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.
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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
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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
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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
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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