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CHAPTER 10
ONE SAMPLE TESTS OF HYPOTHESIS
1.
(a)
Since the values of population mean, , corresponding to H1 (  50) lie on both sides of
the values of  corresponding to H0, ( = 50), this is a two-tailed test.
(b)
Since the sample size is large enough, and the value of the population standard deviation,
, is known, we shall use as test statistic Z 
X  0

n

X  50
. This is a two-tailed Z5 36
test. The decision rule is: reject H0 in favour of H1 if the computed z-value is less than -z =
-z0.025 = -1.96 or if it is greater than z0.025 = 1.96.
3.
49  50
 - 1.2.
5 36
(c)
The value of the test statistic is z 
(d)
The computed z-value (= -1.2), lies between –1.96 and 1.96. Hence, we shall conclude that
we do not have sufficient evidence at  = 0.05, to reject H0 in favour of H1.
(e)
P-value = 2 ( P (Z < -1.2 ) = 2 ( 0.5 – 0.3849 ) = 0.2302.
Our decision will be to reject H0 in favour of H1 for any value of  greater than 0.2302.
(a)
Since the values of population mean, , corresponding to H1 ( > 20) are greater than the
values of  corresponding to H0, ( ≤ 20) this is an upper- tailed test.
(b)
The sample size (n = 49) is large enough for normal approximation and the value of
population standard deviation, , is known. Hence, we shall use as test statistic Z =
X  0

n

X  20
.
6 49
The value of  is given as 0.05. The decision rule is: reject H0 in favour of H1 if the
computed z-value is greater than z = z0.05 = 1.645.
5.
x  0
The value of the test statistic is z 
(d)
The computed z-value (= 1.167) is less than 1.645. Hence, we conclude that we do not
have sufficient evidence, at  = 0.05, to reject H0 in favour of H1.
(e)
The P-value = P (Z > 1.167)  (0.5 – 0.3784) = 0.1216.
Our decision would be to reject H0 in favour of H1 for any value of  greater than 0.1216.
a.

n
The null and the alternative hypotheses are:
Ho:   80000
H1:  < 80000
10-1

21  20
 1.167.
6 49
(c)
b.
Since the values of  corresponding to H1 (≤ 80000) are less than the values of 
corresponding to H0 ( 80000), this is a lower-tailed test.
We shall use as test statistic
X  0


n
X  80000
.
8000 48
The sample size (n = 48) is large enough for normal approximation. Hence, this is a lowertailed Z-test. .
The value of  is given as 0.05. The decision rule is: reject H0 in favour of H1 if the
computed z-value is less than -z = -z0.05 = -1.645.
(c)
7.
The value of the z-statistic is z 
79200  80000
8000
48
 0.693 .
(d)
Since, the computed z-value (= -0.693) is greater than –1.645, we conclude that we do not
have sufficient evidence, at  = 0.05, to reject H0 in favour of H1, that is, to conclude the
manufacturer’s claim is wrong.
(e)
The P-value is P (Z < -0.693)  (0.5 – 0.2558) = 0.2442.
Our decision would be to reject H0 in favour of H1 for any value of  greater than 0.2442.
(a)
The null and the alternative hypotheses are :
H0 :  = 6.0
H1 :  ≠ 6.0
(b)
The probability of type I error is  = 0.05.
(c)
We shall use as test statistic
X  0

n

X 6
.
0.5 n
Since the population distribution is approximately normal, this is a Z statistic.
(d) The test is a two-tailed Z-test. Our decision rule is: reject H0 in favour of H1 if the computed zvalue is less than -z/2 = -z0.025 = –1.96 or greater than z0.025 = 1.96.
(e) (i) The computed z-value is z 
5.84  6
0.5
64
 2.56 . The z-value is less than –1.96. Hence, we
conclude that there is sufficient evidence, at  = 0.05, to reject H0, that is, to infer that the mean
turnover rate has changed.
10-2
(ii) The P-value is 2 (P (Z > 2.56) = 2 (0.5 – 0.4948) = 0.0104. We would reject H0 for any
value of  greater than 0.0104.
(iii) The decision rule could be equivalently stated as: reject H0 if X is less than
 0.5 
 0.5 
6.0  (1.96)
  5.8775 or if X is greater than 6.0  (1.96)
  6.1225
 64 
 64 


 
0.5 
  Normal  5.9,

If the true value of  is 5.9, then X ~ Normal   ,
n
64 


z-value corresponding to x = 5.8775, is z 
z-value corresponding to x = 6.1225, is z 
5.8775  5.9
0.5 64
6.1225  5.9
0.5
64
 0.36
 3.56
Probability of type II error =  = P(-0.36 ≤ Z ≤ 3.56) = P(-0.36 ≤ Z ≤ 0) + P(0 ≤ Z ≤ 3.56)
= (approximately) 0.1406 + 0.5 = 0.6406
Power of the test = (1 - ) = (approximately) (1 – 0.6406) = 0.3594.
9.
(a)
This is an upper-tailed test.
The value of  is given as 0.05.
 X   0   X  10 

 . Since the population distribution
 S n   S 10 
is approximately normal, the distribution of the test statistic, when  = 10, is approximately
We shall use as test statistic T  
student’s t-distribution with df = (n –1). Thus, this is an upper-tailed t-test.
For df = 9, t = t0.05 = 1.833.
Hence, our decision rule is: reject H0 in favour of H1 if the computed t-value is greater than
1.833.
11.
12  10
 2.108.
3 10
(b)
The value of test statistic is t 
(c)
The computed t-value (= 2.108) is greater than 1.833. Hence, we shall conclude that there
is sufficient evidence, at  = 0.05, to reject H0 in favour of H1.
(a)
Let the mean number of videos college students watched last month be . Then, the null
and the alternative hypotheses are:
H0 :   6.8
H1 :  < 6.8
(b)
The value of  is given as 0.05.
10-3
 X   0   X  6.8 

 . Since the population
S
n
S
36

 

distribution is approximately normal, the distribution of the test statistic, when  = 6.8, is
We shall use as test statistic T  
approximately student’s t-distribution with df = (n –1) = 35. Thus, this is a lower-tailed ttest.
For df = 35, t = t0.05 = (approximately)1.69. Hence, our decision rule is: reject H0 in
favour of H1 if the computed t-value is less than -t0.05  -1.69.
(c)
(d)
13.
The computed t-value is t 
6.2 - 6.8
0.5
36
 7.2 .
Since the computed t-value (= -7.2) is less than –1.69, we conclude that there is sufficient
evidence, at  = 0.05, to reject H0 in favour of H1, that is, to infer that the college students,
on average, watched less than 6.8 videos last month.
Let the mean life of the modified battery be . Then, the null and the alternative
hypotheses are:
H0:   305
H1:  > 305
The value of  is given as 0.05.
 X   0   X  305 

 . Since the population
 S n   S 20 
We shall use as test statistic T  
distribution is approximately normal, the distribution of the test statistic, when  = 305, is
approximately student’s t-distribution with df = (n – 1) = 19. Thus, this is an upper-tailed ttest.
For df = 19, t = t0.05 = 1.729.
Hence, our decision rule is: reject H0 in favour of H1 if the computed t-value is greater than
t0.05 = 1.729.
The computed t-value is t 
311 - 305
12
20
 2.236 .
Since 2.236 > 1.729, there is sufficient evidence, at  = 0.05, to reject H0, that is, to infer
that the modification increased the mean battery life to more than 305 days.
15.
Let the population mean of assembly times (in minutes) using the new method be . Then,
the null and the alternative hypotheses are:
H0 :   42.3
H1 :  < 42.3
We have selected  = 0.1.
 X   0   X  42.3 

 . Since the population
 S n   S 24 
distribution is approximately normal, the distribution of the test statistic, when  = 42.3, is
We shall use as test statistic T  
10-4
approximately student’s t-distribution with df = (n – 1) = 23. Thus, this is a lower-tailed ttest.
For df = 23, t = t0.1 = 1.319.
Hence, our decision rule is: reject H0 in favour of H1 if the computed t-value is less than t0.1 = -1.319.
The computed t-value is t 
40.6 - 42.3
2.7
24
 3.085
Since –3.085 < –1.319, we conclude that there is sufficient evidence, at  = 0.1, to reject
H0 in favour of H1, that is, to infer that the new assembling method is faster.
17.
Let the population mean of service times (in minutes) be . Then, the null and the
alternative hypotheses are:
H0 :   15
H1 :  > 15
This is an upper-tailed test.
We have selected  = 0.05.
 X   0   X  15 

 . Since the population distribution
 S n   S 21 
is approximately normal, the distribution of the test statistic, when  = 15, is approximately
We shall use as test statistic T  
student’s t-distribution with df = (n –1) = 20. Thus, this is an upper-tailed t-test.
For df = 20, t = t0.05 = 1.725.
Hence, our decision rule is: reject H0 in favour of H1 if the computed t-value is greater than
t0.05 = 1.725.
The computed t-value is t 
18 - 15
1
21
 13.748 .
Since t > 1.725, we conclude that there is sufficient evidence, at  = 0.05, to infer that the
advertised claim is incorrect.
19.
Let the mean weight (in kg) of the chickens after the special additive is added to the feed,
be . Then, the null and the alternative hypotheses are:
H0 :   1.9
H1:  > 1.9
This is an upper-tailed test.
We have selected  = 0.01.
 X   0   X  1.9 

 . Since the population
 S n   S 10 
distribution is approximately normal, the distribution of the test statistic, when  = 1.9, is
We shall use as test statistic T  
approximately student’s t-distribution with df = (n –1) = 9. Thus, this is an upper-tailed ttest.
For df = 9, t = t0.01 = 2.821.
Hence, our decision rule is: reject H0 in favour of H1 if the computed t-value is greater than
t0.01 = 2.821.
For the given sample data,
10-5
x
s
2.01  1.92    1.98
 1.934
10
2.01  1.9342  1.92  1.9342    1.98  1.9342
The computed t-value is t 
9
1.934 - 1.9
0.05 10
 0.05
 2.15 .
Since t < 2.821, we conclude that there is insufficient evidence, at  = 0.01, to infer that
the special additive increased mean weight of chickens.
P-value is the area under the t-curve (corresponding to df = 9) to the right of 2.15. From
the t-table, we see that the P-value is between 0.025 and 0.05 (using Minitab, we get Pvalue = 0.03).
21.
Let the population mean of number of trout caught per day be . Then, the null and the
alternative hypotheses are:
H0 :   4.0
H1 :  > 4.0
This is an upper-tailed test.
We have selected  = 0.05.
 X  0   X  4 

 . Since the population distribution
S
n
S
12

 

is approximately normal, the distribution of the test statistic, when  = 4, is approximately
We shall use as test statistic T  
student’s t-distribution with df = (n –1) = 11. Thus, this is an upper-tailed t-test.
For df = 11, t = t0.05 = 1.796.
Hence, our decision rule is: reject H0 in favour of H1 if the computed t-value is greater than
t0.05 = 1.796.
For the given sample data,
x
s
4  4   6
 4.5
12
4  4.52  4  4.52    6  4.52
11
The computed t-value is t 
4.5 - 4
2.68 12
 2.68
 0.65
Since t < 1.796, 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 population mean is greater than 4.0.
P-value is the area under the t-curve (corresponding to df = 11) to the right of 0.65.
From the t-table, we see that the P-value is greater than 0.1 (using Minitab, we get P-value
= 0.2645).
10-6
23.

 

pˆ  p0
pˆ  0.7
  
 .
ˆ
ˆ
(
p
)(1

p
)
/
n
(0.7)(0.3)
/100

 

We shall use as test statistic 

n p0 = (100)(0.7) > 5 and n (1- p0) = (100)(0.3) > 5.
Hence, if p = 0.7, the distribution of the test statistic is approximately standard normal. We
shall therefore apply an upper-tailed Z-test.
We have selected  = 0.05. Our decision rule is: reject H0 in favour of H1 if the computed
value of the test statistic (the z-value) is greater than z = z0.05 = 1.645.
0.75  0.70
The computed z-value is z 
= 1.09.
0.70(0.30)
100
Since the computed z-value (= 1.09) is less than 1.645, we conclude that we do not have
sufficient evidence, at  = 0.05, to reject H0 in favour of H1.
25.
The null and the alternative hypotheses are:
H0 : p  0.52
H1 : p > 0.52
For the sample size n = 300, n p0 = (300)(0.52) > 5 and n (1- p0) = (300)(0.48) > 5. Hence,
the sample size is large enough for normal approximation. We shall use as test statistic Z =


pˆ  0.52

 .
 (0.52)(0.48) / 300 
This is an upper-tailed Z-test.
We have selected  = 0.01.
Our decision rule is: reject H0 in favour of H1 if the computed z-value is greater than z =
z0.01 = 2.326.
pˆ 
170
 0.567.
300
The computed z-value is z =
pˆ  0.52

(0.52)(0.48) / 300
0.567-0.52
 0.52 0.48
= 1.62.
300
Since the computed z-value is less than 2.326, we conclude that we do not have sufficient
evidence, at  = 0.01, to reject H0 in favour of H1, that is, to infer that proportion of men,
driving yesterday on the Highway 401, was larger than 0.52.
27.
The null and the alternative hypotheses are:
H0 : p  0.90
H1 : p < 0.90
For the sample size n = 100, n p0 = (100)(0.9) > 5 and n (1- p0) = (100)(0.1) > 5. Hence, the
sample size is large enough for normal approximation. We shall use as test statistic Z =


pˆ  0.9

 .
 (0.9)(0.1) /100 
This is a lower-tailed Z-test.
We have selected  = 0.1; z = z0.1 = (approximately) 1.281.
Our decision rule is: reject H0 in favour of H1 if the computed z-value is less than -z = -z0.1
 -1.281.
10-7
pˆ 
82
 0.82.
100
The computed z-value is z =
0.82-0.9
 0.9  0.1 100
= -2.67.
Since the computed z-value (-2.67) is less than –1.281, we conclude that we have sufficient
evidence at  = 0.1 to reject H0 in favour of H1, that is, to infer that proportion of orders
delivered in less than 10 minutes is less than 0.9.
29.
Let the average weight loss (in kg) in the first 2 weeks among those who join the new
weight reduction program be . Then, the null and the alternative hypotheses are:
H0 :   5
H1 :  < 5
We have selected  = 0.05.
 X  0   X  5 

 . Since the population
 S n   S 50 
We shall use as test statistic T  
distribution is approximately normal and the sample size (n = 50) is large enough, the
distribution of the test statistic, when  = 5, is approximately student’s t-distribution with
df = (n –1) = 49. Thus, this is a lower-tailed t-test.
For df = 49, t = t0.05 = approximately 1.678.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is less than -t0.05  1.678.
The computed t-value is t 
4.6 - 5
1.3
50
 2.176 .
Since the computed t-value (= -2.176) is less than -1.677, we conclude that there is
sufficient evidence at  = 0.05, to reject H0 in favour of H1, that is to infer that the average
weight loss in first two weeks among those who join the weight reduction program is less
than 5 kg.
The P-value is the area under the t-curve (corresponding to df = 49) to the left to –2.176.
From the t-table, we find that the P-value is between 0.01 and 0.025. Using Minitab, we
get the P-value = 0.0172.
31.
Let the population mean of current selling times of farm property in Nova Scotia be .
Then, the null and the alternative hypotheses are:
H0 :   90
H1 :  > 90
This is an upper-tailed test.
We have selected  = 0.1.
 X   0   X  90 

 . Since the population
 S n   S 100 
We shall use as test statistic T  
distribution is approximately normal and the sample size (n = 100) is large enough, the
distribution of the test statistic, when  = 90, is approximately student’s t-distribution with
df = (n –1) = 99. Thus, this is an upper-tailed t-test.
For df = 99, t = t0.1 = (approximately) 1.291.
10-8
Our decision rule is: reject H0 in favour of H1 if the computed t-value is greater than t0.1 
1.291.
The computed t-value is t 
94 - 90
22 100
 1.818 .
Since the computed t-value (= 1.818) is greater than 1.291, we conclude that we have
sufficient evidence, at  = 0.1, to reject H0 in favour of H1, that is to infer that the
population mean of current selling times is greater than 90 days.
33.
Let the mean of the amount of time per day spent by all the Canadian men above the age of
15 on paid work be . Then, the null and the alternative hypotheses are:
H0 :   3.1
H1 :  < 3.1
This is a lower-tailed test.
We have selected  = 0.05.
 X   0   X  3.1 

 . Since the population
 S n   S 60 
We shall use as test statistic T  
distribution is approximately normal and the sample size (n = 60) is large enough, the
distribution of the test statistic, when  = 3.1, is approximately student’s t-distribution with
df = (n –1) = 59. Thus, this is a lower-tailed t-test.
For df = 59, t = t0.05 = approximately 1.671.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is less than –1.671.
The computed t-value is t 
2.95 - 3.1
1.2
60
 0.968 .
Since the computed t-value (= -0.968) is greater than –1.671, 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
population mean is less than 3.1 hours.
P-value = P (T < -0.968). From the t-table, we see that the P-value is greater than 0.1.
Using Minitab, we get P (T < -0.968) = 0.1685. So, the P-value = 0.1685.
We would reject H0 for any value of  greater than 0.1685.
35.
(a)
Let the population mean of weights of bags produces today be . Then, the null and the
alternative hypotheses are:
H0 :   25
H1 :  < 25
This is a lower-tailed test.
We have selected  = 0.01.
 X   0   X  25 

.
  n   1.2 10 
We shall use as test statistic Z  
z = z0.01 = 2.326.
Our decision rule is: reject H0 in favour of H1 if the computed z-value is less than –2.326.
For the given sample data,
10-9
x
24.2  24.1    24.5
 24.56
10
The computed z-value is z 
24.56 - 25
1.2 10
 1.1595
Since the computed z-value (= -1.1595) is greater than -2.326, we conclude that we do not
have sufficient evidence, at  = 0.01, to reject H0 in favour of H1, that is to infer that the
population mean of weights of bags produced today is less than 25 kg.
(b)
The population distribution is approximately normal and  is known. Hence, for  = 25,
the distribution of
X  25 is approximately standard normal distribution. We shall,

10
assume it to be Z and apply the Z-test.
(c)
37.
P-value = (Z ≤ -1.1595) = approximately (0.5 – 0.377) = 0.123.
Let the population mean of fuel consumption rate be  litres/100 km. Then, the null and
the alternative hypotheses are:
H0 :   2.6
H1 :  < 2.6
This is a lower-tailed test.
We have selected  = 0.05.
 X   0   X  2.6 

 . Since the population
 S n   S 8 
distribution is approximately normal, the distribution of the test statistic, when  = 2.6, is
We shall use as test statistic T  
approximately student’s t-distribution with df = (n – 1) = 7. Thus, this is a lower-tailed ttest.
For df = 7, t = t0.05 = 1.895. Our decision rule is: reject H0 in favour of H1 if the computed
t-value is less than -t0.05 = -1.895.
For the given sample data,
x
s
2.5  2.4    2.4
 2.488
8
2.5  2.4882  2.4  2.4882    2.4  2.4882
The computed t-value is t 
7
2.488 - 2.6
0.155
8
10-10
 2.044
 0.155
Since the computed t-value (= -2.044) is less than –1.895, we conclude that we have
sufficient evidence, at  = 0.05, to reject H0 in favour of H1, that is, to infer that the
population mean of gasoline consumption rate is less than 2.6 litres/100 km.
39.
Let the average monthly coffee consumption (in litres) of students at the University of
Windsor be . Then, the null and the alternative hypotheses are:
H0 :  = 4.8
H1 :   4.8
This is a two-tailed test.
We have selected  = 0.05.
 X   0   X  4.8 

 . Since the population
 S n   S 12 
distribution is approximately normal, the distribution of the test statistic, when  = 4.8, is
We shall use as test statistic T  
approximately student’s t-distribution with df = (n – 1) = 11. Thus, this is a two-tailed ttest.
For df = 11, t/2 = t0.025 = 2.201.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is less than –2.201 or
if it is greater than 2.201.
For the given sample data,
x
s
4.75  4.96    5.26
 4.978
12
4.75  4.9782  4.96  4.9782    5.26  4.9782
The computed t-value is t 
11
4.978 - 4.8
0.374 12
 0.374
 1.649
Since the computed t-value (= 1.649) is between –2.201 and +2.201, 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 average monthly coffee consumption of University of Windsor students differs
from the national average of 4.8 litres.
41.
Let the population mean of number of returns per day from online shoppers of Egolf.com
be . Then, the null and the alternative hypotheses are:
H0 :   6.5
H1 :  < 6.5
This is a lower-tailed test.
We have selected  = 0.01.
 X   0   X  6.5 

 . Since the population
S
n
S
12

 

distribution is approximately normal, the distribution of the test statistic, when  = 6.5, is
We shall use as test statistic T  
approximately student’s t-distribution with df = (n – 1) = 11. Thus, this is a lower-tailed ttest.
10-11
For df = 11, t = t0.01 = 2.718.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is less than -t0.05 = 2.718.
For the given sample data,
x
s
0  4    10
 5.1667
11
0  5.1667 2  4  5.1667 2    10  5.1667 2
The computed t-value is t 
11
5.1667  6.5
3.1575 12
 3.1575
 1.463
Since the computed t-value (= -1.463) is greater than –2.718, we conclude that we do not
have sufficient evidence, at  = 0.01, to reject H0 in favour of H1, that is, to infer that the
population mean of number of returns per day is less than 6.5.
43.
Let the population mean of gain in time per week for the watches be  seconds. Then, the
null and the alternative hypotheses are:
H0 :  = 0
H1 :   0
This is a two-tailed test.
We have selected  = 0.05.
 X  0   X  0 

 . Since the population distribution
 S n   S 18 
is approximately normal, the distribution of the test statistic, when  = 0, is approximately
We shall use as test statistic T  
student’s t-distribution with df = (n –1) = 17. Thus, this is a two-tailed t-test.
For df = 17, t/2 = t0.025 = 2.11.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is less than –2.11 or
if it is greater than 2.11.
For the given sample data,
x
s
0.38  0.2    0.05
 0.2322
18
0.38  0.23222   0.2  0.23222    0.05  0.23222
The computed t-value is t 
17
- 0.2322 - 0
0.312 18
 0.312
 3.158
Since the computed t-value (= -3.158) is less than –2.11, we conclude that we have
sufficient evidence, at  = 0.05, to reject H0 in favour of H1, that is, to infer that the
population mean of gain in time per week is not zero.
P-value = 2P (T > 3.158). From the t-table, we see that the P-value is between 2 (0.005) =
0.01 and 2 (0.0005) = 0.001 (Using Minitab, we get P-value = 2 (0.0029) = 0.0058).
10-12
45.
(a)
  
 10 
  250  1.645 

 n
 36 
The lower control limit should be   z0.05 
= 247.26 ml.
  
 10 
  250  1.645 

 n
 36 
The upper control limit should be   z0.05 
= 252.74 ml.
(b)
If the value of  shifts to 245, then the distribution of sample mean X will be

 
10 

  Normal  245,
.
n
36 


The change will not be detected if the value of X obtained is greater than the lower
approximately Normal   ,
control limit, which is 247.26. Thus, we want area under the normal curve to right of
247.26.
z-value corresponding to 247.26 is z 
247.26-245
 1.356 .
10 36
From the Z-table, area under the Z-curve between 0 and 1.356 is approximately 0.4125.
Hence area to the right of 1.356 is approximately (0.5 – 0.4125) = 0.0875.
The probability that the change will not be detected is 0.0875.
(c)
If the value of  shifts to 253, then the distribution of sample mean X will be

 
10 

  Normal  253,
.
n
36 


The change will not be detected if the value of X obtained is less than the upper control
approximately Normal   ,
limit, which is 252.74. Thus, we want area under the normal curve to left of 252.74.
z-value corresponding to 252.74 is z 
252.74-253
 0.156 .
10 36
From the Z-table, area under the Z-curve between 0 and 0.156 is approximately 0.062.
Hence area to the left of –0.156 is approximately (0.5 – 0.062) = 0.438.
The probability that the change will not be detected is 0.438.
47.
This is an upper-tailed test.
Since the population distribution is approximately normal, the distribution of X is
approximately normal. Furthermore, the value of population standard deviation is known.
Hence, we shall use Z-test.
We have chosen the value of probability type I error, , as 0.01.
  
 10 
 = 50  2.326 
.
 n
 n
Hence, critical value of sample mean, X , is 0  z0.01 
We want the probability of type II error, when  = 55, to be 0.3.

10 
 .
If  = 55, then X ~ Normal  55,
n

 10 
 to be 0.3.
 n
Suppose we want area under the normal curve to the left of 50  2.326 
10-13
z0.3 = approximately 0.525.
 10 
 10 
 10 
  55  z0.3 
  55  0.525 
.
 n
 n
 n
Hence, we want 50  2.326 
This gives us,
 2.326  0.52510 
n
55  50  5 , or
  2.85110 
n
  32.49 .
5


So, for  = 0.01, and   0.3 when  = 55, the sample size, n, should be at least 33.
2
49.
Let the population mean of waiting times at local Farmer Jack’s be  minutes. Then, the
null and the alternative hypotheses are:
H0 :   8.0
H1 :  < 8.0
This is a lower-tailed test.
We have selected  = 0.05.
 X  0   X  8 

 . Since the population
 S n   S 24 
distribution is approximately normal, the distribution of the test statistic, when  = 8, is
We shall use as test statistic T  
approximately student’s t-distribution with df = (n – 1) = 23. Thus, this is a lower-tailed ttest.
For df = 23, t = t0.05 = 1.714. Our decision rule is: reject H0 in favour of H1 if the computed
t-value is less than -t0.05 = -1.714.
The computed t-value is t 
7.5  8
 -0.77.
3.2 24
Since the computed t-value (= -0.77) is greater than –1.714, 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
population mean of waiting times at local Farmer Jack’s is less than the national average of
8.0 minutes.
51.
Let the mean fare to fly from Halifax to Toronto be . Then, the null and the alternative
hypotheses are:
H0 :   367
H1 :  > 367
This is an upper-tailed test.
We have selected  = 0.01.
 X  0   X  367 

.
 S / n   S / 13 
We shall use as test statistic, T = 
The population distribution is approximately normal. Hence, for  = 367, the distribution
of the test statistic is approximately student’s t-distribution with df = (n - 1) = 12. So, this
is an upper-tailed t-test.
For df = 12, t = t0.01 =2.681.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is greater than 2.681.
10-14
For the given sample data,
x
s
421  386 
13
 381
 388.31
(421  388.31) 2 
 (381  388.31) 2
12
 22.46.
 388.31  367 
  3.421.
 22.46 / 13 
The computed t-value is t  
Since the computed t-value (=3.421) is greater than 2.681, we have sufficient evidence, at
 = 0.05, to reject H0 in favour of H1, that is, to infer that the mean fare has increased.
The P-value = P(T > 3.421). From the t-table, we see that the P-value is between 0.005
and 0.0005. (Using Minitab, we get the P-value equal to 0.003.)
53.
Let p the fraction of accounts that are in arrears for more than 3 months. Then, the null and
the alternative hypotheses are:
H0 : p  0.6
H1 : p > 0.6
This is an upper-tailed test.
We have chosen  = 0.01.

 

pˆ  p0
pˆ  0.6
  
 .
 ( p )(1  p ) / n 
0
0

  (0.6)(0.4) / 200 
We shall use as test statistic 
n p0= (200)(0.6) > 5 and n (1- p0) = (200)(0.4) > 5.
Hence, the sample size is large enough for normal approximation. We shall apply an
upper-tailed Z-test.
z = z0.01 = approximately 2.326.
Our decision rule is: reject H0 in favour of H1 if the computed z-value is greater than 2.326.
pˆ 
140
 0.7.
200
The computed z-value is z =
0.7-0.6
 0.6 0.4
= 2.89.
200
Since the computed z-value (= 2.89) is greater than 2.326, we conclude that we have
sufficient evidence, at  = 0.01, to reject H0 in favour of H1, that is, to infer that more than
60 percent of the accounts are in arrears for more than 3 months.
55.
Let p the fraction of persons, who wanted the agency to plan a vacation for them, who
wanted to go to Europe. Then, the null and the alternative hypotheses are:
H0 : p  0.44
H1 : p > 0.44
This is an upper-tailed test.
We have chosen  = 0.05.
10-15

 

pˆ  p0
pˆ  0.44
  
 .
 ( p )(1  p ) / n 
0
0

  (0.44)(0.56) /1000 
We shall use as test statistic 
n p0 = (1000)(0.44) > 5 and n (1- p0) = (1000)(0.56) > 5.
Hence, the sample size is large enough for normal approximation. This is an upper-tailed
Z-test.
z = z0.05 = 1.645.
Our decision rule is: reject H0 in favour of H1 if the computed z-value is greater than 1.645.
pˆ 
480
 0.48.
1000
The computed z-value is z =
0.48-0.44
 0.44 0.56 1000
= 2.55.
Since the computed z-value (= 2.55) is greater than 1.645, we conclude that we have
sufficient evidence, at  = 0.05, to reject H0 in favour of H1, that is, to infer that there has
been an upward shift in the percentage of persons who want to go to Europe.
57.
The null and the alternative hypotheses are:
H0 :   1483949
H1 :  > 1483949
This is an upper-tailed test.
We have chosen  = 0.05.
 X  0   X  1483949 

 . Since the population
 S n   S 20 
distribution is approximately normal, when  = 1483949, the distribution of the test
We shall use as test statistic T  
statistic is approximately student’s t-distribution with df = (n – 1) = 19. Thus, this is an
upper-tailed t-test.
For df = 19, t = t0.05 = 1.725.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is greater than 1.725.
Select a random sample of size 20, compute x , s and t 
x - 1483949
s
20
If t > 1.729, conclude that there is sufficient evidence, at  = 0.05, to reject H0. Else,
conclude that there is insufficient evidence to reject H0. Answer will vary according to the
sample chosen.
59.
The null and the alternative hypotheses are:
H0 :  = 18.6
H1 :   18.6
Since the population distribution is approximately normal and the population standard
deviation is unknown, we shall use two-tailed t-test. The Megastat output is given below.
10-16
Hypothesis Test: Mean vs.
Hypothesized Value
34.600000 hypothesized value
38.199000 mean BCE
4.523419 std. dev.
1.011467 std. error
20 n
19 df
3.56 t
.0021 p-value (two-tailed)
We see from computer output that P-value is 0.0021. Hence, our conclusion would be “we
have sufficient evidence to reject H0” if the selected  value is greater than 0.0021. Since
the given value  = 0.05 > 0.0021, we reject H0, that is we infer that the mean value of the
end of the week prices of the BCE stock during the year 2000 was different from 18.6.
10-17