CHAPTER 12
ANALYSIS OF VARIANCE
1.
The null and the alternative hypotheses are:
H0: 1 = 2 = 3
H1: The population means are not all equal.
We have selected  = 0.05.
We shall use as test statistic F =
MST
.
MSE
Since the population distributions are
approximately normal with equal variance, the distribution of the test statistic, under H0, is
F-distribution with df = (k-1, n-k) = (2, 9). This is an upper-tailed F-test.
For df = (2,9), F = F0.05 = 4.26.
The decision rule is: Reject H0 is favour of H1 if the computed F-value is greater than 4.26.
For the given sample data:
86  4
= 5.083
12
7 9
3 3
3  4
x.1 
= 8.25 ; x.2 
= 3 ; x.3 
=4
4
4
4
2
2
SSTotal =  8  5.083    4  5.083  74.92 .
The overall mean is x 
SST
= 4  8.25  5.083   4 3  5.083   4 4  5.083   62.1 .
2
2
2
SSE = SS Total – SST = 74.92 – 62.17 = 12.75.
ANOVA Table
Source of Variation Sum of Squares
F
Treatment
Error
Total
Degrees of Freedom
2.17
12.75
74.92
2
9
11
Mean Square
31.085
1.4167
21.94
The computed F-value is 21.94, which is greater than the critical value of 4.26. So, there is
sufficient evidence, at  = 0.05, to reject H0, that is, to conclude that the population means
corresponding to the three treatments are not all equal.
3.
Let 1, 2, and 3 be the population means of family incomes in the area #1, area #2 and
area #3, respectively. Then, the null and the alternative hypotheses are:
H0: 1 = 2 = 3
H1: ’s are not all the same.
We have selected  = 0.05.
12-1
We shall use as test statistic F =
MST
.
MSE
Since the population distributions are
approximately normal with equal variance, the distribution of the test statistic, under H0, is
F-distribution with df = (k-1, n-k) = (2, 9). This is an upper-tailed F-test.
For df = (2, 9), F = F0.05 = 4.26.
The decision rule is: Reject H0 is favour of H1 if the computed F-value is greater than 4.26.
For the given sample data:
64  68   78
= 71.25,
12
64   60
74   70
75   78
x.1 
= 65.5, x.2 
= 71, x.3 
= 77.25
4
4
4
The overall mean is x 
SSTotal =  64  71.25  
2
SST
  78  71.25  = 364.25.
2
= 4  65.5  71.25  
2
 4  77.25  71.25  = 276.50.
2
SSE = SS Total – SST = 364.25 – 87.75 = 87.75.
Thus, we get the following ANOVA table:
Source of Variation Sum of Squares
F
Treatment
Error
Total
Degrees of Freedom
276.50
87.75
364.25
2
9
11
Mean Square
138.25
9.75
14.18
The computed F-value is 14.18, which is greater than the critical value of 4.26. So, there is
sufficient evidence, at  = 0.05, to reject H0, that is, to conclude that the population means
are not all equal.
5.
Let 1, 2, and 3 be the population means corresponding to the three treatments. Then, the
null and the alternative hypotheses are:
H0: 1 = 2 = 3
H1: The population means are not all equal.
We have selected  = 0.05.
We shall use as test statistic F =
MST
.
MSE
Since the population distributions are
approximately normal with equal variance, the distribution of the test statistic, under H0, is
F-distribution with df =( k-1, n-k ) = (2, 9). This is an upper-tailed F-test.
For df = (2, 9), F = F0.05 = 4.26.
The decision rule is: Reject H0 is favour of H1 if the computed F-value is greater than 4.26.
For the given sample data:
12-2
The overall mean = x 
64  68 
12
 78
3
2
The sample means are:
x.1 
8
 10
= 9.667, x.2 
3
SSTotal = (8.0  4.667) 2 
= 4.667.
= 2.2, x.3 
3
4
5
4
2
  4.0  4.667  = 116.67.
SST
= 3  9.667  4.667  
SSE
= SS Total – SST = 116.67 - 107.20 = 9.47.
2
= 4.0.
 4  4.0  4.667  = 107.20.
2
ANOVA Table
Source of Variation
F
Treatments
Error
Total
Sum of Squares
107.20
9.47
116.67
Degrees of Freedom
2
9
11
Mean Square
53.600
1.052
50.96
The computed F-value is 50.96, which is greater than the critical value of 4.26. So, there is
sufficient evidence, at  = 0.05, to reject H0, that is, to conclude that the population means
corresponding to the three treatments are not equal.
Using Fisher’s LSD, we get a 95% confidence interval estimate for (1-2) as:
( x.1  x.2 )  t
2C
MSE (1/ n1  1/ n2 )
(9.667  2.20)   2.9358  1.052(1/ 5  1/ 4)
 7.467  2.02  (5.447,9.487)
Since the interval does not contain 0, there is significant evidence to conclude that the population
means of treatments 1 and 2 are different.
7.
(a)
Let 1, 2, 3 be the population means of number of month before a raise in salary was
granted for CPA Inc., AB Intl., Acct ltd., and Pfisters, respectively. Then, the null and the
alternative hypotheses are:
H0: 1 = 2 = 3
H1: The population means are not all equal.
We have selected  = 0.05.
We shall use as test statistic F =
MST
.
MSE
Since the population distributions are
approximately normal with equal variance, the distribution of the test statistic, under H0, is
F-distribution with df = (k-1,n-k) = (3,10). This is an upper-tailed F-test.
For df = (3, 10), F = F0.05 = 3.71.
The decision rule is: Reject H0 is favour of H1 if the computed F-value is greater than 3.71.
12-3
For the given sample data:
The overall mean = x 
14.3  18.1 
16
The sample means are:
x.1 
x.3 
12 
18 
 12
4
3
 16
 12.0 ; x.2 
 13.2
14 
 15.33; x.4 
12 
 10
4
 11.5 ;
 16
3
 13.0.
 14.0
 16  13  78.0.
2
SSTotal = (12  13) 2 
SST
= 4 12.0  13.0  
SSE
= SS Total – SST = 78.0 - 32.33 = 45.67.
2
 3 14.0  13.0   32.33.
2
Thus, we get the following ANOVA table:
Source of Variation Sum of Squares
Degrees of Freedom
Mean Square
F
Treatments
Error
Total
3
10
13
10.78
4.567
2.36
32.33
45.67
78.00
The computed F-value is 2.36, which is less than the critical value of 3.71. So, we do not
have sufficient evidence, at  = 0.05, to reject H0, that is, to infer that the population means
of number of months before a salary raise was granted of the four CPA firms are not all the
same.
(b)
9.
Since the null hypothesis is not rejected in part (a), we do not need to perform the Tukey’s
test.
Let 1 and 2 be the population means corresponding to the treatments 1 and 2
respectively. Then, the null and the alternative hypotheses for treatments are:
H0: 1 = 2
H1: 1 ≠ 2
We shall use as test statistic: F =
MST
MSE
The population distributions are approximately normal with equal variance. Hence, under
H0, the test statistic has approximately an F-distribution with df = (k-1, n-k-r+1) = (1, 2).
This is an upper-tailed F-test.
We have selected a significance level of 0.05. For df = (1, 2), F0.05 = 18.5.
The decision rule is: reject H0 in favour of H1 if the computed F-value is greater than 18.5.
12-4
Let b1, b2, and b3 be the population means corresponding to the blocks A, B, and C,
respectively. Then the null and the alternative hypotheses for blocks are:
H0: b1 = b2 = b3
H1: the block means are not all equal.
We shall use as test statistic: FB =
MSB
.
MSE
The population distributions are approximately normal with equal variance. Hence, under
H0, the test statistic has approximately an F-distribution with df = (r-1, n-k-r+1 ) = (2, 2).
This is an upper-tailed F-test.
We have selected a significance level of 0.05. For df = (2, 2), F0.05 = 19.0.
The decision rule is: reject H0 in favour of H1 if the computed F-value is greater than 19.0.
The overall mean is x 
46  37 
6
The sample means are:
x.1 
46 
 44
= 42.3 ; x.2 
 35
31 
= 36.5.
 35
= 30.7;
3
3
46  31
37  26
44  35
x1. 
= 38.5 ; x2. 
= 31.5; x3. 
= 39.5.
2
2
2
2
2
SSTotal =  46  36.5     35  36.5   289.5.
SST= 3  42.33  36.5   3  30.667  36.5   204.167.
2
2
SSB = 2  38.5  36.5   2  31.5  36.5   2  39.5  36.5   76.
2
2
2
SSE = SSTotal – SST – SSB = 289.5 – 204.167 – 76 = 9.333.
We get the following ANOVA table:
Source of Variation SS
Treatment 204.167
Blocks
76.000
Error
9.333
Total
289.5000
df
1
2
2
5
MS
204.167
38.000
4.667
F
43.75
8.14
Our conclusions are as follows:
Test for treatment means:
43.75 is greater than the critical value 18.5. Hence, there is sufficient evidence, at  = 0.05,
to reject H0, that is, to infer that the treatment means are not all the same.
Test for block means:
8.14 is less than 19.0. Hence, we do not have sufficient evidence, at  = 0.05, to reject H0,
that is, to infer that the block means are not all the same.
12-5
11.
Let 1, 2, and 3 be the population means of number of units produced per employee on
day, afternoon and night shifts, respectively. Then the null and the alternative hypotheses
are:
H0: 1 = 2 = 3
H1: At least one of the  ’s is different.
MST
.
MSE
We shall use as test statistic: F =
The population distributions are approximately normal with equal variance. Hence, under
H0, the test statistic has approximately an F-distribution with df = (k-1,n-k-r+1) = (2, 8).
This is an upper-tailed F-test.
We have selected a significance level of 0.05. For df = (2, 8), F0.05 = 4.46.
The decision rule is: reject H0 in favour of H1 if the computed F-value is greater than 4.46.
Let b1, b2, b3, b4, and b5 be the population means of number of units produced by Mehta,
Lum, Clark, Kurz, and Morgan, respectively. Then the null and the alternative hypotheses
for blocks are:
H0: b1 = b2 = b3 =b4 = b5 ;
H1: the means are not all equal.
We shall use as test statistic: FB =
MSB
.
MSE
The population distributions are approximately normal with equal variance. Hence, under
H0, the test statistic has approximately an F-distribution with df = (r-1, n-k-r+1) = (4, 8).
This is an upper-tailed F-test.
We have selected a significance level of 0.05. For df = (4, 8), F0.05 = 3.84.
The decision rule is: reject H0 in favour of H1 if the computed F-value is greater than 3.84.
31  33   27
= 28.867.
15
31   28
25   26
35   27
x.1 
= 30.0; x.2 
= 26.0; x.3 
= 30.6;
5
5
5
31  25  35
33  26  33
28  24  30
x1. 
= 30.3; x2. 
= 30.7; x3. 
=27.3;
3
3
3
30  29  28
28  26  27
x4. 
= 29.0; x5. 
= 27.0.
3
3
The overall mean is x 
SSTotal =  31  28.867  
  27  28.867   139.73 .
2
2
SST
= 5  30.0  28.867  
 5  30.6  28.867   62.53
SSB
= 3  30.3  28.867  
 3  27.0  28.867   33.73
SSE
= SSTotal – SST – SSB = 139.73 – 62.53 – 33.73 = 43.47.
2
2
2
2
We get the following ANOVA table :
Source of Variation SS
Treatment
Blocks
Error
Total
df
MS
F
62.53 2
31.265 5.75
33.73 4
8.4325 1.55
43.47 8 12-6 5.4338
139.73
Our conclusions are as follows:
Test for treatment means:
5.75 is greater than the critical value, 4.46. Hence, there is sufficient evidence, at  = 0.05,
to reject H0, that is, to infer that the mean production rates during different shifts are not all
the same.
Test for block means:
1.55 is less than 3.84. Hence, we do not have sufficient evidence, at  = 0.05, to reject H0,
that is, to infer that mean production rates of different employees are not all the same.
13.
The null and the alternative hypotheses are:
H0: 1 = 2 = 3 = 4
H1: Treatment means are not all equal.
Since the population distributions are approximately normal, the distribution of the test
statistic, F 
MST
is approximately F-distribution, with df = (k – 1, n – k) = (4-1, 24-4)
MSE
= (3, 20).
For df = (3, 20), F = F0.05 = 3.10.
The decision rule is: reject H0 in favour of H1 if the computed F-value is greater 3.10.
MSE =
SSE
SSE SSE
.


df ( for error ) n  k
20
Hence, SSE = 20 (MSE) = 20(10) = 200.
SST = SS total – SSE = 250 - 200 = 50.
MST =
F=
SST
50

 16.67.
df ( for treatment ) 3
MST 16.67

 1.667.
MSE
10
This gives us the following ANOVA table.
Source of Variation
SS
df
MS
Treatment
Error
Total
50
200
250
3
20
23
16.67 1.667
10
12-7
F
Since the computed F-value (=1.667) is less than 3.10, we conclude that we do not have
sufficient evidence, at  = 0.05, to reject H0 in favour of H1, that is, to infer that the
treatment means are not all equal.
15.
Let 1, 2, and 3 be the population means of the prices of the toy at discount stores, variety
stores and department stores, respectively. Then, the null and the alternative hypotheses
are:
H0: 1 = 2 = 3
H1: The population means are not all equal.
We have selected  = 0.05.
We shall use as test statistic F =
MST
.
MSE
Since the population distributions are
approximately normal with equal variance, the distribution of the test statistic, under H0, is
F-distribution with df=(k-1,n-k) = (2, 12). This is an upper-tailed F-test.
For df = (2, 12), F = F0.05 = 3.89.
The decision rule is: Reject H0 is favour of H1 if the computed F-value is greater than 3.89.
For the given sample data:
The overall mean = x 
23 5
 15.867.
15
The sample means are:
x.1 
12 
 15
 13.2 ; x.2 
5
SSTotal = (12.0  15.867) 2 
15 
 17
 16.2 ; x.3 
5
2
 19  15.867   91.73.
SST
= 5 13.2  15.867  
SSE
= SS Total – SST = 91.73 - 63.33 = 28.40.
2
19 
 19
5
 18.2.
 5 18.2  15.867   63.33.
2
Thus, we get the following ANOVA table:
SOURCE OF VARIATION SS
Treatment
Error
Total
63.33 2
28.40 12
91.73 14
DF
MS
F
31.667 13.38
2.367
The computed F-value is 13.38, which is greater than the critical value of 3.89. So, there is
sufficient evidence, at  = 0.05, to reject H0, that is, to infer that the mean prices of the toy,
in different types of stores, are not all equal.
12-8
17.
Let 1, 2, 3, and 4 be the population means of number of crimes committed per day in
Rec Centre, Key Street, Monclova and Whitehouse, respectively.
The null and the alternative hypotheses are:
H0: 1 = 2 = 3 = 4
H1: The population means are not all equal.
We have selected  = 0.05.
MST
.
MSE
We shall use as test statistic F =
Since the population distributions are
approximately normal with equal variance, the distribution of the test statistic, under H0, is
F-distribution with df = (k-1, n-k) = (3, 20). This is an upper-tailed F-test.
For df = (3, 20), F = F0.05 = 3.10.
The decision rule is: Reject H0 is favour of H1 if the computed F-value is greater than 3.10.
For the given sample data:
The overall mean = x 
13  15   18
 15.792.
24
The sample means are:
x.1 
13 
 15
21 
 19
 18.0 ;
6
12   15
16   18
x.3 
 13.5 ; x.4 
 17.333.
6
6
6
 14.333 ; x.2 
 18.0  15.792   151.96.
2
2
SSTotal = (13.0  15.792) 
SST
= 6 14.333  15.792  
SSE
= SS Total – SST = 151.96 - 87.79 = 64.17.
2
 6 17.333  15.792   87.79.
2
Thus, we get the following ANOVA table:
Source of Variation SS
Factor
Error
Total
df
87.79 3
64.17 20
151.96 23
MS
F
29.26 9.12
3.21
The computed F-value is 9.12, which is greater than the critical value of 3.10. So, there is
sufficient evidence, at  = 0.05, to reject H0, that is, to infer that the mean number of
crimes per day is not the same in all the districts.
19.
Let 1, and 2 be the population means of the number of correct answers obtained by the
two groups of students. Then, the null and the alternative hypotheses are:
H0: 1 = 2
H1: 1  2.
12-9
(a)
We have selected  = 0.05.
We shall use as test statistic F =
MST
.
MSE
Since the population distributions are
approximately normal with equal variance, the distribution of the test statistic, under H0, is
F-distribution with df = (k-1, n-k) = (1, 12). This is an upper-tailed F-test.
For df = (1, 12), F = F0.05 = 4.75.
The decision rule is: Reject H0 is favour of H1 if the computed F-value is greater than 4.75.
For the given sample data:
The overall mean is x 
19  17   25
 23.57.
14
The sample means are:
x.1 
19 
 16
6
 19.00 ; x.2 
SSTotal = (19.0  23.57) 2 
32 
 25
8
 27.00.
  25  23.57   333.43.
2
SST = 6 19.0  23.57   8  27.0  23.57   219.43.
2
2
SSE = SS Total – SST = 333.43 – 219.43 = 114.00.
Thus, we get the following ANOVA table:
Source of Variation
Treatment
Error
Total
SS
df
219.43 1
114.00 12
333.43 13
MS
F
219.43 23.097
9.50
The computed F-value is 23.097, which is greater than the critical value of 4.75. So, there
is sufficient evidence, at  = 0.05, to reject H0, that is, to infer that the two means are
different.
b.
We shall use two-tailed t-test.
For df = (n1+n2-2)=12, t = t0.025 = 2.179. Hence, the decision rule is: reject H0 if t < 2
2.179 or if t > 2.179.
t
19  27
 1 1
9.5   
 6 8
 4.806 .
Since –4.806 < -2.179, there is sufficient evidence, at  = 0.05, to reject H0, that is, to infer
that the two means are different.
c.
21.
(4.806)2  23.097. So, t2 = F. Also (2.179)2  4.75. Hence, both the tests give us the same
result: there is sufficient evidence, at  = 0.05, to reject H0.
For colour:
12-10
Let 1, 2, 3 and 4 be the population means of ratings given by the magazine readers for
four different colours of advertisements. Then, the null and the alternative hypotheses are:
H0: 1 = 2 = 3 = 4
H1: At least one of the  ’s is different
We shall use as test statistic: F =
MST
MSE
The population distributions are approximately normal with equal variance. Hence, under
H0, the test statistic has approximately an F-distribution with df = (k-1,n-k-r+1) = (3, 6).
This is an upper-tailed F-test.
We have selected a significance level of 0.05. For df = (3.6), F0.05 = 4.76.
The decision rule is: reject H0 in favour of H1 if the computed F-value is greater than 4.76.
For size:
Let b1, b2, and b3 be the population means of ratings given by the magazine readers for
three different sizes of advertisements. Then, the null and the alternative hypotheses are:
H0: b1 = b2 = b3
H1: the means are not all equal.
We shall use as test statistic: F =
MST
MSE
The population distributions are approximately normal with equal variance. Hence, under
H0, the test statistic has approximately an F-distribution with df = (r-1,n-k-r+1) = (2, 6).
This is an upper-tailed F-test.
We have selected a significance level of 0.05. For df = (2, 6), F0.05 = 5.14.
The decision rule is: reject H0 in favour of H1 if the computed F-value is greater than 5.14.
For the given sample data :
The overall mean is x 
23 8
 5.5.
12
The sample means are:
23 6
35 7
 3.67 ; x.2 
 5.0;
3
3
3 68
878
x.3 
 5.67 ; x.4 
 7.67.
3
3
x.1 
x1. 
2
8
4
 4.0, x2. 
3
SS(Color)  3  3.67  5.5  
2
SS(Size)  4  4.0  5.5  
2
SStotal   2  5.5  
2
7
4
 5.25, x3. 
6
8
4
 7.25.
 3  7.67  5.5   25.0
2
 4  7.25  5.5  21.5
2
  8  5.5   55.0
2
SSE = SStotal - SS(Color) - SS(Size) = 55.0 – 25.0 – 21.5 = 8.5.
Thus, we get the following ANOVA table:
12-11
Source of Variation SS
df
MS
Color
Size
Error
Total
3
2
6
11
8.33 5.88
10.75 7.59
1.42
25.0
21.5
8.5
55.0
F
Our conclusions are as follows:
Test for color:
5.88 is greater than the critical value, 4.76. Hence, there is sufficient evidence, at  = 0.05,
to reject H0, that is, to infer that the means for different colors are not all the same.
Test for size:
7.59 is greater than the critical value, 5.14. Hence, there is sufficient evidence, at  = 0.05,
to reject H0, that is, to infer that the means for different sizes are not all the same.
23.
Let 1, 2, 3 and 4 be the population means of the values of homes assessed by the four
different assessors. Then, the null and the alternative hypotheses are:
H0: 1 = 2 = 3 = 4
H1: the means are not all equal.
We shall use as test statistic: F =
MST
MSE
The population distributions are approximately normal with equal variance. Hence, under
H0, the test statistic has approximately an F-distribution with df = (k-1,n-k-r+1) = (3, 12).
This is an upper-tailed F-test.
We have selected a significance level of 0.05. For df = (3, 12), F0.05 = 3.49.
The decision rule is: reject H0 in favour of H1 if the computed F-value is greater than 3.49.
For the given sample data :
The overall mean = x 
153  150 
20
The sample means are:
x.1 
x.3 
x1. 
x3. 
x5. 
153 
149 
153 
148 
184 
 184
5
5
4
4
4
 192
 145
 153
 186
 161.0 ; x.2 
 161.0 ; x.4 
 150.5; x2. 
 150.0; x4. 
 186
155 
145 
150 
170 
 187.8.
12-12
 161.30.
 189
5
5
4
4
 186
 153
 164
 163.0;
 160.2
 151.5 ;
 166.8 ;
SST  5 161.0  161.3 
 5 160.2  161.3  21.4
SSB  4 150.5  161.3 
 4 187.8  161.3  4278.7
2
2
2
2
SS total  153.0  161.3 
 186.0  161.3  4428.2
2
2
SSE = 4428.2 – 21.4 – 4278.7 = 128.1.
Thus, we get the following ANOVA table:
SOURCE
Home
Assessor
Error
Total
4
3
12
19
DF
4278.7
21.4
128.1
4428.2
SS
1069.7
7.1
0.67
10.7
MS
F
100.20
F = 0.67 is less than the critical value, 3.49. Hence, we do not have sufficient evidence, at
 = 0.05, to reject H0, that is, to infer that the population means are not all the same.
25.
For Gasoline :
Let 1, 2, and 3 be the population means of automobile fuel efficiencies using three
different grades of gasoline – unleaded regular, mid-grade and super premium gasoline,
respectively. Then, the null and the alternative hypotheses are:
H0: 1 = 2 = 3
H1: The mean fuel efficiencies of the three grades of gasoline are not all the same.
We shall use as test statistic: F =
MST
MSE
The population distributions are approximately normal with equal variance. Hence, under
H0, the test statistic has approximately an F-distribution with df = (k-1,n-k-r+1) = (2, 12).
This is an upper-tailed F-test.
We have selected a significance level of 0.05. For df = (2, 12), F0.05 = 3.89.
The decision rule is: reject H0 in favour of H1 if the computed F-value is greater than 3.89.
For Automobiles:
Let b1, b2, b3, …, b be the population means of fuel efficiencies of the seven different
automobiles. Then, the null and the alternative hypotheses are:
H0: b1  b2   b7
H1: The mean fuel efficiencies of the seven automobiles are not all the same.
For df = (6,12), F0.05 = 3.00.
The decision rule is: reject H0 in favour of H1 if the computed F-value is greater than 3.00.
For the given sample data:
The overall mean = x 
7.8  8.0 
21
 9.0
The sample means are:
12-13
 8.2714.
x.1 
x.3 
7.8 
8.3 
 8.3
7
 8.143 ; x.2 
8.0 
 8.5
7
 9.0
 8.186;
 8.486 ;
7
7.8  8.0  8.3
8.0  7.9  8.2
x1. 
 8.033 ; x2. 
 8.033 ;
3
3
8.1  8.2  8.5
8.1  8.1  8.3
x3. 
 8.267 ; x4. 
 8.167 ;
3
3
8.3  8.4  8.7
8.4  8.2  8.4
x5. 
 8.467 ; x6. 
 8.333 ;
4
3
8.3  8.5  9.0
x7. 
 8.6.
3
2
2
SS (Gasoline)  7 8.143  8.2714    7  8.486  8.2714   0.4886
SS ( Auto)  3  8.033  8.2714  
2
SS total   7.8  8.2714  
2
 3  8.6  8.2714   0.8229
2
  9.0  8.2714   1.4829
2
SSE = 1.4829-0.4886-0.8229=0.1714.
We get the following ANOVA table:
Autos
Gasoline
Error
Total
Source of Variation
DF
MS
F
6
0.8229
0.1371
2
0.4886
0.2443
12
0.1714
0.0143
20
1.4829
SS
9.60
17.10
Our conclusions are as follows:
Test for gasoline:
17.1 is greater than the critical value, 3.89. Hence, there is sufficient evidence, at  = 0.05,
to reject H0, that is, to infer that the mean fuel efficiencies, for the different grades of
gasoline, are not all the same.
Test for automobiles: 9.6 is greater than the critical value, 3.0. Hence, there is sufficient
evidence, at  = 0.05, to reject H0, that is, to infer that the mean fuel efficiencies of the
automobiles are not all the same.
27.
Let 1 and  2 be the population means of one-year percent returns on Canadian Balanced
Funds and Health Care Companies, respectively. Then the null and the alternative
hypotheses are:
H0: 1 = 2
H1: 1 ≠ 2
12-14
Use as test statistic: F =
MST
MSE
The population distributions are approximately normal with equal variance. Hence, under
H0, the test statistic has approximately an F-distribution with df = (k-1,n-k-r+1) = (1, 4).
This is an upper-tailed F-test.
We have selected a significance level of 0.05. For df = (1, 4), F0.05 = 7.71.
The decision rule is: reject H0 in favour of H1 if the computed F-value is greater than 7.71.
Collect the random sample data, compute x , x.1 , x.2 , x1. , x2. , x3. , x4. , x5. , SSTotal, SST,
SSB, and SSE, and construct the ANOVA table.
If the computed F-value is greater than 7.71, conclude that there is sufficient evidence, at
α = 0.05, to reject H0 in favour of H1. Else, conclude that we do not have sufficient
evidence, at α = 0.05, to reject H0 in favour of H1.
29.
(a)
Let  12 and  22 be the population variances of the number of stolen bases among
the teams that play their home games on natural grass and artificial turf, respectively. Then
the null and the alternative hypotheses are:
H0:
 12
=1
 22
H1:
 12
1
 22
This is a two-tailed test.
Since the population distributions are approximately normal, we can use two-tailed F-test.
The Minitab and Excel outputs are given below.
MINITAB OUTPUT
Test for Equal Variances
Level1
C1
Level2
C2
F-Test (normal distribution)
Test Statistic: 1.073
P-Value: 0.984
F-Test Two-Sample for Variances
Variable 1 Variable 2
Mean
196.8696 188.2857
Variance
1911.937 1781.905
Observations
23
7
df
22
6
F
1.072974
P(F<=f) one-tail 0.508137
F Critical one-tail
3.8564
Both the outputs give us a P-value of 0.984. (This is directly given in Minitab output. From
the Excel output we find P-value = min{2(0.508137), 2(1-0.508137)} = 0.983746  0.984.
Since the value of α (=0.1) is smaller than the P-value, we do not have sufficient evidence
to reject H0, that is, to infer that the two population variances are different.
(b)
Let 1, 2, and 3 be the population means of number of games won in the three groups.
Then the null and the alternative hypotheses are:
H0: 1 = 2 = 3
H1: The ’s are not all equal
12-15
The Minitab and Excel outputs are given below.
From both the outputs, we get P-value = 0.039. Since the selected value of α (= 0.05) is
greater than the P-value, we have sufficient evidence to reject H0, that is, to infer that the
population means are not all the same.
MINITAB OUTPUT
Analysis of Variance
Source
DF
SS
Factor
2
618.5
Error
27
2273.4
Total
29
2891.9
MS
309.3
84.2
F
3.67
P
0.039
MICROSOFT EXCEL OUTPUT
ANOVA
Source of Variation
SS
Between Groups
618.502
Within Groups
2273.365
Total
2891.867
(c)
df
MS
F
P-value
F crit
2 309.251 3.672872 0.038826 3.354131
27 84.19869
29
Let 1, 2, and 3 be the population means of team batting averages for the three groups.
Then the null and the alternative hypotheses are:
H0: 1 = 2 = 3
H1: The ’s are not all equal
The Minitab and Excel outputs are given below.
From both the outputs, we get P-value = 0.231. Since the selected value of α (= 0.1) is less
than the P-value, we do not have sufficient evidence to reject H0, that is, to infer that the
population means are not all the same.
MINITAB OUTPUT
Analysis of Variance
Source
DF
SS
Factor
2
0.000401
Error
27
0.003494
Total
29
0.003894
MS
0.000200
0.000129
F
1.55
P
0.231
MICROSOFT EXCEL OUTPUT
ANOVA
Source of Variation
SS
Between Groups
0.000401
Within Groups
0.003494
Total
0.003894
31.
(a)
df
MS
F
P-value
2
0.0002 1.548684 0.230821
27 0.000129
29
F crit
2.51061
Let  12 and  22 be the population variances of the selling prices of the homes that do not
have a pool and that have a pool, respectively. Then, the null and the alternative
hypotheses are:
12-16
H0:
 12
=1
 22
H1:
 12
1
 22
This is a two-tailed test.
Since the population distributions are approximately normal, we can use two-tailed F-test.
The Minitab and Microsoft Excel outputs are given below.
MICROSOFT EXCEL OUTPUT
MINITAB OUTPUT
Test for Equal Variances
F-Test Two-Sample for Variances
Level1
Level2
ConfLvl
Mean
Variance
Observations
df
F
P(F<=f) one-tail
F Critical one-tail
C1
C2
95.0000
F-Test (normal distribution)
Test Statistic: 0.444
P-Value
: 0.009
Variable 1 Variable 2
202.7974 231.4851
1136.031 2557.258
38
67
37
66
0.444238
0.004381
0.488537
Both the outputs give us a P-value of approximately 0.009. (From Excel output, we get the
P-value for two-tailed test as 2(P(F<=f)) = 2(0.004381) = 0.008762  0.009.) Since the
chosen value of α (=0.02) is greater than the P-value, we have sufficient evidence to reject
H0, that is, to infer that the population variances are not equal.
(b)
Let  12 and  22 be the population variances of the selling prices of the homes that do not
have and that have an attached garage, respectively. Then, the null and the alternative
hypotheses are:
H0 :
 12
1
 22
H1 :
 12
1
 22
This is a two-tailed test.
Since the population distributions are approximately normal, we can use two-tailed F-test.
The Minitab and Microsoft Excel outputs are given below.
MINITAB OUTPUT
Test for Equal Variances
Level1
Level2
C1
C2
F-Test (normal distribution)
Test Statistic: 0.389
P-Value
: 0.004
MICROSOFT EXCEL OUTPUT
F-Test Two-Sample for Variances
Variable 1 Variable 2
Mean
185.45 238.1761
Variance
784.2571 2013.896
Observations
34
71
df
33
70
F
0.389423
P(F<=f) one-tail 0.001871
12-17
F Critical one-tail 0.47369
Both the outputs give us a P-value of approximately 0.004. Since the chosen value of
α (=0.02) is greater than the P-value, we have sufficient evidence to reject H0, that is, to
infer that the population variances are not equal.
(c)
Let 1, 2, and 3 be the population means of selling prices among the five townships.
Then, the null and the alternative hypotheses are:
H0: 1 = 2 = 3
H1: the population means are not all equal.
Since the population distributions are approximately normal with equal variances, we shall
use ANOVA test. The Minitab and Microsoft Excel outputs are given below.
MINITAB OUTPUT
Analysis of Variance
Source
DF
SS
Factor
4
13263
Error
100
217505
Total
104
230768
MS
3316
2175
F
1.52
P
0.201
MICROSOFT EXCEL
ANOVA
Source of Variation
SS
Between Groups
13262.85
Within Groups
217504.7
Total
230767.6
df
MS
F
P-value
F crit
4 3315.711 1.524432 0.200824 3.061984
100 2175.047
104
Both the outputs give a P-value approximately equal to 0.201. Since the selected value of 
(= 0.02) is less than the P-value, we do not have sufficient evidence to conclude that the
population means are not all equal.
12-18