Chapter 9
Hypothesis Testing
Solutions:
12. a.
z
x  0
/ n

78.5  80
12 / 100
 1.25
Lower tail p-value is the area to the left of the test statistic
Using normal table with z = -1.25: p-value =.1056
p-value > .01, do not reject H0
b.
z
x  0
/ n

77  80
12 / 100
 2.50
Lower tail p-value is the area to the left of the test statistic
Using normal table with z = -2.50: p-value =.0062
p-value  .01, reject H0
c.
z
x  0
/ n

75.5  80
12 / 100
 3.75
Lower tail p-value is the area to the left of the test statistic
Using normal table with z = -3.75: p-value ≈ 0
p-value  .01, reject H0
d.
z
x  0
/ n

81  80
12 / 100
 .83
Lower tail p-value is the area to the left of the test statistic
Using normal table with z = .83: p-value =.7967
p-value > .01, do not reject H0
17. a.
H0:   125,500
Ha:   125,500
b.
z
x  0
/ n

118, 000  125,500
30, 000 / 40
 1.58
9-1
Chapter 9
Because z < 0, p-value is two times the lower tail area
Using normal table with z = -1.58: p-value = 2(.0571) = .1142
c.
p-value > .05, do not reject H0. We cannot conclude that the year-end bonuses paid by Jones & Ryan
differ significantly from the population mean of $125,500.
d.
Reject H0 if z  -1.96 or z  1.96
z = -1.58; cannot reject H0
24. a.
b.
t
x  0
s/ n

17  18
4.5 / 48
 1.54
Degrees of freedom = n – 1 = 47
Because t < 0, p-value is two times the lower tail area
Using t table: area in lower tail is between .05 and .10; therefore, p-value is between .10 and .20.
Exact p-value corresponding to t = -1.54 is .1303
c.
p-value > .05, do not reject H0.
d.
With df = 47, t.025 = 2.012
Reject H0 if t  -2.012 or t  2.012
t = -1.54; do not reject H0
34. a.
H0:  = 2
H a:   2
b.
x
c.
s
d.
t
xi 22

 2.2
n
10
  xi  x 
n 1
x  0
s/ n

2
 .516
2.2  2
.516 / 10
 1.22
Degrees of freedom = n - 1 = 9
Because t > 0, p-value is two times the upper tail area
Using t table: area in upper tail is between .10 and .20; therefore, p-value is between .20 and .40.
Exact p-value corresponding to t = 1.22 is .2535
9-2
e.
p-value > .05; do not reject H0. No reason to change from the 2 hours for cost estimating purposes.
Chapter 10
Comparisons Involving Means
Solutions:
2.
a.
z
 x1  x2   D0

2
1
n1
3.

n2
(25.2  22.8)  0
(5.2) 2 62

40
50
b.
p-value = 1.0000 - .9788 = .0212
c.
p-value  .05, reject H0.
a.
z
 x1  x2   D0

2
1
n1
8.


2
2


2
2

n2
(104  106)  0
(8.4) 2 (7.6) 2

80
70
b.
p-value = 2(.0630) = .1260
c.
p-value > .05, do not reject H0.
a.
z
 x1  x2   0

2
1
n1


2
2

(69.95  69.56)  0
n2
2.52 2.52

112
84
 2.03
 1.53
 1.08
b.
p-value = 2(1.0000 - .8599) = .2802
c.
p-value > .05; do not reject H0. Cannot conclude that there is a difference between the population
mean scores for the two golfers.
54
9
6
x1 
b.
s1 
( xi  x1 ) 2
 2.28
n1  1
s2 
( xi  x2 ) 2
 1.79
n2  1
c.
x2 
42
7
6
11. a.
x1  x2 = 9 - 7 = 2
Chapter 9
2
d.
2
 s12 s22 
 2.282 1.792 

  


6 
 n1 n2 
 6
df 

 9.5
2
2
2
2
1  2.282  1  1.792 
1  s12 
1  s22 
  
 

  

5 6  5 6 
n1  1  n1  n2  1  n2 
Use df = 9, t.05 = 1.833
2.282 1.792

6
6
x1  x2  1.833
2 2.17
(-.17 to 4.17)
H0: 1 - 2 = 0
40.
Ha: 1 - 2  0
z
( x1  x2 )  D0

2
1
n1


2
2
n2

(4.1  3.4)  0
(2.2)2 (1.5) 2

120
100
 2.79
p-value = 2(1.0000 - .9974) = .0052
p-value  .05, reject H0. A difference exists with system B having the lower mean checkout time.
Chapter 12
Simple Linear Regression
Solutions:
a.
16
14
12
10
y
1.
8
6
4
2
0
0
1
2
3
x
9-4
4
5
6
b.
There appears to be a positive linear relationship between x and y.
c.
Many different straight lines can be drawn to provide a linear approximation of the
relationship between x and y; in part (d) we will determine the equation of a straight line
that “best” represents the relationship according to the least squares criterion.
d.
x
xi 15

3
n
5
y
( xi  x )( yi  y )  26
b1 
yi 40

8
n
5
( xi  x ) 2  10
( xi  x )( yi  y ) 26

 2.6
10
( xi  x )2
b0  y  b1 x  8  (2.6)(3)  0.2
yˆ  0.2  2.6 x
e.
a.
2500
2000
Price ($)
5.
yˆ  0.2  2.6(4)  10.6
1500
1000
500
0
1
1.5
2
2.5
3
3.5
4
4.5
Baggage Capacity
b.
Let x = baggage capacity and y = price ($).
There appears to be a positive linear relationship between x and y.
c.
Many different straight lines can be drawn to provide a linear approximation of the
relationship between x and y; in part (d) we will determine the equation of a straight line
that “best” represents the relationship according to the least squares criterion.
d.
x
xi 29.5

 3.277778
n
9
y
( xi  x )( yi  y )  2909.888891
yi 11,110

 123.444444
n
9
( xi  x ) 2  4.555559
Chapter 9
b1 
( xi  x )( yi  y ) 2909.888891

 638.755615
4.555559
( xi  x )2
b0  y  b1 x  1234.4444  (638.7561)(3.2778)  859.254658
yˆ  859.26  638.76 x
e.
A one point increase in the baggage capacity rating will increase the price by approximately $639.
f.
yˆ  859.26  638.76 x  859.26  638.76(3)  $1057
15. a.
The estimated regression equation and the mean for the dependent variable are:
yi  0.2  2.6xi
y 8
The sum of squares due to error and the total sum of squares are
SSE  ( yi  yi ) 2  12.40
SST  ( yi  y ) 2  80
Thus, SSR = SST - SSE = 80 - 12.4 = 67.6
b.
r2 = SSR/SST = 67.6/80 = .845
The least squares line provided a very good fit; 84.5% of the variability in y has been explained by
the least squares line.
c.
18. a.
rxy  .845  .9192
The estimated regression equation and the mean for the dependent variable are:
yˆ  1790.5  581.1x
y  3650
The sum of squares due to error and the total sum of squares are
SSE  ( yi  yˆi ) 2  85,135.14
SST  ( yi  y ) 2  335, 000
Thus, SSR = SST - SSE = 335,000 - 85,135.14 = 249,864.86
b.
r2 = SSR/SST = 249,864.86/335,000 = .746
We see that 74.6% of the variability in y has been explained by the least squares line.
c.
26. a.
rxy  .746  .8637
In solving exercise 18, we found SSE = 85,135.14
s2 = MSE = SSE/(n - 2) = 85,135.14/4 = 21,283.79
s  MSE  21,28379
.  14589
.
9-6
( xi  x ) 2  0.74
sb1 
t
s
( xi  x )
2

145.89
0.74
 169.59
b1
581.1

 3.43
sb1 169.59
Using t table (4 degrees of freedom), area in tail is between .01 and .025
p-value is between .02 and .05
Using Excel or Minitab, the p-value corresponding to t = 3.43 is .0266.
Because p-value   , we reject H0: 1 = 0
b.
MSR = SSR/1 = 249,864.86/1 = 249.864.86
F = MSR/MSE = 249,864.86/21,283.79 = 11.74
Using F table (1 degree of freedom numerator and 4 denominator), p-value is between .025 and .05
Using Excel or Minitab, the p-value corresponding to F = 11.74 is .0266.
Because p-value   , we reject H0: 1 = 0
c.
Source
of Variation
Regression
Error
Total
Sum
of Squares
249864.86
85135.14
335000
Degrees
of Freedom
1
4
5
Mean
Square
249864.86
21283.79
F
11.74
p-value
.0266