STAT 350 – Fall 2008
Homework #9 –
Solution
through Lecture #13
Text Exercises
Chapter 9, Problems #6(af), 10, 12
1,2
, 14, 16
1
, 22
2
, 24
3,4
1
You may conduct the analysis using SAS or Minitab, but must present the ANOVA table by
hand.
DO not paste in extraneous computer output.
State your conclusion in a clear and
complete sentence!
2
You must also give the ANOVA table for this analysis
3
You may use Minitab to construct the probability plot
4
Be sure to construct the probability plot using the
deviations
, not on the original observations.
To get the deviations, take each observation and subtract the average for that group.
That is, if
x
i,j
is the
j
th observation in the
i
th group, then the deviation is
( )
,
ij
i
x
x
−
.
9.6 a. 3.69
b. 4.82
c. 6.63
d. 10.29
e. 3.10
f. 0.95
9.10
Source
df
SS
MS
F
Brand
4
39.2192
9.8048
15.32
Error
45
28.8
0.64
Total
49
68.0192
9.12
Source
DF
SS
MS
F
P
Factor
3
456.50
152.17
17.11
0.000
Error
14
124.50
8.89
Total
17
581.00
The
p
value is clearly < 0.05, so we can conclude that there is a statistically significant
difference in the yield of tomatoes among the soil salinity levels
Homework #9 – Solution
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The overall mean,
()
( ) ( ) ( ) ( )
6 2.58
6 2.63
6 2.13
6 2.41
6 2.49
2.448
30
x
++++
==
SSTr = 6(2.582.448)
2
+ 6(2.632.448)
2
+ 6(2.132.448)
2
+ 6(2.412.448)
2
+ 6(2.492.448)
2
= 0.92928
SSE = SST – SSTr = 3.619 – 0.92928 = 2.68972
a.
Source df
SS
MS
F
Sugar
4
0.92928
0.23232
2.15933
Error
25
2.68972
0.107589
Total
29
3.619
b. For df
1
= 4, and df
2
= 25, the critical value of
F
for an
α
=0.05 test is 2.76
The
F
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 Fall '08
 Staff
 Statistics, Normal Distribution, Statistical hypothesis testing, Statistical significance, statistically significant difference

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