CHAPTER 2: Solutions to Selected Exercises
2.1
a.
y
32.3  30.5  31.7  31.6  31.4  32.6 190.1

6
6
= 31.68
b.
s
2
2
2
2

32.3  31.68  30.5  31.68  31.7  31.68

6 1
31.6  31.68  31.4  31.682  32.6  31.682

6 1
2.7084

 0.54168
5
2
2.2
2.3
2.4
2.5
0.54168  0.74
c.
s=
a.
y  103
b.
s 2 = 673.6666667
c.
s = 25.96
a.
y  65.24
b.
s 2 = 0.13125
c.
s = 0.36
a.
y  76.44
b.
s 2 = 21.52777778
c.
s = 4.64
a.
y  31.413
b.
s 2 = 0.769
c.
s = 0.877
d.
To determine the number of intervals, we find the smallest k such that 2 k  30 . Since
2 4  16  30 , and 2 5  32  30 , we choose k = 5 intervals.
4
The length of each interval is
Largest mileage
- Smallest mileage
k
33.3  29.8 3.5


 .7
5
5

Therefore, we have the following frequency distribution.
Interval
Frequency
[29.8, 30.4]
[30.5, 31.1]
[31.2, 31.8]
[31.9, 32.5]
[32.6, 33.2]
[33.3, 33.9]
5
8
7
8
1
1
Note that we have added an interval to include 33.3. The histogram is
8
6
4
2
29.7
e.
30.4
a.
31.8
32.5
33.2
The stem-and-leaf display is as follows:
29
30*
30
31*
31
32*
32
33*
2.6
31.1
y
8
1434
89586
303401
7685
104104
558
3
50.6  49.8  50.8  ...  50.5
 50.433
36
5
33.9
c.
s
50.6  50.4332  49.8  50.4332    50.5  50.4332
36  1
=0.888
d.
Since 2 5  32  n  36 and 2 6  64  n  36 , we employ k  6 classes. Since
52.2  46.8 5.4

 0.9 , and since the data is recorded to the nearest tenth of a unit, we use
6
6
classes that contain nine measurement values. These are listed below along with class
boundaries. Note that since the largest value (52.2) is not in the sixth class, we add a seventh
class.
Measurement
Values
46.8, 46.9, …, 47.6
47.7, 47.8, …, 48.5
48.6, 48.7, …, 49.4
49.5, 49.6, …, 50.3
50.4, 50.5, …, 51.2
51.3, 51.4, …, 52.1
52.2, 52.3, …, 53.0
Class
1
2
3
4
5
6
7
Class
Boundaries
46.75 – 47.65
47.65 - 48.55
48.55 – 49.45
49.45 – 50.35
50.35 – 51.25
51.25 – 52.15
52.15 – 53.05
Frequency
1
0
2
9
21
2
1
The histogram is as shown below:
Freq.
20
15
10
5
46.75
47.65
48.55
49.45
50.35
6
51.25
52.15
53.05
e. Stem-and-leaf display:
46*
46
47*
47
48*
48
49*
49
50*
50
51*
51
52*
52
2.7
a.
8
0
8
1
5
1
1
8
2
5
2
8 9
2 3 3 4 4
6 6 6 6 6 6 7 7 8 8 8 8 8 8 8
4
0 2
We wish to find P30.7  y  32.3 .
30.7  31.5  .8

 1
.8
.8
32.3  31.5 .8
z 32.3 
 1
.8
.8
P30.7  y  32.3  P 1  z  1
z 30.7 
 P 1  z  0  P0  z  1
 .3413  .3413
 .6826
b.
We wish to find P29.1  y  33.9
29.1  31.5  2.4

 3
.8
.8
33.9  31.5 2.4


3
.8
.8
z 29.1 
z 33.9
P29.1  y  33.9  P 3  z  3
 P 3  z  0  P0  z  3
 .4987  .4987
 .9974
c.
We wish to find P29.5  y  32.3
7
29.5  31.5  2

 2.5
.8
.8
32.3  31.5 .8

 1
.8
.8
z 29.5 
z 32.3
P29.5  y  32.3  P 2.5  z  1
 P 2.5  z  0  P0  z  1
 .4938  .3413
 .8351
d.
We wish to find P31.0  y  31.3
31.0  31.5  .5

 63
.8
.8
31.3  31.5  .2


 .25
.8
.8
z 31.0 
z 31.3
P .63  z  0  P .25  z  0
 .2357  .0987  .1370
e.
We wish to find P y  29.5
z 29.5 
29.5  31.5  2

 2.5
.8
.8
P y  29.5  P z  2.5
 1  P 2.5  z  0   1  .4938  .5  .0062
f.
We wish to find P y  29.5
g.
We wish to find P y  33.4
P y  29.5  P z  2.5
 .5  P 2.5  z  0   .5  .4938  .9938
z 33.4 
33.4  31.5 1.9

 2.38
.8
.8
P y  33.4  P z  2.38
 .5  P0  z  2.38  .5  .4913  .0087
2.8
h.
P  y  33.4  P z  2.38
 .5  P 0  z  2.38  .5  .4913  .9913
a.
z .05  1.645
b.
z .02  2.054
8
2.9
2.10
2.11
2.12
2.13
c.
z .01  2.327
d.
z .005  2.575
a.
t.705   1.895
b.
t.701   2.998
c.

t.7005
  3.499
a.
F.052,5  5.79
b.
F.055, 2   19.30
a.
x 2 .05 3  7.81473
b.
x 2 .01 2  9.21034
a.
P y  250,000  Pz  2.5
 .5  .4938  .0062
b.
P260,000  y  330,000  P 2  z  1.5
 .4772  .4332  .9104
c.
P y  346,000  Pz  2.3
 .5  .4893  .0107
a.
A 100 1    % confidence interval for  is
s
y  t n/12 
n
32.3  30.5  31.7  31.4  32.6
y
 31.7
5
2
2
2

32.3  31.7   30.5  31.7   31.7  31.7 
2
s 
5 1
2
31.4  31.7   32.6  31.7 2

5 1
2
.
7
s2 
 .675 and s  .675  .822
4
A 90% confidence interval for  is
y  t .405 
s
n

 .822 
  30.916, 32.484
31.7  2.132
 5 

A 95% confidence interval for  is
9

 .822 
 4   s  
  31.7  2.776

 y  t .025 
n
5





 
 30.68, 32.72
A 98% confidence interval for  is
y  t .401 
s
n

 .822 
  30.32, 33.08
31.7  3.747
5



A 99% confidence interval for  is

 .822 
  30.01, 33.39
31.7  4.604
 5 

For the first sample, we have y  31.2 and s  .7517 .
The 99% confidence interval for  is
2.14

y  t .4005

s
n

 .7517 
  29.65, 32.75
31.2  4.604
 5 

For the second sample, we have y  31.7 and s  .8093 .
The 99% confidence interval for  is

 .8093 
  30.03, 33.37
31.7  4.604
 5 

For the third sample, we have y  32.7 and s  .8276 .
The 99% confidence interval for  is

 .8276 
  31.0, 34.4
32.7  4.604
 5 

All three intervals contain  . If we consider the population of all possible samples of
size 5 that can be drawn from the population and the population
of all 99% confidence intervals for  calculated using these samples, 99% of the
confidence intervals in the population of all possible 99% confidence intervals
for  contain  .
2.15
a.
A 99% confidence interval for the mean assembly
time  is y  t n / 12
s
n
10

 2.19 
  10.69, 17.90 .
 6 
= 14.29  4.032

b.
2.16
Zenex Radio can be 99% confident that the mean assembly time is between 10.68
minutes and 17.90 minutes. Therefore, Zenex can be very sure (at least 99% confident)
that  is less than 20 minutes.
We have n  81, y  57.8, and s  6.02
a.
A 90% confidence interval for  is

 6.02 
s  

 y  z  / 2 
  57.8  1.645
n 

 81 
 57.8  1.645 .6689  57.8  1.10
 56.70, 58.90
A 95% confidence interval for  is
 57.8  1.96 .6689  57.8  1.31
 56.49, 59.11
A 98% confidence interval for  is
 57.8  2.33 .6689  57.8  1.56 or
 56.24, 59.36
A 99% confidence interval for  is
57.8  2.58 .6689  57.8  1.73
 56.07, 59.53
2.17
b.
National Motors can be at least 95% confident that  is less than 60 feet because the
upper endpoint of the 95% interval is less than 60.
c.
National Motors can be at least 98% confident that  is less than 60 feet because the
upper endpoint of the 98% interval is less than 60.
a.
An appropriate t-statistic is
t
y  16
s/ n

15.7665  16
.1524 / 6
 3.75
If we wish to test H 0 :   16 versus H a :   16 with   .05, the appropriate
rejection point is

tn/ 12  t.6051/ 2  t.5025
  2.571

Since | t | 3.75  t.5025
  2.571, we reject H 0 :   16 with   .05. The process
should be readjusted on the basis of this test.
11
b.
If we wish to test H 0 :   16 versus H a :   16 with   .01, the appropriate
rejection point is

tn/ 12  t.6011/ 2  t.5005
  4.032 . Since

| t | 3.75  t.5005
  4.032, we cannot reject
H 0 :   16 with   .01. The process should not be readjusted on the basis of this
test.
2.18
The appropriate t-statistic is
t
y  20
s/ n

14.29  20
2.19 / 6
 6.39
Since t  6.39  tn1  t.6051   2.015, we can
reject H 0 :   20 in favor of H a :   20 with   .05.
2.19
The appropriate z-statistic is
z
y  60
s/ n

57.8  60
6.02 / 81
 3.29
Since z  3.29   z     z .05  1.645, we can reject
H 0 :   60 in favor of H a :   60 with   .05.
The p-value is the area under the standard normal curve to the left of -3.29. Since the area to the
left of -3.09 is .5 - .499 = .001, the p–value is less than .001.
2.20
a.
H 0 :   3.5
H a :   3.5
b.
7467549
6
7
2
2
2
2
2
2
2

7  6   4  6   6  6   7  6   5  6   4  6   9  6 
2
s 
6
20

 3.3333
6
s  3.3333  1.8257
y
t
y  3.5
s/ 7

6  3.5
1.8257 / 7
 3.62
Since t  3.62  t.n011  t.701.1  3.143, we can
reject H 0 :   3.5 in favor of H a :   3.5
12