1. Lecture 9 TWO GROUP MEANS TESTS EPSY 640 Texas A&M University

2. Two independent groups experiments Randomization distributions. 6 scores (persons, things) can be randomly split into 2 groups 20 ways: 1 2 3  4 5 6 1 2 4  3 5 6 1 2 5  3 4 6 1 2 6  3 4 5 1 3 4  2 5 6 1 3 5  2 4 6 1 3 6  2 4 5 1 4 5  2 3 6 1 4 6  2 3 5 1 5 6  2 3 4 2 3 4  1 5 6 2 3 5  1 4 6 2 3 6  1 4 5 2 4 5  1 3 6 2 4 6  13 5 2 5 6  1 3 4 3 4 5  1 2 6 3 4 6  1 2 6 3 5 6  1 2 4 4 5 6  1 2 3

3. Two independent groups experiments Differences between groups can be arranged as follows: -3 -1 1 3 -5 -3 -1 1 3 5 -9 -7 -5 -3 -1 1 3 5 7 9 look familiar?

4. t-distribution Gossett discovered it similar to normal, flatter tails different for each sample size, based on N-2 for two groups (degrees of freedom) randomization distribution of differences is approximated by t-distribution

5. t-distribution assumptions NORMALITY –(W test in SPSS) HOMOGENEITY OF VARIANCES IN BOTH GROUPS’ POPULATIONS –Levene’s test in SPSS INDEPENDENCE OF ERRORS –logical evaluation –Durbin-Watson test in serial data

6. Null hypothesis for test of means for two independent groups H 0 :  1 -  2 =0 H 1 :  1 -  2  0. fix a significance level, . Then we select a sample statistic. In this case we choose the sample mean for each group, and the test statistic is the sample difference d = y 1 – y 2.

7. Variance and Standard deviation of differences in the Population The variance in the POPULATION of a difference of two independent scores is:  2 d =  2 ( y 1 – y 2 ) =  2 y 1 +  2 y 2  d =  2 ( y 1 – y 2 ) = standard error of difference Example,  2 1 = 100,  2 2 = 100,  2 (y 1 -y 2 ) = 100 + 100 = 200  (y 1 -y 2 ) = 14.14

8. Variance and Standard deviation of difference in means The variance of the difference in POPULATION MEANS is the variance of score difference divided by the sample sizes:  2 d =  2 ( y 1 – y 2 ) = (  2 y 1 /n 1 +  2 y 2 /n 2 )  d =   2 ( y 1 – y 2 ) Example,  2 1 = 100,  2 2 = 100, n 1 = 16, n 2 =16  2 (y 1 -y 2 ) = 100 + 100 = 200 s (y 1 -y 2 ) = 14.14  2 d = 100/16 + 100/16 =12.5  d = 3.54, standard deviation of mean difference

9. MEANING OF VARIANCE OF POPULATION MEAN DIFFERENCE WE ASSUMED EQUAL VARIANCES FOR THE TWO POPULATIONS THUS, VARIANCE OF DIFFERENCE IS EQUAL TO SINGLE VARIANCE (AVERAGE OF THE TWO VARIANCES) TIMES SUM OF 1/SAMPLE SIZE:  2 d = (  2 y 1 /n 1 +  2 y 2 /n 2 ) =  2 (1/n 1 + 1/n 2 )

10. Mean difference = 0 Null Hypothesis t-distribution, df=16+16- 2 = 30 Critical t(30) = 2.042 SD=3.54 2.042 * 3.54 = 7.22 points needed for significance from difference=0

11. Standard error of mean difference score for unequal sample size standard error of the sample difference. It consists of the square root of the average variance of the two samples, [(n 1 –1)s 2 1 + (n 2 – 1)s 2 2 ] / (n 1 + n 2 –2) multiplied by the sum of 1/sample size ( 1/n 1 + 1/n 2 ). Same as previous slide, only difference is adjusting for difference sample sizes in the two groups

12. Null hypothesis for test of means for two independent groups t = d / s d _____________________________________________ = (y 1 – y 2 )/  { {[(n 1 –1)s 2 1 + (n 2 – 1)s 2 2 ] / (n 1 + n 2 –2)} { 1/n 1 + 1/n 2 } Weighted average variance of two groups Sampling weights

13. Boy-Girl differences on Sense of Inadequacy on BASC for a nonrandom sample

14. REGRESSION APPROACH ANOVA Model Sum of Squares dfMean SquareFSig. Regression185.97 1185.97 1.802.18 Residual 40454.07 392103.20 Total 40640.04 393 a Predictors: (Constant), SEX b Dependent Variable: SENSE OF INADEQUACY Model Summary Model RR Square Adjusted R Square Std. Error of the Estimate 1.068.005.00210.16 Predictors: (Constant), SEX

15. REGRESSION COEFFICIENTS FOR SEX PREDICTING SENSE OF INADEQUACY

16. VENN DIAGRAM OF REGRESSION Ssresidual = 40454.07 Sense of Inadequacy sex SSregression = 185.97 R 2 =.005 = 185.97 40640.04

17. Path Diagram for Group Mean Difference SEX SENSE OF INADEQUACY -.068 ERROR  1-.005 =.997

18. Correlation representation of the two independent groups experiment r 2 pb t 2 =   (1 – r 2 pb )/ (n-2) t 2 r 2 pb =  t 2 + n - 2

19. Correlation representation of the two independent groups experiment t 2 r pb =  t 2 + n - 2 1/2

20. Test of point biserial=0 H 0 :  pb = 0 H 1 :  pb  0 is equivalent to t-test for difference for two means.

21. POINT-BISERIAL CORRELATION M F Y X XXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXX m m

22. Example Willson (1997) studied two groups of college freshman engineering students, one group having participated in an experimental curriculum while the other was a random sample of the standard curriculum. One outcome of interest was performance on the Mechanics Baseline Test, a physics measure (Hestenes & Swackhammer, 1992). The data for the two groups is shown below. A significance level of.01 was selected for the hypothesis that the experimental group performed better than the standard curriculum group (a directional test): GroupMeanSDSample size Exper471575 Std Cur371650 __________________________________________ t = (47 – 37) /  [(74 y 15 2 ) + (49 y 16 2 ) / (75 + 50 – 2)][1/75 + 1/50] _______________________________ = (10) /  [(16650 + 12554) / (123)][1/75 + 1/50] = 1.947 The t-statistic is compared with the tabled value for a t-statistic with 123 degrees of freedom at the.01 significance level, 2.358. The observed probability of occurrence is 1 - 0.97309 =.02691, greater than the intended level of significance. The conclusion was that the experimental curriculum group, while performing better than the standard, did not significantly outperform them.

23. Confidence interval around d d  t   {{[(n 1 –1)s 2 1 + (n 2 – 1)s 2 2 ] / (n 1 + n 2 –2)} { 1/n 1 + 1/n 2 } Thus, for the example, using the.01 significance level the confidence interval is 1.393  2.588 (1.037) = 1.393  2.684 = (-1.281, 4.077) This includes 0 (zero) so we do not reject the null hypothesis.

24. Wilcoxon rank sum test for two independent groups. While the t-distribution is the randomization distribution of standardized differences of sample means for large sample sizes, for small samples it is not the best procedure for all unknown distributions. If we do not know that the population is normally distributed, a better alternative is the Wilcoxon rank sum test.

25. Wilcoxon rank sum test for two independent groups.

26. Dependent groups experiments d = y 1 – y 2 for each pair. Now the hypotheses about the new scores becomes H 0 :  = 0 H 1 :   0 The sample statistic is simply the sample difference. The standard error of the difference can be computed from the standard deviation of the difference scores divided by n, the number of pairs

27. Standard deviation of differences in related (dependent) data s 2 (y 1 -y 2 ) = s 2 1 + s 2 2 -2r 12 s 1 s 2 Example, s 2 1 = 100, s 2 2 = 144, r 12 =.7 s 2 (y 1 -y 2 ) = 100 + 144 -2(.7)(10)(12) = 244 - 168 = 76 s (y 1 -y 2 ) = 8.72

28. Dependent groups experiments _________________ s d =  [s 2 1 + s 2 2 –2r 12 s 1 s 2 ]/n. Then the t-statistic is _ t = d / s d

29. Dependent groups experiments In a study of the change in grade point average for a group of college engineering freshmen, Willson (1997) recorded the following data over two semesters for a physics course: Variable N Mean Std Dev PHYS1 128 2.233333 1.191684 PHYS2 128 2.648438 1.200983 Correlation Analysis: r 12 =.5517 To test the hypothesis that the grade average changed after the second semester from the first, for a significance level of.01, the dependent samples t-statistic is ________________________________________ t = [2.648 – 2.233]/ [ 1.192 2 + 1.201 2 – 2 (.5517) x 1.192 x 1.201]/128 =.415 /.1001 = 4.145 This is greater than the tabled t-value  t(128-1) = 2.616. Therefore, it was concluded the students averaged higher the second semester than the first.

30. VENN DIAGRAM SS Between pairs SS Within pairs SS Treatment SS Treatment*Pair Design for pairs of persons, each assigned to experimental or control condition

31. VENN DIAGRAM SS Between Persons SS Within Persons SS Time SS Time*Person Time 1 vs. Time 2 comparison of achievement for a group of persons

32. Nonparametric test of difference in dependent samples. sign test. A count of the positive (or negative) difference scores is compared with a binomial sign table. This sign test is identical to deciding if a coin is fair by flipping it n times and counting the number of heads. Within a standard error of.5n 1/2 the number should be equal to n/2.As n becomes large, the distribution of the number of positive difference scores divided by the standard error is normal. An alternative to the sign test is the Wilcoxon signed rank test or symmetry test

33. DIFFERENCE BETWEEN SENSE OF INADEQUACY AND ANXIETY IN BASC SAMPLE- NONPARAMETRIC

35. Summary of two group experimental tests of hypothesis Table below is a compilation of last two chapters: –sample size –one or two groups –normal distribution or not –known or unknown population variance(s)

36. One or IndependentNormalHypothesesPopulation varianceTest StatisticDistribution Two orDistributionknown? GroupsDependentAssumed? _ Onenot applicable YesH0:  = a  2 Known y. - anormal H1:   az =  [  2 /n ]1/2 Onenot applicable YesH0:  = a  2 unknown y. - at with n-1 df H1:   at =  [ s2 /n ]1/2 Onenot applicable NoH0:  = a  2 unknownS =  R+i, yi > aWilcoxon rank sum H1:   a orn+ =  i+, i+ =1 if yi > a, 0 elsebinomial (sign test)

37. One or IndependentNormalHypotheses Population varianceTest StatisticDistribution Two orDistribution known? GroupsDependentAssumed? _ _ TwoIndependentYesH0:  0 -  1 = 0  2 0 =  2 1 =  2,y0. – y1. H1:  0 -  1  0 known z =  normal [  2 (1/n0 + 1/n1) ] 1/2 _ _ TwoIndependentYesH0:  0 -  1 = 0  2 0 =  2 1, y0. – y1. H1:  0 -  1  0 unknown t =  t with n0 + n1 –2 df [ s2 (1/n0 + 1/n1) ] 1/2 s 2 = (n0 –1)s 2 0 + (n1 –1)s 2 1 n0 + n1 –2 TwoIndependentNoH0:  0 -  1 = 0  2 0 =  2 1, S =  R+i Wilcoxon rank sum H1:  0 -  1  0 unknown for one of the groups _ _ TwoDependentYesH0:  0 -  1 = 0  2 0 =  2 1=  2, y0. – y1. H1:  0 -  1  0 Known z =  normal [ 2  2 ( 1 -  ) /n ] 1/2  = population correlation between y0 and y1 __ __ TwoDependentYesH0:  0 -  1 = 0  20 =  21=  2, y0. – y1.t with n-1 df H1:  0 -  1  0 inknown t =  [ 2 s 2 ( 1 -  ) /n ] 1/2 r = sample correlation between y0 and y1 s 2 = s 2 0 + s 2 1 – 2r 12 s 0 s 1 TwoDependentNoH0:  0 -  1 = 0  2 0 =  2 1=  2 S =  R+i Wilcoxon Ranks sum H1:  0 -  1  0 unknownfor positive differences