can either calculate the differences ahead of time (after - before) and put the differences into a list or you can put the after data into a first list and the before data into a second list. Then go to a third list and arrow up to the name. Enter 1st list name - 2nd list name. The calculator will do the subtraction and you will have the differences in the third list.TI-83+ and TI-84: Use your list of differences as the data. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 0 for μ0, the name of the list where you put the data, and 1 for Freq:. Arrow down to μ: and arrow over to μ0. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is 0.0094 and the test statistic is -3.04. Do these instructions again except arrow to Draw (instead of Calculate). Press ENTER. A college football coach was interested in whether the college's strength development class increased his players' maximum lift (in pounds) on the bench press exercise. He asked 4 of his players to participate in a study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows: Weight (in pounds) Player 1 Player 2 Player 3 Player 4 Amount of weighted lifted prior to the class 205 241 338 368 Amount of weight lifted after the class 295 252 330 360 The coach wants to know if the strength development class makes his players stronger, on average. Record the differences data. Calculate the differences by subtracting the amount of weight lifted prior to the class from the weight lifted after completing the class. The data for the differences are: {90, 11, -8, -8}Using the differences data, calculate the sample mean and the sample standard deviation. x d = 21.3 s d = 46.7 Using the difference data, this becomes a test of a single __________ (fill in the blank).Define the random variable: X d = average difference in the maximum lift per player.The distribution for the hypothesis test is t 3 . H o : μ d ≤ 0 H a : μ d > 0 Graph:

Calculate the p-value: The p-value is 0.2150Decision: If the level of significance is 5%, the decision is to not reject the null hypothesis because α < p-value .What is the conclusion? means; At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped to make the players stronger, on average. Seven eighth graders at Kennedy Middle School measured how far they could push the shot-put with their dominant (writing) hand and their weaker (non-writing) hand. They thought that they could push equal distances with either hand. The following data was collected. Distance (in feet) using Student 1 Student 2 Student 3 Student 4 Student 5 Student 6 Student 7 Dominant Hand 30 26 34 17 19 26 20 Weaker Hand 28 14 27 18 17 26 16 Conduct a hypothesis test to determine whether the differences in distances between the children's dominant versus weaker hands is significant. use a t-test on the difference data. The test statistic is 2.18 and the p-value is 0.0716.What is your conclusion?H0: μd equals 0; Ha: μd does not equal 0; Do not reject the null; At a 5% significance level, from the sample data, there is not sufficient evidence to conclude that the differences in distances between the children's dominant versus weaker hands is significant (there is not sufficient evidence to show that the children could push the shot-put further with their dominant hand). Alpha and the p-value are close so the test is not strong.