# Ch 9/11 ```SOLUTIONS FOR HYPOTHESIS TESTING HOMEWORK
Problems from Chapter 9 and 11
Chapter 9 Homework
2 . a. Problem about mean sales volume, therefore we need hypothesis about a
population mean. Increasing sales volume implies volume is greater (&gt;).
Ho: 14
Ha: &gt;14
b. If Ho is not rejected, we conclude that sales volume is not greater.
c. If Ho is rejected, we conclude that sales volume is greater.
4.
a. Problem about reducing mean costs, therefore we need a hypothesis about a
population mean. Reducing costs implies cost is less (&lt;).
Ho: 220
Ha: &lt;220
b. If Ho is not rejected, we conclude that costs were not reduced.
c. If Ho is rejected, we conclude that costs have been reduced.
6.
a. Problem is about the average amount of fat being no more than () one gram.
Ho: 1
Ha: &gt;1
b. Type I error is the probability of concluding that the average amount of fat is
greater than one gram when it isn’t. If this error is made the company could be
unfairly criticized.
c. Type II error is the probability of concluding that the average amount of fat is ≤1,
when it is actually greater. If this error is made, the company will not be
criticized when it should be.
8. a. Problem is about mean operating cost.
Ho: 220
Ha: &lt;220
b. Type I error is the probability of concluding that the mean cost is less than 220
per hour, when it isn’t. If this error is made the company will implement the new
production method when it shouldn’t.
c. Type II error is the probability of concluding that the mean cost is not less than
220 when it really is. If this error is made, the company will not implement the
new production method when it should, thereby failing to save money.
14. Problem is about mean hours of television viewing. The population standard
deviation is unknown. Assume the population is normally distributed. The question
is “ Has the mean viewing time increased?”
1. Ho: ≤6.7, Ha: &gt;6.7
2.  =0.01
x 
s
, Sx 
sx
n
3.
t=
4.
Reject Ho if t &gt;2.358
2.5
2.5
Sx 

 0.1768
200 14.1421
2.
t=
3.
4.
7.25  6.7
0.55

 3.11
0.1768
0.1768
Reject Ho
It appears that mean viewing time has increased.
18. Problem about mean expenditure at Burger King. Population standard deviation ()
is unknown. Assume the population is normally distributed.
1. Ho: 5.72, Ha: &gt;5.72
2.  = 0.05
3. t =
x 
s
, Sx 
sx
n
4. df = 101 so use 60 for t table. t60,0.05=1.671 , Reject Ho if t&gt;1.671
1.24
1.24
5. S x 

 .1228 ,
102 10.0995
5.98  5.72
.26

 2.12
t=
.1228
.1228
6. Reject Ho.
7. It appears that the conclusion that the mean is greater than 5.72 is correct.
26. Problem about mean waiting time. Population standard deviation () is unknown.
Assume the population is normally distributed.
1. Ho: =8, Ha: ≠8
2.  = 0.05
3. t =
x 
s
, Sx 
sx
n
4. t119,0.025=1.98 , (use Df=120, since so close to 119)
Reject Ho if t&lt;-1.699
3.2
3.2
5. S x 

 0.2921 ,
120 10.9545
7.5  8
 0.5

 1.711
t=
0.2921 0.2921
6. Do not reject Ho
7. It appears that the population mean does not differ from the 8 minute
waiting time assumption.
27 c. Problem is about a mean filling weight.  is known.
1. Ho: =16, Ha: 16
2. α = 0.05
3. z=

5.
6.
7.
x 
x
x 

n
known, n≥ 30, therefore use normal sampling distribution.
Reject Ho if Z&gt;1.96 or Z&lt;-1.96

0.8
0.8
x 

 0.14606
30 5.4772
15.82  16
 0.18

 1.23
Z=
0.14606
0.14606
Do not reject Ho
It appears that the line does not need to be shut down and adjusted.
31. b. Problem is about a mean exam score. The population standard deviation is known.
1. Ho: =900, Ha: ≠90
5.  =0.05
 known, n≥30, therefore use normal sampling distribution
z=
x 
x
x 

n


Reject Ho if Z&gt;1.96 or Z&lt;-1.96
180
180
5.  x 

 12.728
200 14.1421
935  900
35

 2.75
Z=
12.728
12.728
6. Reject Ho.
It appears that exam scores are not the same as in previous years.
42.
Problem about mean tree planting time. Population standard deviation () is
unknown. Assume the population is normally distributed.
1. Ho: ≤ 2, Ha: &gt;2
2.  = 0.05
3. t =
x 
s
, Sx 
sx
n
4. t 9,0.05=1.699 , Reject Ho if t&gt;1.833
5. Calculations from statistical calculator:
x =2.4, S=0.5164
0.5164
 0.1633 ,
3.1623
10
2.4  2
0.4

 2.45
t=
0.1633 0.1633
6. Reject Ho.
7. It appears that the mean tree planting time exceeds 2 hours. We would
conclude that they are underestimating their costs and should come up with a
new cost structure.
Sx 
0.5164

50. Problem is about the proportion of employees preferring 4 day 10 hour work
schedule. The sample is large enough for a normal sampling distribution.
a. 1. Ho: p= 0. 5, Ha: p0. 5
2.  = 0.10
p p
3.
z=
p
, p 
p(1  p)
n
4.
Reject Ho if z &gt;1.282
(0.5)(0.5)
5.  p 
 0.0488 ,
105
67
p
 0.6381
105
0.6381  0.5
z
 2.83
0.0488
6. Reject Ho.
7. It appears that the 4 day 10 hour work schedule is preferred.
b. p value =2(0.5-0.4977)=0.0046. Since smaller than , reject Ho.
52. Problem is about the proportion of e-mail users using Outlook. The sample is large
enough to use a normal sampling distribution.
1. Ho: p≥ 0.75, Ha: p&lt;0.75
2.  = 0.05
p p
3. z =
p
, p 
p(1  p)
n
4. Reject Ho if z&lt;-1.645
5. p  .72
p 
z
(0.75)(0.25)
 0.025 ,
300
0.72  0.75  0.03

 1.20
0.025
0.025
6. Do not reject Ho.
7. No. The statistical evidence appears to indicate that the executive is correct.
58. The problem is about type II error in a problem about mean time.
a. Ho:   15,
Ha:  &gt; 15
Type II error is concluding that   15 when this is not true. Fowle would
not charge the premium rate even though the rate should be charged.
b. Reject Ho: if z &gt; 2.33
4
) = 15 + (2.33)(.6761) = 15 + 1.58 = 16.58
x = 15 + (2.33)(
35
16.58  17
 .62
For  = 17: z =
0.6761
 = .5 - .2324 = .2676
16.58  18
 2.10
c. For  = 18: z =
0.6761
 = .5 - .4821 = .0179
60.
The problem is about type II error in a problem about mean age. The population
standard deviation is known.
a. A type II error would be the conclusion that the average age is 28 when it
isn’t.
b. Type II error if population mean is actually 26,
*
x is value of x at critical values
*
x = &plusmn; z 
x
= 28 &plusmn; 1.96 (
6
100
)
= 28 &plusmn; 1.176
= 28.176 and 26.824
26.824  26
 1.37
Z=
0.6
28.176  26
 3.63
Z=
0.6
 = 0.5-0.4147 = 0.0853
A=0.4147
A=0.5000
c. power= 1-  = 1-0.0853 = 0.9147
Chapter 11 Variance Hypothesis Tests
10. Problem is about the standard deviation of useful life of a VCR. Assume
population has normal distribution.
1. Ho: 2≤(0.75)2 , Ha: 2&gt;(0.75)2
2.  = 0.05
x 
2
3.
(n  1) S 2
2
x 2 29,0.05 = 42.5569
2
Reject Ho if x &gt; 42.5569
29(2) 2 (29)( 4)
2

 206.22
5. x 
0.752
0.5625
4.
6. Reject Ho.
7. It appears that the variance of TV sets is greater than for VCRs.
12. Problem is about the variance in number of vehicles owned. Assume the population is
normally distributed.
a. S2= 0.8106 using standard deviation key and square key on calculator.
b. 1. Ho: 2= 0.94, Ha: 2‡0.94
2.  = 0.05
2
3. x 
(n  1) S 2
2
x 2 11,0.025 = 21.9200
x 2 11,0.975 = 3.81575
2
2
Reject Ho if x &gt;21.92 or x &lt; 3.81575
(11)(0.8106) 2
2
 9.4857
5. x 
0.94
4.
6. Do not reject Ho.
7. It appears that the variance is the same for both magazines.

```