Hypothesis Tests: t-test for Two Sample Data PDF

Hypothesis Tests: t-test for Two Sample Data (Chapter 11) Kyung Sam Park Professor of LSOM Korea University Business School [email protected] Contents  Two sample hypothesis tests: t-tests  Mean difference tests (Today’s Group A mean vs. Group B...

Hypothesis Tests: t-test for Two Sample Data (Chapter 11) Kyung Sam Park Professor of LSOM Korea University Business School [email protected] Contents  Two sample hypothesis tests: t-tests  Mean difference tests (Today’s Group A mean vs. Group B mean) Paired (or dependent) data (쌍체자료) Independent two group data (독립된 두 그룹 자료)  Proportion difference tests 2 Examples of paired data  Paired sample data (or dependent sample data):  Samples characterized by a “before” and “after” study, where we want to measure the difference.  Samples characterized by matching or pairing observations. Program of weight reduction House prices offered by two real estate agents Name Before After Home A B Sam Park 68 66 1 135 128 David 120 110 2 110 105 Christina 55 50 3 131 119 Larry 70 70 4 142 140 Bill 150 120 5 105 98 Jihee 60 55 6 130 123 Jiwhan 70 75 7 131 127 Wang 100 80 8 110 115 9 125 122 10 149 145 mean mean Question: Is there a significant difference between the two means? 3 t-test for paired data (쌍체비교) d 0  Test statistics: T sd / n  d-bar = the mean of the differences between the paired observations.  sd = the standard deviation of the differences. sd   (d  d ) 2  n = the number of paired observations. n 1  d-bar = 8.38 Program of weight reduction  sd = 11.45 Name Before After Difference(d)  n=8 Sam Park 68 66 2  T = 2.07 David 120 110 10 Christina 55 50 5  P{T  2.07} = 0.039 Larry 70 70 0  p-value for one-tail test = 0.039 Bill 150 120 30 (Hypothesis: H0: 1  2 H1: 1 > 2) Jihee 60 55 5  p-value for two-tail test = 0.078 Jiwhan 70 75 -5 (Hypothesis: H0: 1 = 2 H1: 1  2) Wang 100 80 20 4 Paired t-test Name Before After Difference(d) Sam Park 68 66 2 n=8 David 120 110 10 Mean (d-bar) = 8.375 Christina 55 50 5 Standard deviation (sd) = 11.451 Larry 70 70 0 Bill 150 120 30 Standardize the normal distribution Jihee 60 55 5 T = (8.375 – 0) / [11.451/root(8)] Jiwhan 70 75 -5 = 2.068 Wang 100 80 20 Comparing 1(Before) & 2(After) Groups (H0: 1 = 2) or (H0: d-bar = 0) standardization Probability= 0.0387 0 8.375 = d-bar 0 = Xtoday T = 2.068 5 Using Excel: t-test for paired data  For the dataset of the weight reduction program  For the dataset of the house prices 6 t-test for independent data  Two (independent) group data: Woman Man  Test scores are divided into two groups, 71 87 78 Woman and Man. We want to know that 91 78 77 there is a significant mean difference 91 47 85 between the two groups. 77 74 63 91 78 81 X1  X 2  Test statistics: T 78 67 88 1 1 82 84 96 s 2p     n1 n2  90 92 91 93 81 82  X1-bar and X2-bar are the two sample 81 63 58 means. 95 58 79  s12 and s22 are the two sample variances. 87 96  n1 and n2 are the two sample sizes. 92 89  sp2 is the pooled variance of two groups. 92 40 90 94 (n1  1) s12  (n2  1) s 22 89 87 s  2 n1  n2  2 96 80 p 7 Concept of the t-test for independent data  Subtract the two normal distribution functions, and then standardize the resulting normal distribution, which results in the t-distribution. Group 1’s mean forms Group 2’s mean forms H0: 1 = 2 a normal distribution: a normal distribution: X-bar X-bar T X1-bar X2-bar 0 X1  X 2 where T  1 1 s 2p     n1 n2  8 Note for the mean difference tests  Example 1:  50, 52, 53, 55, 100 (Mean = 62)  10, 50, 52, 53, 55 (Mean = 44) (by t-test, p-value = about 20%)  Example 2:  51, 52, 53, 54, 55 (Mean = 53)  56, 57, 58, 59, 60 (Mean = 58) (by t-test, p-value = about 0.1%) Group 1 Group 2 Group 1 Group 2 X-bar X-bar 44 62 53 58 9 Proportion difference test p1  p2  Test statistics: Z 1 1 pc (1  pc )    n1 n2   p1 = X1/n1 : The success rate in Group 1, where n1 = no. of data, X1 = no. of successes in Group 1.  p2 = X2/n2 : The success rate in Group 2.  pc = (X1 + X2)/(n1 + n2): The pooled proportion of the two groups  Example  In the women group, 19 persons out of 100 support candidate A.  In the men group, 62 out of 200 support the same candidate.  Compare the two groups’ supporting rate; Is there a big gap?  H0: 1 = 2 H1: 1  2  p1 = 19/100 = 0.19, p2 = 62/200 = 0.31, pc = 0.27. Therefore, Z = 2.21.  The p-value = 2  P{Z < 2.21} = 0.0272  The men group’s supporting rate is significantly higher than the women group’ supporting rate. 10