Unformatted text preview: STAT 350 – Spring 2009 Homework #8 – SOLUTION
covers through Lecture L Note: to obtain a pvalue for an Fstatistic for this assignment, you should use Excel.
For example, if your Fstatistic is 12.39 and df1 = 3 and df2 = 5.
In Excel, type: =Fdist(12.39,3,5)
Excel will return: 0.009434
Your pvalue is 0.009434
Text Exercises
Chapter 9, Problems #6(af), 10, 24(b)
9.6 a. 3.69
d. 10.29 b. 4.82
e. 3.10 c. 6.63
f. 0.95 9.10
Source
Brand
Error
Total df
4
45
49 SS
39.2192
28.8
68.0192 MS
9.8048
0.64 F
15.32 9.24
b. The variances for the three groups are: 5.708, 2.799, 7.990. The ratio of the largest to
smallest is 7.990/2.799 = 2.855. This ratio is less than 4, so the assumption of equal
variances is reasonable. Homework #8  Solution Page 1 of 1 Then answer the following problems:
1. For this problem, use the data summarized below.
Group ni
xi
5
59.2
A
5
64.9
B
5
65.8
C
5
69.5
D
5
70.8
E si
5.2
5.1
6.1
5.9
5.5 a. Give the ANOVA table for this data (including the pvalue). Be sure to show your work.
5
⎛n ⎞
= ∑ ⎜ i ⎟ xi
i =1 ⎝ n ⎠
⎛5⎞
= ⎜ ⎟ 59.2
⎝ 25 ⎠
= 66.04 x ⎛5⎞
⎛5⎞
⎛5⎞
⎛5⎞
+ ⎜ ⎟ 64.9 + ⎜ ⎟ 65.8 + ⎜ ⎟ 69.5 + ⎜ ⎟ 70.8
⎝ 25 ⎠
⎝ 25 ⎠
⎝ 25 ⎠
⎝ 25 ⎠ 5 SSTr = ∑ ni ( xi − x ) 2 i =1 = 5 ( 59.2 − 66.04 ) + 5 ( 64.9 − 66.04 ) + 5 ( 65.8 − 66.04 )
2 2 +5 ( 69.5 − 66.04 ) + 5 ( 70.8 − 66.04 )
2 2 2 = 413.86
5 SSE = ∑ ( ni − 1) si2
i =1 = ( 5 − 1) 5.22 + ( 5 − 1) 5.12 + ( 5 − 1) 6.12 + ( 5 − 1) 5.92 + ( 5 − 1) 5.52
= 621.28
ANOVA Table
Source Trt error total df 4 20 24 SS MS F p 413.86 103.465 3.330704 0.030317147
621.28 31.064 1035.14 b. What is the critical value of the test statistic?
The numerator degrees of freedom are 4
The denominator degrees of freedom are 20
For α = 0.05, the critical value is 2.87 (see Appendix Table VIII, page 575) Homework #8  Solution Page 2 of 2 c. Conduct Tukey’s Multiple Comparison Procedure and present your results using the line
method as given in class.
qα = 4.23 (see Table IXa on p. 577; there are 5 groups and the error df = 20) T = qα MSE
31.064
= 4.23
= 10.5435
ni
5 The following table gives the absolute value of the differences in the sample means for
each pair of groups. I have highlighted those cells that exceed 10.5435 (T)
GROUP A B C D E Mean 59.2 64.9 65.8 69.5 70.8 A 59.2 0 5.7 6.6 10.3 11.6 B 64.9 5.7 0 0.9 4.6 5.9 C 65.8 6.6 0.9 0 3.7 5 D 69.5 10.3 4.6 3.7 0 1.3 E 70.8 11.6 5.9 5 1.3 0 Answer:
A 59.2 Homework #8  Solution B 64.9 C 65.8 D 69.5 E 70.8 Page 3 of 3 2. For this problem, use the data summarized below.
Group ni
xi
4
59.2
A
6
64.9
B
5
65.8
C
7
69.5
D
7
70.8
E si
5.2
5.1
6.1
5.9
5.5 a. Give the ANOVA table for this data (including the pvalue). Be sure to show your work.
5
⎛n ⎞
= ∑ ⎜ i ⎟ xi
x
i =1 ⎝ n ⎠
⎛4⎞
⎛6⎞
⎛5⎞
⎛7⎞
⎛7⎞
= ⎜ ⎟ 59.2 + ⎜ ⎟ 64.9 + ⎜ ⎟ 65.8 + ⎜ ⎟ 69.5 + ⎜ ⎟ 70.8
⎝ 29 ⎠
⎝ 29 ⎠
⎝ 29 ⎠
⎝ 29 ⎠
⎝ 29 ⎠
= 66.80345
5 SSTr = ∑ ni ( xi − x ) 2 i =1 = 4 ( 59.2 − 66.80345 ) + 6 ( 64.9 − 66.80345 ) + 5 ( 65.8 − 66.80345 )
2 2 +7 ( 69.5 − 66.80345 ) + 7 ( 70.8 − 66.80345 )
2 2 2 = 420.7297
5 SSE = ∑ ( ni − 1) si2
i =1 = ( 4 − 1) 5.22 + ( 6 − 1) 5.12 + ( 5 − 1) 6.12 + ( 7 − 1) 5.92 + ( 7 − 1) 5.52
= 750.37
ANOVA Table
Source Trt error total df 4 24 28 SS MS F p 420.7297 105.1824 3.364178 0.025431136
750.37 31.26542
1171.1 b. What is the critical value of the test statistic?
The numerator degrees of freedom are 4
The denominator degrees of freedom are 24
For α = 0.05, the critical value is 2.78 (see Appendix Table VIII, page 575) Homework #8  Solution Page 4 of 4 c. Assume that Group A is a control. Conduct Dunnett’s Multiple Comparison Procedure to
determine whether any of the other 4 treatments have significantly different means than
the control (Group A).
Critical value for Dunnett’s Method is 2.61 (Appendix Table X, page 579, k=5, n = 29) Group n xi T xi − x A Significantly
Different A 4 59.2 n/a n/a 5.7 6.6 10.3 *** 11.6 *** ⎛1 1⎞
2.61 31.26542 ⎜ + ⎟ ⎝6 4⎠
B 6 64.9 = 9.420349 ⎛1 1⎞
2.61 31.26542 ⎜ + ⎟ ⎝5 4⎠
C 5 65.8 = 9.789914 ⎛1 1⎞
2.61 31.26542 ⎜ + ⎟ ⎝7 4⎠
D 7 69.5 = 9.147237 ⎛1 1⎞
2.61 31.26542 ⎜ + ⎟ ⎝7 4⎠
E 7 70.8 = 9.147237 Based on Dunnett’s Multiple Comparison Procedure, the means for Groups D and E are
significantly different from the mean of Group A. Homework #8  Solution Page 5 of 5 ...
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 Spring '08
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 Degrees Of Freedom, Variance, multiple comparison procedure

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