CHAPTER 11
STATISTICAL INFERENCE: TWO POPULATIONS
1.
The null and the alternative hypotheses are:
H0 : 1 = 2;
H1: 1 ≠ 2
This is a two-tailed test.
We have chosen α = 0.01.
The samples are chosen independently, the population variances are unknown and they are


_
 _

(X 1 X 2 )  0 

not known to be equal. Hence, we shall use as test statistic,

.
2
2
S
S
1

 2 

n
n2 
1

The populations are approximately normal. Hence, this is approximately a two -tailed ttest.
df =
s
2
1

/ n1  s22 / n2
 

2

2
 s2 / n 2
s22 / n2 
1
1



 n1  1
n2  1 


=

5
2
/ 40  6 2 / 50
 5 2 / 40

 39

  6
2
2

2
/ 50
49

2




= 87.835
For df ≈ 88, t0.005 = approximately 2.64.
Decision rule : reject H0 in favour of H1 if t < -2.64 or if t > 2.64.
t=
(102  99)  0
52 62

40 50
= 2.59.
Since 2.59 is between –2.64 and 2.64, there is insufficient evidence, at α = 0.01, to reject
H0 in favour of H1.
3.
The null and alternate hypotheses are:
H0 : 1 = 2;
H1: 1 ≠ 2
This is a two-tailed test.
We have chosen α = 0.05.
Samples are chosen independently, the population variances are unknown, but we are
assuming them to be equal. Hence, we shall use as test statistic
11-1


 _

_
 (X 1 X 2 )  0 
T= 
.
 S2  1  1  

p

 n1 n2  

The population distributions are approximately normal. Hence, this is a two-tailed t-test.
df = n1+n2-2 = 10 + 8 - 2= 16.
For df = 16, tα/2 = t0.025 = 2.12.
Hence, decision rule is: reject H0 in favour of H1 if t > 2.12 or if t < 2.12.
Pooled estimate of the population variance is,
(n1  1) s12  (n2  1) s22 (10  1)(4) 2  (8  1)(5) 2
s 

 19.9375
(n1  n2  2)
10  8  2
2
p
t
(23  26)  0
 1 1
19.9375  
 10 8 
 1.416.
Since –1.416 is between –2.12 and +2.12, we do not have sufficient evidence, at α = 0.05,
to reject H0.
5.
Let μ1 and μ2 be the population means of weight gains in infants using Gabbs’ products and
the competitors’ products, respectively, during first three month after birth. Then, the null
and alternative hypotheses are:
H0 : 1  2;
H1: 1 < 2
This is a lower-tailed test.
We have chosen α = 0.05.
The samples are chosen independently, the population variances are unknown and they are


_
 _

(X 1 X 2 )  0 

not known to be equal. Hence, we shall use as test statistic,

.
2
2
S
S
1
2





n
n
1
2


The populations are approximately normal. Hence, this is approximately a lower-tailed ttest.
df =
s
2
1

/ n1  s22 / n2
 

2

2
 s2 / n 2
s22 / n2 
1
1



 n1  1
n2  1 


=

(1.01)
2
/ 30  (1.3) 2 / 20
 (1.01) 2 / 30


29

  (1.3)
2
2

2
/ 20
19

2




= 33.784.
For df ≈ 34, tα = t0.05  1.691.
Decision rule : reject H0 in favour of H1 if the computed t-value is less than –1.691.
11-2
t=
(3.5  3.7)  0
=-0.581.
(1.01) 2 (1.3) 2

30
20
Since –0.581 > -1.691, we do not have sufficient evidence, at α = 0.05, to reject H0, that is
to accept the claim that the babies using the Gibbs brand gain less weight.
α = 1 – 0.99 = 0.01. For df = 34, t = t0.005  2.73. A 99% confidence interval estimate for
2
(μ1- μ2) is (3.5  3.7)  (2.73)
7.
(1.01) 2 (1.3) 2
= -0.2 + (2.73)(0.3442) = (-1.14, 0.74).

30
20
Let μ1 and μ2 be the population means of turnover rates of oil related stocks and other
stocks, respectively. Then, the null and alternative hypotheses are:
H0 : 1 = 2;
H1: 1 ≠ 2
This is a two-tailed test.
We have chosen α = 0.01.
The two samples are independent, the population variances are unknown and they are not
known to be equal. Hence, we shall use as test statistic,


_
 _

 (X 1 X 2 )  0  .


S12 S 22 



n1 n2 

The populations are approximately normal. Hence, this is a two-tailed t-test.
df =
s
2
1

/ n1  s22 / n2
 

2

2
 s2 / n 2
s22 / n2 
1
1



 n1  1
n2  1 


=

(5.1)
2
/ 32  (6.7) 2 / 49
 (5.1) 2 / 32


31

  (6.7)
2
2

2
/ 49
48

2




= 77.048
For df ≈ 77, tα/2 = t0.005  2.641.
Decision rule : reject H0 in favour of H1 if the computed t-value is less than -2.641 or if it
is greater than 2.641.
t=
(31.4  34.9)  0
(5.1) 2 (6.7) 2

32
49
= -2.66
Since -2.66 is less than -2.641, there is sufficient evidence, at α = 0.01, to reject H0, that is,
to infer that there is a difference in the mean turnover rates.
9.
Let the population means of grades of men and women in the examination be μm and μf,
respectively. Then, the null and alternative hypotheses are:
11-3
H0 : m ≥ f ;
H1: m < f
This is a lower-tailed test.
We have chosen α = 0.01.
Samples are chosen independently, and the population variances are unknown, but are
known to be equal. Hence, we shall use as test statistic


 _

_
 (X m X f )  0 
T= 
 .


 S2 1  1 

p

 n1 n2  

The population distributions are approximately normal. Hence, this is a lower-tailed t-test.
df = 9 + 7 - 2 = 14. For df = 14, tα = t0.01 = 2.624.
Hence, decision rule is: reject H0 in favour of H1 if the computed t-value is less than 2.624.
For the given sample data,
72  69    77
(72  78) 2    (77  78) 2
xm 
 78; s m 
 9.49
9
8
81  67    76
(81  79) 2    (76  79) 2
xf 
 79; s f 
 6.88
7
6
  n1  1 sm2   n2  1 s 2f  (9  1)(9.49) 2  (7  1)(6.88) 2
s 2p  
 71.749
 

n

n

2
9

7

2
1
2


(78  79)  0
t
 0.234.
1
1


71.749  
9 7
Since -0.234 is greater than -2.624, we do not have sufficient evidence, at α = 0.01,
to reject H0 in favour of H1, that is, to conclude that the mean grade of women is
higher.
11.
Let 1 and  2 be the population mean values of number of defective units on the day
shift and the afternoon shift, respectively. Let  D  1   2 .
The null and the alternative hypotheses are:
H0 : μD ≤ 0
H1 : μD > 0
This is an upper-tailed test.
We have chosen  = 0.05.
11-4
 D0   D 
  
 as the test statistic.
S
/
n
S
/
4
 D
  D

We shall use T = 

The population distributions are approximately normal. Hence, for D = 0, the distribution
of T is approximately student’s t-distribution with df = (n - 1) = 3.
So, this is an upper-tailed t-test.
For df = 3, tα = t0.05 = 2.353.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is greater than 2.353.
Day
Afternoon
d
10
8
2
12
9
3
15
12
3
19
15
4
233 4
(2  3) 2    (4  3) 2
d 
 3; s D 
 0.816.
4
3
3
t
 7.35
0.816 / 4
Since the computed t-value (=7.35) is greater than 2.353, there is sufficient evidence, at 
= 0.05, to reject H0 in favour of H1, that is, to conclude that there are less defective units
produced in the afternoon shift.
Let 1 and  2 be the population mean values of student weights (in lbs.) upon arrival on
13.
campus, and one year later, respectively. Let  D  1   2 .
The null and the alternative hypotheses are :
H0 : μD ≥ 0
H1 : μD < 0
This is a lower-tailed test.
We have chosen  = 0.01.
 D0  
D 
  
 .
 S D / n   S D / 11 
We shall use as test statistic, T = 

The population distributions are approximately normal. Hence, for D = 0, the distribution
of T is approximately student’s t-distribution with df = (n - 1) = 10.
So, this is a lower-tailed t-test.
For df = 10, tα = t0.01 = 2.764.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is less than –2.764.
For the given sample data, the sample values of D are:
On
arrival
After 1
year
d
124
157
98
190
103
135
149
176
200
180
256
142
157
96
212
116
134
150
184
209
180
269
-18
0
2
-22
-13
1
-1
-8
-9
0
-13
11-5
d
t
18  0 
11
 13
 7.3636; sD 
(18  7.3636) 2 
 (13  7.3636) 2
10
 8.37;
7.3636
 2.92
8.37 / 11
Since the computed t-value (=-2.92) is less than -2.764, there is sufficient evidence, at  =
0.01, to reject H0 in favour of H1, that is, to conclude that there is a weight gain in the first
year.
α = 1 – 0.95 = 0.05. For df = 10, t = t0.025 = 2.228. A 95% confidence interval estimate
2
for mean increase in weight during the first year is 7.3636  (2.228)(8.37 / 11) =
(1.74, 12.99).
15.
Let 1 and  2 be the population mean values of strength before and after the use of
special vitamins, respectively. Let  D  1   2 .
The null and the alternative hypotheses are
H0 : μD ≥ 0;
H1 : μD < 0.
This is a lower-tailed test.
We have chosen  = 0.01.
 D0  

D
  
 .
S
/
n
S
/
10
 D
  D

We shall use as test statistic, T = 

The population distributions are approximately normal. Hence, for D = 0, the distribution
of T is approximately student’s t-distribution with df = (n - 1) = 9.
So, this is a lower-tailed t-test
For df = 9, tα = t0.01 = 2.821.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is less than -2.821.
For the given sample data, the sample values of D are:
11-6
Before
After
d
86
89
-3
115
114
1
156
156
0
96
97
-1
52
51
1
57
58
-1
85
87
-2
53
54
-1
89
88
1
57
56
1
 3 11
(3  0.4) 2    (1  0.4) 2
d 
 0.4; s D 
 1.43
10
9
0.4
t
 0.885
1.43 / 10
Since the computed t-value (=-0.885) is greater than -2.821, we do not have sufficient
evidence, at  = 0.01, to reject H0 in favour of H1, that is, to conclude that there the special
vitamin increases the strength.
17.
The null and the alternative hypotheses are :
H0 : p1 ≤ p2
H1 : p1 > p2
This is an upper-tailed test.
We have selected  = 0.05.
The samples are selected independently. Therefore, we shall use as test statistic:
( pˆ1  pˆ 2 )  0
1 1
pˆ (1  pˆ )   
 n1 n2 
70
90
 0.7; pˆ 2 
 0.6.
100
150
n1 pˆ1  (100)(0.7)  5, n1 (1  pˆ1 )  (100)(0.3)  5,
n2 pˆ 2  (150)(0.6)  5, n2 (1  pˆ 2 )  (150)(0.4)  5.
pˆ1 
Hence, the sample sizes are large enough for the use of normal approximation. This is an
upper-tailed Z-test.
The decision rule is: reject H0 in favour of H1 if the computed z-value is greater than
z  z0.05  1.645 .
The pooled estimate of proportion is pˆ 
70  90
 0.64 .
100  150
70
90
 0.7; pˆ 2 
 0.6;
100
150
( pˆ1  pˆ 2 )
0.7  0.6
z

 1.614 .
1 
1 1
 1
(0.64)(0.36) 

pˆ (1  pˆ )   

 100 150 
 n1 n2 
pˆ1 
11-7
Since the computed z-value (=1.614) is less than 1.645, we do not have sufficient
evidence,  = 0.05, to reject H0 in favour of H1, that is to conclude that p1 is greater than
p2.
A 98 percent confidence interval estimate for (p1-p2) is:
^
^
^
( p1  p 2 )  z0.01
^
^
^
p1 (1  p1 ) p 2 (1  p 2 )

n1
n2
(0.7)(0.3) (0.7)(0.3)

100
150
 (0.1  0.141)  (0.041, 0.241).
 (0.7  0.6)  (2.326)
19.
Let p1 and p2 be, respectively, population fractions of vines treated with Action and Pernod
5, respectively, that get infested. Then, the null and alternative hypotheses are:
H0 : p1- p2 ≤ 0.02
H1 : p1-p2 > 0.02
This is an upper-tailed test.
We shall use as test statistic
( pˆ1  pˆ 2 )  0.02
pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )

n1
n2
40
24
 0.1; pˆ 2 
 0.06;
400
400
n1 pˆ1  (400)(0.1)  5; n1 (1  pˆ1 )  (400)(0.9)  5;
n2 pˆ 2  (400)(0.6)  5; n2 (1  pˆ 2 )  (400)(0.4)  5.
pˆ1 
Hence, the sample sizes are large enough for normal approximation. This is an upper-tailed
Z-test.
We have chosen α = 0.05; zα = z0.05 = 1.645.
The decision rule is: reject H0 in favour of H1 if the computed z-value is greater than
1.645.
z
(0.1  0.06)  0.02
= 1.045
(0.1)(0.9) (0.06)(0.94)

400
400
Since the computed z-value (= 1.045) is less than 1.645, we do not have sufficient
evidence, at α = 0.05, to conclude that (p1-p2) is greater than 0.02.
21.
Let ps and pm be, respectively, population proportions of single and married drivers insured
by Cambridge Insurance that had an accident during the three years period. Then, the null
and alternative hypotheses are:
11-8
H0 : ps - pm = 0
H1 : ps - pm ≠ 0
This is a two-tailed test.
We have selected  = 0.05.
We assume that the samples are selected independently, and therefore, we shall use as test
statistic:
( pˆ s  pˆ m )  0
1
1 
pˆ (1  pˆ )   
 ns nm 
120
150
 0.3; pˆ 2 
 0.25.
400
600
ns pˆ s  (400)(0.3)  5; ns (1  pˆ s )  (400)(0.7)  5;
nm pˆ m  (600)(0.25)  5; nm (1  pˆ m )  (600)(0.75)  5.
pˆ s 
Hence, the sample sizes are large enough for the use of normal approximation. This is a
two-tailed Z-test.
Zα/2 = z0.025 = 1.96.
The decision rule is: reject H0 in favour of H1 if the computed z-value is less than -1.96 or
if it is greater than 1.96.
The pooled estimate of proportion is: pˆ 
xs  xm 120  150

 0.27.
ns  nm 400  600
The value of the test statistic is
z
( pˆ s  pˆ m )
1 1 
pˆ (1  pˆ )   
 ns nm 

(0.3  0.25)
1 
 1
(0.27)(0.73) 


 400 600 
 1.745.
Since the computed z-value (= 1.745) is between -1.96 and 1.96, we do not have sufficient
evidence,  = 0.05, to reject H0 in favour of H1, that is to conclude that the values of the
two population proportions are different.
A 95% confidence interval estimate for the difference between the two population
proportions is: (0.3  0.25)  (1.96)
23.
(0.3)(0.7) (0.25)(0.75)
= (-0.0067, 0.1067).

400
600
Let 1 and 2 be the population means of useful life times (in months) of Cooper paint and
King paint, respectively. Then, the null and alternative hypotheses are:
11-9
H0 : 1 = 2
H1 : 1 ≠ 2
This is a two-tailed test.
We have chosen  = 0.01.
The samples are chosen independently, the population variances are unknown and they are




( X1  X 2 )  0 

not known to be equal. Hence, we shall use as test statistic,

.
2
2
S
S
1

 2 

n
n2 
1

The population distributions are approximately normal and the sample sizes are large
enough. Hence, this is a two-tailed t-test
df =
s
2
1

/ n1  s22 / n2
 

2

2
 s2 / n 2
s22 / n2 
1
1



 n1  1
n2  1 


=

(1.14)
2
/ 35  (1.3) 2 / 40
 

2
 (1.14) 2 / 35 2 (1.3) 2 / 40



34
39


2




= 72.9988
For df ≈ 73, tα = t0.005  2.651.
Decision rule : reject H0 in favour of H1 if the computed t-value is less than -2.651 or if it
is greater than 2.651.
t=
(36.2  37.0)  0
(1.14) 2 (1.3) 2

35
40
= -2.839
Since –2.839 is less than –2.651, there is sufficient evidence, at  = 0.01, to reject H0, that
is, to infer that the mean useful lives of the two paints are different.
α = 1 – 0.95 = 0.05. For df = 73, tα/2 = t0.025 = approximately 1.996.
A 95 percent confidence interval estimate for the difference ( 1  2 ) is
_
_
( x1  x 2 )  t 2
 s12 s22 
(1.14)2 (1.3)2

    (36.2  37.0)  (1.996)
35
40
 n1 n2 
= (-0.8 + 0.562) = (-1.362, -0.238).
25.
Let 1 and 2 be the population means of costs of health benefit packages (in terms of
percentage of salary), for employees, offered by large and small firms, respectively. Then,
the null and alternative hypotheses are:
11-10
H0 : 1 = 2
H1: 1 ≠ 2
This is a two-tailed test.
We have chosen  = 0.05.
The samples are chosen independently, the population variances are unknown and they are




( X1  X 2 )  0 

not known to be equal. Hence, we shall use as test statistic,

.
2
2
S
S
1

 2 

n
n2 
1

The populations are approximately normal and the sample sizes are large enough. Hence,
this is a two-tailed t-test
df =

(2.6)
2
/ 15  (3.3) 2 / 12
 

2
 (2.6) 2 / 15 2 (3.3) 2 / 12



14
11

For df ≈ 21, t = t0.025 = 2.08.
2

2




= 20.6389
Decision rule is : reject H0 in favour of H1 if the computed t-value is less than -2.08 or if it
is greater than 2.08.
t=
(17.6  16.2)  0
(2.6) 2 (3.3) 2

15
12
= 1.2013
Since 1.2013 is between –2.08 and 2.08, we do not have sufficient evidence, at α = 0.05, to
conclude that the population means of percent of employee salaries, spent by large and
small firms on health benefits, are different.
27.
Let 1 and 2 be the population mean values of number of hamburgers sold per day at
northside site and the southside site, respectively. Then, the null and alternative hypotheses
are:
H0 : 1 = 2;
H1 : 1 ≠ 2.
This is a two-tailed test.
We have chosen  = 0.05.
11-11
The samples are chosen independently, the population variances are unknown and they are




( X1  X 2 )  0 

not known to be equal. Hence, we shall use as test statistic,

.
2
2
S
S
1

 2 

n
n2 
1

The populations are approximately normally distributed and the sample sizes are large
enough. Hence, this is a two-tailed t-test.
df =
s
2
1
/ n1  s22 / n2

 

2

2
 s2 / n 2
s22 / n2 
1
1



 n1  1
n2  1 


=

(10.5)
2
/ 10  (14.25) 2 / 12
 (10.5) 2 / 10


9

 
2

2

(14.25) / 12 


11

2
2
=19.7541
For df ≈ 20, tα/2 = t0.025 = 2.086.
Decision rule is: reject H0 in favour of H1 if the computed t-value is less than -2.086 or if it
is greater than 2.086.
t=
(83.55  78.8)  0
2
(10.5)
(14.25)

10
12
= 0.8985
2
Since 0.8985 is between –2.086 and 2.086, we do not have sufficient evidence, at α = 0.05,
to conclude that the mean number of hamburgers sold at the two locations are different.
29.
Let 1 and 2 be the population means of amounts purchased on impulse per customer at
stores on Peach street and Plum street, respectively. Then, the null and alternative
hypotheses are:
H0 : 1 = 2
H1 : 1 ≠ 2
This is a two-tailed test.
We have chosen  = 0.01.
The two samples are independent, and the population variances are unknown, but they are




 ( X1  X 2 )  0 
known to be equal. Hence, we shall use as test statistic, 
.


 S2 1  1 

p

 n1 n2  

The populations are approximately normally distributed. Hence, this is a two-tailed t-test.
11-12
For df = 10 + 14 - 2 = 22, tα/2 = t0.005 = 2.819.
Decision rule is : reject Ho if the computed t-value is less than -2.819 or if it is greater than
2.819 .
For the given sample
data,
x1 
x2 
17.58 
 15.85
10
18.19 
 15.87; s1 
(17.58  15.87)2 
 14.83
 18.29; s2 
14
9(2.33) 2  13(2.55) 2
s 2p 
 6.06.
10  14  2
(15.87  18.29)  0
t
 2.374.
1 1
6.06  
 10 14 
 (15.85  15.87)2
9
(18.19  18.29) 2 
 (14.83  18.29) 2
13
 2.33
 2.55
Since –2.374 is between –2.819 and +2.819, we do not have sufficient evidence, at α =
0.01, to reject H0, that is, to infer that the mean amounts, purchased on impulse at the two
stores, are different.
31.
Let 1 and 2 be the population means of number of units sold at discount and regular
prices, respectively. Then, the null and alternative hypotheses are:
H0 : 1 ≤ 2
H1 : 1 > 2
This is an upper-tailed test.
We have chosen  = 0.01.
The two samples are independent, and the population variances are unknown, but they are




 ( X1  X 2 )  0 
known to be equal. Hence, we shall use as test statistic, 
.
 S2  1  1  

p

 n1 n2  

The populations are approximately normally distributed. Hence, this is an upper-tailed ttest.
For df = 8 + 7 - 2 = 13, tα = t0.01 = 2.65.
Decision rule is: reject Ho in favour of H1 if the computed t- value is greater than 2.65.
For the given sample data,
11-13
x1 
x2 
128 
 120
8
138 
 125.125; s1 
 96
(128  125.125) 2 
 (120  125.125) 2
8
(138  117.714)2 
 117.714; s2 
7
7(15.094) 2  6(19.914) 2
s 2p 
 305.708.
872
(125.125  117.714)  0
t
 0.819.
1 1
305.708  
8 7
 (96  117.714)2
6
 15.094.
 19.914
Since the computed t-value (=0.819) is less than 2.65, we do not have sufficient evidence,
at α = 0.01, to reject H0, that is, to infer that price reduction resulted in an increase in mean
value of sales.
33.
Let 1 and  2 be the population mean values of student grades under the quarter system
and the semester system, respectively. Let  D  1   2 .
Then, the null and the alternative hypotheses are:
H0 : μD ≤ 0
H1 : μD > 0
This is an upper-tailed test.
We have chosen  = 0.05.
We have chosen a matched paired sample. Hence, we shall use T =
 D0  

D

  
 as the test statistic.
 S D / n   S D / 10 
The population distributions are approximately normal. Hence, for D = 0, the distribution
of T is approximately student’s t-distribution with df = (n - 1) = 9.
So, this is an upper-tailed t-test.
For df = 9, t = t0.05 = 1.833.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is greater than 1.833.
For the given sample data, the sample values of D are:
last fall
this fall
d
2.98
3.17
-0.19
2.34
2.04
0.30
3.68
3.62
0.06
3.13
3.19
-0.06
11-14
3.34
2.90
0.44
2.09
2.08
0.01
2.45
2.88
-0.43
2.96
3.15
-0.19
2.80
2.49
0.31
4.00
2.98
0.02
0.19   0.02
 0.027;
10
d
(0.19  0.027) 2 
sD 
t
 (0.02  0.027) 2
9
0.027  0
0.2661 / 10
 0.2661.
 0.321
Since the computed t-value (=0.321) is less than 1.833, we do not have sufficient evidence,
at α = 0.05, to conclude that student grades declined after the conversion.
35.
Let 1 and  2 be the population mean values of annual family incomes (in thousands of
dollars) of potential buyers at the first and second developments, respectively. Let
 D  1  2 .Then, the null and the alternative hypotheses are:
H0 : 1 = 2
H1: 1 ≠ 2
This is a two-tailed test.
We have chosen  = 0.05.
The samples are chosen independently, the population variances are unknown and they are
not known to be equal. Hence, we shall use as test statistic,




 ( X 1  X 2 )  ( 1   2 )  .


S12 S22





n
n
1
2


The population distributions are approximately normal. Hence, this is a two-tailed t-test.
df =
s
2
1

/ n1  s22 / n2
 

2

2
 s2 / n 2
s22 / n2 
1
1



 n1  1
n2  1 


=

(40)
2
/ 75  (30) 2 / 120
 (40) 2 / 75


74

  (30)
2
2

2
/ 120
119

2




= 125.5294
For df ≈ 126, tα/2 = t0.025  1.979.
Decision rule is: reject H0 in favour of H1 if the computed t-value is less than -1.979 or if it
is greater than 1.979.
t=
(150  180)  0
(40) 2 (30) 2

75
120
= -5.587.
Since the computed t-value (= –5.587) is less than -1.979, there is sufficient evidence, at α
= 0.05, to reject H0 in favour of H1, that is, to infer that the two population means are
different.
11-15
37.
Let 1 and  2 be the population mean values of contamination measurements, before and
after the use of new soap, respectively. Let  D  1   2 .
The null and the alternative hypotheses are:
H0 : μD ≤ 0
H1 : μD > 0
This is an upper-tailed test.
We have chosen  = 0.05.
Since the sample selected is a matched pair sample, we shall use T =
 D0   D 

  
 as the test statistic.
S
/
n
S
/
2
 D
  D

The population distributions are approximately normal. Hence, for D = 0, the distribution
of T is approximately student’s t-distribution with df = (n - 1) = 7.
So, this is an upper-tailed t-test.
For df = 7, t = t0.05 = 1.895.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is greater than 1.895.
For the given sample data, the sample values of D are:
Before
After
d
6.6
6.8
-0.2
6.5
2.4
4.1
9.0
7.4
1.6
10.3
8.5
1.8
11.2
8.1
3.1
8.1
6.1
2.0
6.3
3.4
2.9
11.6
2.0
9.6
 0.2    9.6
(0.2  3.11) 2    (9.6  3.11) 2
d 
 3.11; s D 
 2.91.
8
7
3.11  0
t
 3.02.
2.91 / 8
Since the computed t-value (= 3.02) is greater than 1.895, there is sufficient evidence, at α
= 0.05, to reject H0 in favour of H1, that is, to infer that the contamination measurements
are lower after the use of the new soap.
39.
Let p1 and p2 be the population proportions of current Advil users who, when given the
new drug and the old drug, respectively, would indicate that the drug given is more
effective. Then, the null and alternative hypotheses are:
H0 : p1-p2 ≤ 0
H1 : p1-p2 > 0
This is an upper-tailed test.
We have selected  = 0.05.
11-16
The samples are selected independently, and therefore, we shall use as test statistic:
( pˆ1  pˆ 2 )
1 1
pˆ (1  pˆ )   
 n1 n2 
180
261
 0.9; pˆ 2 
 0.87.
200
300
n1 pˆ1  (200)(0.9)  5, n1 (1  pˆ1 )  (200)(0.1)  5,
n2 pˆ 2  (300)(0.87)  5, n2 (1  pˆ 2 )  (300)(0.13)  5.
pˆ1 
Hence, the sample sizes are large enough for the use of normal approximation. This is an
upper-tailed Z-test.
The decision rule is: reject H0 in favour of H1 if the computed z-value is greater than z =
z0.05 = 1.645.
The pooled estimate of proportion is: pˆ 
z 
( pˆ1  pˆ 2 )
1 1
pˆ (1  pˆ )   
 n1 n2 
x1  x2 180  261

 0.882 .
n1  n2 200  300
0.9  0.87
1 
 1
(0.882)(0.118) 


 200 300 
 1.019
Since 1.019 < 1.645, we do not have sufficient evidence, at α = 0.05, to reject H0 in favour
H1, that is, to infer that the new drug is more effective.
41.
Let p1 and p2 be the population proportions of college women and men, respectively, who
smoke at least a pack of cigarettes a day.
Then, the null and alternative hypotheses are:
H0 : p1-p2 = 0
H1 : p1-p2 ≠ 0
This is a two-tailed test.
We have selected  = 0.05.
The samples are selected independently, and therefore, we shall use as test statistic:
( pˆ1  pˆ 2 )
1 1
pˆ (1  pˆ )   
 n1 n2 
72
70
 0.18; pˆ 2 
 0.14.
400
500
n1 pˆ1  (400)(0.18)  5, n1 (1  pˆ1 )  (400)(0.82)  5,
n2 pˆ 2  (500)(0.14)  5, n2 (1  pˆ 2 )  (500)(0.86)  5.
pˆ1 
11-17
Hence, the sample sizes are large enough for the use of normal approximation. This is a
two-tailed Z-test.
The decision rule is: reject H0 in favour of H1 if the computed z-value is less than -z/2 = z0.025 = -1.96 or if it is greater than 1.96.
The pooled estimate of proportion is: pˆ 
z 
( pˆ1  pˆ 2 )
1 1
pˆ (1  pˆ )   
 n1 n2 
x1  x2
72  70

 0.1578
n1  n2 400  500
0.18  0.14
1 
 1
(0.1578)(0.8422) 


 400 500 
 1.636.
Since 1.636 is between –1.96 and +1.96, we do not have sufficient evidence, at α = 0.05, to
reject H0, that is, to infer that the two population proportions are different.
Let 1 and  2 be the population mean values of stock prices of the top 100 companies on
43.
November 2001 and at present, respectively. Let  D  1   2 .
The null and the alternative hypotheses are:
H0 : μD = 0
H1 : μD ≠ 0
This is a two-tailed test.
We have chosen  = 0.05.
 D0  
D 
  

 S D / n   S D / 10 
The sample is a matched-pair sample. Hence, we shall use T = 

as the test statistic.
The population distributions are approximately normal. Hence, for D = 0, the distribution
of T is approximately student’s t-distribution with df = (n - 1) = 9.
So, this is a two-tailed t-test.
For df = 9, t/2 = t0.025 = 2.262.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is less than -2.262 or
if it is greater than 2.262.
Collect the sample data, compute d , s D , and t 
45.
(a)
d
s D / 10
, and draw the conclusion.
Let 1 = mean selling price of homes without pool; 2 = mean selling price of homes with
pool. Then, the null and the alternative hypotheses are :
H0 : 1 = 2
H1 : 1 ≠ 2
11-18
By sorting the data according to the values of the “Pool” variable, separating the data on
houses without pool (value of Pool = 0) from the data on houses with pool (value of
Pool=1), and using Minitab and Microsoft Excel we get the following outputs.
MINITAB OUTPUT
Two-sample T for C1 vs C2
C1
C2
N
38
67
Mean
202.8
231.5
StDev
33.7
50.6
SE Mean
5.5
6.2
Difference = mu C1 - mu C2
Estimate for difference: -28.69
95% CI for difference: (-45.06, -12.32)
T-Test of difference = 0 (vs not =): T-Value = -3.48
P-Value = 0.001
DF = 100
MICROSOFT EXCEL OUTPUT
t-Test: Two-Sample Assuming Unequal Variances
Variable 1 Variable 2
Mean
202.7974 231.4851
Variance
1136.031 2557.258
Observations
38
67
Hypothesized Mean Difference
0
df
100
t Stat
-3.47727
P(T<=t) one-tail
0.000376
t Critical one-tail
1.660235
P(T<=t) two-tail
0.000751
t Critical two-tail
1.983972
From both the outputs, we see that the P-value is approximately 0.001. 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 of selling prices of houses with and without a
pool are different.
(b)
Let 1 = mean selling price of homes without garage; 2 = mean selling price of homes
with garage. Then, the null and the alternative hypotheses are :
H0 : 1 = 2
H1 : 1 ≠ 2
By sorting the data according to the values of the “garage” variable, separating the data on
houses without attached garage (value of garage = 0) from the data on houses with
attached garage (value of garage =1), and using Minitab and Microsoft Excel we get the
following outputs.
11-19
MINITAB OUTPUT
Two-sample T for C1 vs C2
C1
C2
N
34
71
Mean
185.5
238.2
StDev
28.0
44.9
SE Mean
4.8
5.3
Difference = mu C1 - mu C2
Estimate for difference: -52.73
95% CI for difference: (-66.96, -38.49)
T-Test of difference = 0 (vs not =): T-Value = -7.35
P-Value = 0.000
DF = 95
MICROSOFT EXCEL OUTPUT
t-Test: Two-Sample Assuming Unequal Variances
Mean
Variance
Observations
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Variable 1 Variable 2
185.45 238.1761
784.2571 2013.896
34
71
0
96
-7.35212
3.25E-11
1.660883
6.5E-11
1.984986
From both the outputs, we see that the P-value is almost 0. 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 of selling prices of houses with and without an attached garage
are different.
(c)
Let 1 = mean selling price of homes in township 1; 2 = mean selling price of homes in
township 2. Then, the null and the alternative hypotheses are :
H0 : 1 = 2
H1 : 1 ≠ 2
By sorting the data according to the values of the “township” variable, separating the data
on houses in township 1 (value of township = 1) from the data on houses in township 2
(value of township = 2), and using Minitab and Microsoft Excel we get the following
outputs.
11-20
MINITAB OUTPUT
Two-sample T for C1 vs C2
C1
C2
N
15
20
Mean
196.9
227.5
StDev
35.8
44.2
SE Mean
9.2
9.9
Difference = mu C1 - mu C2
Estimate for difference: -30.5
95% CI for difference: (-58.1, -3.0)
T-Test of difference = 0 (vs not =): T-Value = -2.26
P-Value = 0.031
DF = 32
MICROSOFT EXCEL OUTPUT
t-Test: Two-Sample Assuming Unequal Variances
Variable 1 Variable 2
Mean
196.9133
227.45
Variance
1280.498 1953.054
Observations
15
20
Hypothesized Mean Difference
0
df
33
t Stat
-2.25722
P(T<=t) one-tail
0.015368
t Critical one-tail
1.69236
P(T<=t) two-tail
0.030736
t Critical two-tail
2.034517
From the computer outputs, we see that the P-value is approximately 0.031. 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 of selling prices of houses in townships
1 and 2 are different.
47.
Let 1 = the mean value of earnings per share of the top 1000 companies during the year
2000; and let 2 = the mean value of earnings per share of the top 1000 companies during
the year 1999. Then, the null and the alternative hypotheses are:
H0 : 1 ≤ 2
H1 : 1 > 2
The Minitab and Microsoft Excel outputs are given below.
MINITAB OUTPUT
Paired T for C1 - C2
C1
C2
Difference
N
100
100
100
Mean
0.939
0.785
0.155
StDev
1.826
1.466
1.234
SE Mean
0.183
0.147
0.123
95% lower bound for mean difference: -0.050
T-Test of mean difference = 0 (vs > 0): T-Value = 1.25
11-21
P-Value = 0.107
MICROSOFT EXCEL
OUTPUT
t-Test: Paired Two Sample for Means
Variable 1 Variable 2
Mean
0.9394
0.7848
Variance
3.333123 2.14933
Observations
100
100
Pearson Correlation
0.739906
Hypothesized Mean Difference
0
df
99
t Stat
1.253294
P(T<=t) one-tail
0.106525
t Critical one-tail
1.660392
P(T<=t) two-tail
0.21305
t Critical two-tail
1.984217
From the computer outputs, we see that the P-value is approximately 0.107. Since the
selected value of α (= 0.05) is less than the P-value, we do not have sufficient evidence to
reject H0, that is, to infer that the mean value for the year 2000 was higher than that for the
year 1999.
11-22