36.
Sample Mean
17.0
20.4
25.0
Sample Variance
5.01
6.26
4.01
( )
2
1
SSTR k
jj
j
n x x
=
=−
50
60
70
Sample Mean
33
29
28
Sample Variance
32
17.5
9.5
x
= (33 + 29 + 28)/3 = 30
( )
2
1
SSTR k
jj
j
n x x
=
=−
= 5(33 – 30) 2 + 5(29 – 30) 2 + 5(28 – 30) 2 = 70
equal.
38.
Paint 1
Paint 2
Paint 3
Paint 4
Sample Mean
13.3
139
136
144
Sample Variance
47.5
.50
21
54.5
x
= (133 + 139 + 136 + 144)/3 = 138
2
Because p-value
= .05, we reject the null hypothesis that the mean meal prices are the same for
the three types of restaurants.
40. H0:
1 –
2 = 0
Ha:
1 –
2 0
1 2 0
() (4.1 3.4) 0 2.79
x x D
−− −−
p-value = 2(1.0000 – .9974) = .0052
p-value
.05, reject H0. A difference exists with system B having the lower mean checkout time.
41. a.
1
1
6,776,900 225,897
30
i
x
xn
= = =
Mean resale price in 2006
6,839,735 170,993
i
x
c. To answer this question we need to conduct a one-tailed hypothesis test. No value for the level of
significance (α) has been given. But, most people would agree that a p-value
.01 would justify
concluding that prices have declined from 2006 to 2009.
0 1 2
12
:
:
a
H
H
12
2 2 2 2
12
12
54,904 4.45
55,207 44,958
30 40
xx
tss
nn
−
= = =
+
+
For t = 4.45 and df =54, we find p-value
0.00. Thus, we are justified in concluding that existing
home prices have declined between 2006 and 2009.
42. a. H0:
1 –
2 0
Ha:
1 –
2 > 0
b. n1 = 30 n2 = 30
x1
= 16.23
x2
= 15.70
s1 = 3.52 s2 = 3.31
1 2 0
() (16.23 15.70) 0 .60
x x D
−− −−
2
22 22
12
12
2 2 2 2
2 2 2 2
12
1 1 2 2
3.52 3.31
30 30 57.8
1 3.52 1 3.31
11
29 30 29 30
11
ss
nn
df ss
n n n n
++
= = =
+
+
−−
Use df = 57
Using t table, p-value is greater than .20
43. a. n1 = 10 n2 = 8
1
x
= 21.2
2
x
= 22.8
s1 = 2.70 s2 = 3.55
b.
22
22 22
12
12
2 2 2 2
2 2 2 2
12
1 1 2 2
2.70 3.55
10 8 12.9
1 2.70 1 3.55
11
9 10 7 8
11
ss
nn
df ss
n n n n
++
= = =
+
+
−−
Use df = 12, t.05 = 1.782
22
2.70 3.55
1.6 1.782 10 8
− +
January 1
April 30
i
d
()
i
dd−
2
()
i
dd−
10.13
12.21
-2.08
-4.53
20.5209
28.33
25.48
2.85
0.40
0.1600
73.97
66.10
7.87
5.42
29.3764
16.30
19.32
-3.02
-5.47
29.9209
45.27
43.05
2.22
-0.23
0.0529
16.88
15.46
1.42
-1.03
1.0609
2.29
5.98
-3.69
-6.14
37.6996
16.20
12.65
3.55
1.10
1.2100
59.83
52.36
7.47
5.02
25.2004
31.53
33.00
-1.47
-3.92
15.3664
19.44
20.26
-0.82
-3.27
10.6929
17.73
19.34
-1.61
-4.06
16.4836
17.71
13.36
4.35
1.90
3.6100
43.51
36.18
7.33
4.88
23.8144
61.82
49.44
12.38
9.93
98.6049
Sum
36.75
313.7742
36.75 2.45
15
i
d
dn
= = =
The mean price per share declined $2.45 over the four months.
b.
2
()313.7742 4.73
1 14
i
d
dd
sn
−
= = =
−
df = n – 1 = 14, t.05 = 1.761
.05 d
s
dt n
=
4.73
2.45 1.761 15
c. Sample mean price per share January 1:
460.94 $30.73
15
i
x
xn
= = =
Percentage decrease over the 4 months:
2.45 (100) 8%
30.73 =
45.
x
y
z
Sample Mean
92
97
84
Sample Variance
30
6
35.33
x
= (92 + 97 + 44) /3 = 91
( )
2
1
SSTR k
jj
j
n x x
=
=−
= 4(92 – 91) 2 + 4(97 – 91) 2 + 4(84 – 91) 2 = 344
MSTR = SSTR /(k – 1) = 344 /2 = 172
2
1
k
j
=
MSE = SSE /(nT – k) = 213.99 /(12 – 3) = 23.78
F = MSTR /MSE = 172 /23.78 = 7.23
Using F table (2 degrees of freedom numerator and 9 denominator), p-value is between .01 and .025
Using Excel, the p-value corresponding to F = 7.23 is .0134.
Because p-value
= .05, we reject the null hypothesis that the mean absorbency ratings for the
three brands are equal.
46.
Lawyer
Physical
Therapist
Cabinet
Maker
Systems
Analyst
Sample Mean
50.0
63.7
69.1
61.2
Sample Variance
124.22
164.68
105.88
136.62
50.0 63.7 69.1 61.2 61
4
x+ + +
==
( )
SSTR k
jj
n x x
=−
2
1
j
=
2
1
SSE ( 1)
k
jj
j
ns
=
=−
= 9(124.22) + 9(164.68) + 9(105.88) + 9(136.62) = 4,782.60
MSE = SSE /(nT – k) = 4782.6 /(40 – 4) = 132.85
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Rows
903.025
9
100.3361
4.5479
0.0010
2.2501
Columns
160.075
3
53.3583
2.4186
0.0880
2.9604
Error
595.675
27
22.0620
Total
1658.775
39
The label Rows corresponds to the blocks in the problem (Date), and the label column corresponds
to the treatments (City).
Because the p-value corresponding to Columns is .0880 is greater than α = .05, there is no significant
difference in the mean ozone level among the four cites. But, if the level of significance was α = .10,
the difference would have been significant.
SUMMARY
Groups
Count
Sum
Average
Variance
Midwest
16
193.3
12.0813
13.0083
Northeast
16
133.8
8.3625
17.5932
South
25
300.4
12.016
22.2222
West
18
125.8
6.9889
12.4058
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Between Groups
376.87155
3
125.6238
7.4126
0.0002
2.7336
Within Groups
1203.2533
71
16.9472
Total
1580.1248
74
Because the p-value = .0002 is less than α = .05, we reject the null hypothesis that the mean rental
vacancy rate is the same for each geographic region. The mean vacancy rates were highest (over
49.
=
=−
Method A
Method B
Method C
Sample Mean
90
84
81
Sample Variance
98.00
168.44
159.78
x
= (90 + 84 + 81) /3 = 85
( )
2
1
SSTR k
jj
j
n x x
=
=−
= 10(90 – 85) 2 + 10(84 – 85) 2 + 10(81 – 85) 2 = 420
MSTR = SSTR /(k – 1) = 420 /2 = 210
2
1
SSE ( 1)
k
jj
j
ns
=
=−
= 9(98.00) + 9(168.44) + 9(159.78) = 3,836
50.
Nonbrowser
Light Browser
Heavy Browser
Sample Mean
4.25
5.25
5.75
Sample Variance
1.07
1.07
1.36
x
= (4.25 + 5.25 + 5.75) /3 = 5.08
( )
2
1
SSTR k
jj
j
n x x
=
=−
= 8(4.25 – 5.08) 2 + 8(5.25 – 5.08) 2 + 8(5.75 – 5.08) 2 = 9.33
MSTR = SSTR /(k – 1) = 9.33 /2 = 4.67
2
k
Using Excel, the p-value corresponding to F = 3.99 is .0340.
Because p–value
= .05, we reject the null hypothesis that the mean comfort scores are the same
for the three groups.