HOWL13-6.txt Yesterday you worked Exercise 13.6 from Howell's Fundamentals text, an exercise that is especially interesting to me given that I am soon to have sinus surgery for the second time in the past year. Below is Minitab output for the correlated t solution. ---------------------------------------------------------------------- MTB > print c1 c2 ROW 12_hr 10_min 1 10.0 6.5 2 6.5 14.0 3 8.0 13.5 4 12.0 18.0 5 5.0 14.5 6 11.5 9.0 7 5.0 18.0 8 3.5 42.0 9 7.5 7.5 10 5.8 6.0 11 4.7 25.0 12 8.0 12.0 13 7.0 52.0 14 17.0 20.0 15 8.8 16.0 16 17.0 15.0 17 15.0 11.5 18 4.4 2.5 19 2.0 2.0 MTB > let c3 = c2-c1 (create the difference scores) MTB > name c1 '12_hr' c2 '10_min' c3 'diff' MTB > ttest c3 (do the correlated t-test) TEST OF MU = 0.00 VS MU N.E. 0.00 N MEAN STDEV SE MEAN T P VALUE diff 19 7.70 13.52 3.10 2.48 0.023 ------------------------------------------------------------------------ Today we pretended these same data were from an independent samples design, and we conducted a pooled variances independent samples t. After having gone over the variance sum law with you, I had expected this exercise to demonstrate to you that the correlated samples design increased power by lowering the standard error, raising the t, lowering the p. I was surprised when I saw that the t actually was higher with the independent samples design, and commented that the correlation between the two sets of scores must be negative. Here is Minitab output showing that that correlation is negative and showing the results of a pooled variances independent t on the same data. MTB > corr c1 c2 Correlation of C1 and C2 = -0.063 MTB > twos c2 c1; SUBC> pooled. TWOSAMPLE T FOR 12_hr VS 10_min N MEAN STDEV SE MEAN 12_hr 19 8.35 4.40 1.0 10_min 19 16.1 12.5 2.9 95 PCT CI FOR MU 10_min - 12_hr: (1.5, 13.9) TTEST MU 10_mim = MU 12_hr (VS NE): T= 2.53 P=0.016 DF= 36 POOLED STDEV = 9.38 ------------------------------------------------------------------------------ When I noticed the great heterogeneity of variance between the two samples, I introduced the separate variances t. Although we were able to decide to reject the null hypothesis even without knowing exactly what the separate variance df was, note below that Minitab gives us that df (note that it is less than with pooled variances) and an exact p value. MTB > twos c2 c1 TWOSAMPLE T FOR 10_min VS 12_hr N MEAN STDEV SE MEAN 12_hr 19 8.35 4.40 1.0 10_min 19 16.1 12.5 2.9 95 PCT CI FOR MU 10_min - MU 12_hr: (1.4, 14.0) TTEST MU 10_min = MU 12_hr (VS NE): T= 2.53 P=0.019 DF= 22 ------------------------------------------------------------------------- Exercise 13.13 in Howell's text asks you to modify the data to produce a positive correlation between the two columns of scores and note that this does reduce the standard error and make the correlated t more powerful. Of course, one can not do this during a real data analysis. Dr. Howell provided instructors using his text with his own solution, which I replicate below in Minitab. MTB > print c1 c2 ROW 12_hr 10_min 1 10.0 20.0 2 6.5 14.0 3 8.0 13.5 4 12.0 18.0 5 5.0 14.5 6 11.5 9.0 7 5.0 18.0 8 3.5 6.5 9 7.5 7.5 10 5.8 6.0 11 4.7 25.0 12 8.0 12.0 13 7.0 15.0 14 17.0 42.0 15 8.8 16.0 16 17.0 52.0 17 15.0 11.5 18 4.4 2.5 19 2.0 2.0 MTB > let c3 = c2-c1 MTB > ttest c3 TEST OF MU = 0.00 VS MU N.E. 0.00 N MEAN STDEV SE MEAN T P VALUE diff 19 7.70 9.95 2.28 3.37 0.0034