IME 301
HW # 9
Homework Solutions
Winter 2013
Chapter 9- Test of Hypotheses …
Read these Sections and all the solved examples in there: 284-316
Chapter 11- Regression analysis …
Read these Sections and all the solved examples in there: 11-1, 11-2, 11-4.1
Will add a problem
Problem 9-41 on Page 309:
a) covered last HW
b)
  3.25  3  0.25
0.25 * 15
0.25 * 15
)   ( 1.96 
)
0.9
0.9
  (0.883)  ( 3.037)  0.811  0.0012  0.8098
Power = 1-0.8098=0.1902 this power is too small we would like more than 80%.
   (1.96 
c) CL= 95%
Z 0.025  1.96
power = 0.9
Z 0.1  1.28
  0. 1
 1.96  1.28) * 0.9 
n
  15.11
3.75  3


Consider n=16
2
Hypothesis Testing Example on Class Notes web page:
Version 1: n=25, S 2  49 0  162.5
CL=0.99
  0.01
H 0 :    0  162.5
H 1 :    0  162.5
 / 2  0.005
_
X  165.2
t0 
25 (165.2  162.5)
 1.93
7
t / 2,( n 1)  t0.005, 24  2.797
Reject Ho if ( -2.797>1.93 or 1.93>2.797). None of the inequalities are true so do
not reject Ho (accept Ho). That is I am 99% confident that considering the given sample
1
the average height of population is still 162.5. (In spite of sample average of 165.2)
Version 2: n=25, S 2  49 0  162.5
CL=0.99
  0.01
H 0 :    0  162.5
H 1 :   0  162.5
_
X  165.2
t0 
25 (165.2  162.5)
 1.93
7
t ,( n 1)  t0.01, 24  2.492
Reject Ho if ( 1.93<-2.492). The inequality is not true so do not reject Ho (accept Ho).
That is I am 99% confident that considering the given sample the average height of
population has not decreased and is still 162.5.
Version 3: n=25, S 2  49 0  162.5
CL=0.9
  0.1
H 0 :    0  162.5
H 1 :    0  162.5
_
X  165.2
t0 
25 (165.2  162.5)
 1.93
7
t ,( n 1)  t0.05, 24  1.711
Reject Ho if ( -1.711>1.93 or 1.93>1.711). The second inequality is true so reject
Ho. That is I am 90% confident that considering the given sample the average height of
population has changed from 162.5 (We do not know if it has gone up or down.).
Version 4: n=61, S 2  49 0  162.5
CL=0.99
  0.01
H 0 :    0  162.5
H 1 :    0  162.5
 / 2  0.005
_
X  165.2
Z0 
61(165.2  162.5)
 3.009
7
Z 0.005  2.58
Reject Ho if ( -2.58>3.009 or 3.009>2.58). The second inequality is true so reject
Ho. That is I am 99% confident that considering the given sample the average height of
population has changed from 162.5 (We do not know if it has gone up or down.).
Version 5: n=25 S 2  16 0  162.5
CL=0.99
  0.01
H 0 :    0  162.5
H 1 :    0  162.5
_
X  165.2
t0 
25 (165.2  162.5)
 3.375
4
2
t / 2,( n 1)  t0.005, 24  2.797
 / 2  0.005
Reject Ho if ( -2.797>3.375 or 3.375>2.797). The second inequality is true so reject
Ho. That is I am 99% confident that considering the given sample the average height of
population has changed from 162.5 (We do not know if it has gone up or down.).
Version 6: n=61,   6.9
0  162.5
CL=0.99
  0.01
_
X  165.2
H 0 :    0  162.5
H 1 :    0  162.5
Z0 
61(165.2  162.5)
 3.056
6.9
Z 0.005  2.58
 / 2  0.005
Reject Ho if ( -2.58>3.056 or 3.056>2.58). The second inequality is true so reject
Ho. That is I am 99% confident that considering the given sample the average height of
population has changed from 162.5 (We do not know if it has gone up or down.).
Version 7: n=61,
CL=0.99
  6.9
  0.01
H 0 :    0  162.5
H1 :   0  162.5
  0.01
_
0  162.5
X  165.2
Z0 
61(165.2  162.5)
 3.056
6.9
Z 0.01  2.326
Reject Ho if ( 3.056>2.326). The inequality is true so reject Ho. That is I am 99%
confident that considering the given sample the average height of population has
increased from 162.5.
Version 9: n=25, S 2  49 0  162.5
CL=0.99
  0.01
_
X  165.2
t0 
H 0 :    0  162.5
H 1 :    0  162.5
25 (165.2  162.5)
 1.93
7
P  value  2 * P(t24  1.93)
From t-dist
T…
1.711
2.064
prob
0.05
0.025
perform interpolation then
P  value  2 * P(t24  1.93)  2 * 0.0345  0.069
3
For any significance level less than 0.069 (that is for any CL>93.1%) reject Ho. For any
significance level more than 0.069 then accept Ho.
Version 11: n=36,
CL=0.99
0  163
0  162.5
H 0 :    0  162.5
H 1 :    0  162.5
  0.01
Z0 
_
X  165.2
  6.9
36 (165.2  162.5)
 2.35
6.9
Z 0.005  2.58
 / 2  0.005
Reject Ho if (-2.58>2.35 or 2.35>2.58). Both inequalities are not true so fail to reject
Ho (accept Ho).
  163  162.5  0.5
0.5 * 36
0.5 * 36
   (2.58 
)   ( 2.58 
)
6.9
6.9
  (2.15)  ( 3.015)  0.984222  0.001285  0.9829
For -3.015 perform interpolation between -3.02 and -3.01.
That is I am 99% confident that considering the given sample the average height of
population has not changed from 162.5. Moreover, there is a probability of 98.29% to
accept no change in average height if in fact there was a change to average height of 163
cm (type II error is 98.29%).
Power of this test is only about 12% (1-0.9829) which is very low (we would like at least
80%).
Version 12: n=?,
CL=0.99
  20%
  6.9
  0.01
0  163
0  162.5
  163  162.5  0.5
 / 2  0.005 Z 0.005  2.58
Z 0.2  2.055
For two sided test
 2.58  2.055 * 6.9 
n
  4091.27
0.5

2
Therefore select n=4092.
4
Quiz on Mar 5 based on homework Problem 9-41 on Page 309:
The wear of crank shaft is of interest. A random sample of size 16 is tested and the
average wear is 3 with a variance of 4. Test if the average wear of population has
changed from
3.5? CL=90%
_
n=16,
, S=2 we use t-dist
X 3
0  3.5
H 0 :   3.5
H1 :   3.5
  0.1
 / 2  0.05
t0 
n ( X  0 )
s
t0 
16 (3  3.5)
 1
2
t / 2,( n 1)  t0.05,15  1.753
Reject H0 if 1.753  1.....OR....  1  1.753
As none of the inequalities are true, then conclude we cannot reject H0 (accept H0). That
is the average wear of population after 100000 miles has changed from 3.5.
5