17.43 17.14 28.291. Suppose an experimenter is planning an experiment with k= 5 different treatments. How many possible pairwise comparisons need to be tested if the omnibus F test is found significant. What could be the type I error rate if alpha = 0.05 is not adjusted for those comparisons?
2. Find the critical values of F for the following situations:
a. F(4, 30) at alpha =0.05
b. F (2, 120) at alpha 0.01
c. k = 7, n - 10, alpha =0.05
d. k = 3, n-1= 13, n-2 = 18, n-3 = 30, alpha =0.01
3. The effect of three different methods of reinforcement on correct identification of nonsense syllables was studied. Alpha for this problem is .05. The results are showed below:
Method
A B C
----------------------------------------------
n 7 7 7
17.43 17.14 28.29
Summary Table for ANOVA
Source SS df MS F Fcv ----------------------------------------------------------------- Between treatment 19.71 Within treatment 258.00 Total
a) Complete the summary Table for ANOVA and interpret the result of the F test
b) Use the Tukey method to detect the differences among the three groups
c) If the Fisher Test is used, do your conclusions in b) change?
4. Suppose we have an experiment with independent groups of n= 7 subjects randomly assigned to each of 8 treatment conditions. The MSwithin = 58.65. The treatment sums are given below:
a1 a2 a3 a4 a5 a6 a7 a8 316 333 307 373 398 226 123 436
a) Conduct the Tukey test on all pairwise differences, using alpha =0.05
b) Evaluate the same pairwise differences with the Fisher test, alpha =0.05.
5. Assume that we have a control group and seven experimental groups, with n= 16 subjects for each group MSwithin = 28.75. The total for each group are given below:
C E1 E2 E3 E4 E5 E6 E7 289 270 241 279 191 213 205 198
a) Is the overall significant?
b) Use Scheffe procedure to determine which of the treatment means is significantly different from the mean of the control group. (alpha = 0.05)
Answers to Lab 10
1. 10 pairwise comparisons needed. Type I error rate = 0.40
2. a. 2.69
b. the value F(2, 120)= 4.788
c. F ( 6, 63)= 2.244
d. F (2, 58) = 4.992
3. a)
Summary Table for ANOVA
Source SS df MS F Fcv -------------------------------------------------------------- Between treatment 564.88 2 282.44 19.71 3.55 Within treatment 258.00 18 14.33 Total 822.89 20
Decision -------> reject Ho
b)
Group A Group B
Group C (
=28.29)
10.86** 11.15**
Group A (
=17.43
-- 0.29
Group B (
=17.14) --
HSD = (3.61)(1.431) = 5.17 (alpha level =0.05)
HSD = (4.70)(1.431) = 6.73 (alpha level =0.01
** Significant at alpha level 0.01
3. c)
Group A Group B
Group C (=28.29)
10.86** 11.15**
Group A (=17.43
-- 0.29
Group B (=17.14)
LSD = (2.10)(2.02) = 4.25, (alpha level=0.05)
LSD = (2.88)(2.02) = 5.82, (alpha level=0.01)
** Significant at a level 0.01
4. a) Tukey
GroupA5 A4 A2 A1 A3 A6 A7 A8 62.29 5.43 9.00 14.72 17.15 18.43 29.86 44.72 A5 58.86 3.57 9.29 11.72 13.00 24.43 39.29 A4 53.29 5.72 8.15 9.43 20.86 35.72 A2 47.57 2.43 3.71 15.14 30.00 A1 45.14 1.28 12.71 27.57 A3 43.86 11.43 26.29 A6 32.43 14.86 HSD = (4.488 )(2.895) = 12.99 (q = 4.488 using dfwithin = 48) Any difference equal to 12.99 or greater is statistically significant b) Fisher Group
5. a)
Source SS df MS F ----------------------------------------------------------- Between treatment 674.84 7 96.41 3.35** Within treatment 3450.00 120 28.75 Total 4124.84 127 **significant at alpha level 0.01; (Fcv (7,120) = 2.79) -----> Decision: reject Ho
b)
Groups Mean Difference Between Experiment Group & Control Group E1 16.88 1.18 E2 15.06 3.00 E3 17.44 0.62 E4 11.94 6.12 E5 13.31 4.75 E6 12.81 5.25 E7 12.38 5.68 Control 18.06 S = (3.816)(1.895) = 7.24, None of the treatment means differ significantly from the mean of the control group.