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4. A consumer advocacy group compares the proportions of automobiles produced by two different
manufacturers that need major repairs in the first three years of ownership. The results are as below:
Manufacturer 1 Manufacturer 2
x1  45
x2  27
n1  1000
n2  1000
a. Do a 2-sided 99% confidence interval for the proportion produced by manufacturer 1 that need major
repairs. (2)
b. Do a 2-sided 91% confidence interval for the proportion produced by manufacturer 1 that need major
repairs. (2)
c. An automobile columnist states that over 5% of automobiles produced by the first manufacturer need
repairs in the first three years. State the null and alternative hypotheses to be tested in this problem..(1)
d. Do a test of the hypothesis in c) at a 95% confidence level. (3)
e. Repeat d) at the 91% confidence level.(2)
f. If the true proportion that need repairs is 6%, what is the power of the test? (3)
g. If I claim that the proportion needing repairs is 6%, on how large a sample would this have to be based to
be accurate within 0.5%? (2)
Solution: From page 10 of the Syllabus Supplement:
Interval for
Confidence
Hypotheses
Interval
Proportion
p  p  z 2 s p
pq
n
q  1 p
sp 
a) p 
Test Ratio
H 0 : p  p0
H 1 : p  p0
x1
45

 .045 and q  1  p  .955
n1 1000
sp 
p  p  z.005s p  .045  2.576.00655  .045  .017
z
pq

n
p  p0
p
Critical Value
p cv  p 0  z  2 
p 
p
p0 q 0
n
.045 .955   .00655553
1000
b) From page 1 z.045  1.70 . So p  p  z.005 s p  .045  1.70.00655  .045  .011
c)
H 0 : p  .05
H1 : p  .05
p0 q0
.05.95 

 .00689 z  z.05  1.645
n
1000
p  p 0 .045  .05

 .  0.726 . This is below 1.645 so accept H 0 .
(i) Test Ratio: z 
p
.00689
d)  p 
or (ii) Critical Value: pcv  p0  z  p  .05  1.645.00689  .06134 . p  .045 is below this, so
accept H 0 .
or (iii) Confidence Interval: p  p  z.05 s p  .045 1.645.00655  .0342. . This does not contradict H 0 .
e) We want a point z .09 , so that Pz z .09   .09 . From the diagram,
. . Redo one of the
P 0  z  z.09  .41 . The closest we can come is P 0  z  134
.  .4099 . So z.09  134
.
three ways of doing d) with 1.645 replaced by 1.34.
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f)
The power of a test is the probability of rejecting the null hypothesis when it is false. It is the opposite
of the probability of wrongly accepting the null hypothesis. According to d) above, if   .05 , we will
accept H 0 if p is less than or equal to .06134. So






p  .06 
.06134  .06 

power  1  P z  cv
 1  P z 
  1  P z  0.18   1  .5  .0714 

 p 


.
06
.
94





1000 

 1  .5714  .4286
g) From the outline, if we assume a 95% confidence level so that z  1.960 ,
n
pqz
e
2
2

.06 .94 1.96 
.005 2
2
2
 8667
5. A consumer advocacy group compares the proportions of automobiles producer by two different
manufacturers that need major repairs in the first three years of ownership. The results are as below:
Manufacturer 1 Manufacturer 2
x1  45
x2  27
n1  1000
n2  1000
a. Test the hypothesis that the proportions are equal at the 90% level. (5)
b. Do a confidence interval for the difference at the 95% level. (4)
c. Test the hypothesis that the proportion that need repairs is greater for the first manufacturer than the
second at the 90% level. (2)
d. Find p-values for your results in a) and c). (2)
Those who do not learn from the past are doomed to repeat it.
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6. A city sets a requirement that a sewer pipe must have a mean breaking strength exceeding 2500 pounds
per linear foot. A test of 200 sections of sewer pipe yields an average strength of 2530 pounds per linear
foot and a standard deviation of 200. Assume a confidence level of 95%.
a. State null and alternative hypotheses for this problem and a critical value for the sample mean.(2)
b. Given the sample results above, do you accept or reject the null hypotheses?(1)
c. What is the power of the test if the true mean is 2535 pounds per square inch? (3)
d. Do a power curve for this test. (4)
e. If we claim that a population has a Poisson distribution with a mean of 9 , and the actual value found is
15, do a hypothesis test at the 95% level. (3)
f. Find the lowest value above 9 and the highest value below 9 that would lead to a rejection of the null
hypothesis in f). Using these, find the power of the test if the mean is actually 12.5.(5)
Solution:
Interval for
Mean (
known)
a)
H 0 :   2500
Confidence
Interval
Hypotheses
  x  z  x
2
Test Ratio
H0 :   0
z
H1 :    0
x  0
x
Critical Value
x cv   0  z 2  x
H1 :   2500 .   .05, z.05  1.645 , n  200 , x  2530 and s  200 . Since the above
can be used for large samples, replace  x with sx  s  200  14 .142 . If we use the critical value
n
method, xcv  0  z sx  2500  1.645 14.142  2523 .26 .
b) We accept H 0 if x  2523 .26 , so reject H 0 .
c)
200
We accept H 0 if x  2523 .26 , so we need the probability that x is less than 2523.26 when the true
2523  2535 

mean is 2535. P x  2523   2535   P z 
  P z  0.84   .5  .2995  .2005 . This is
14 .142 

the probability of a type II error, so that the power is 1 - .2005 = .7995.
d) Try the following points:
At 1  2500 the power is 5% (the significance level)
2523  2511 

At 1  2511 the power is 1  Px  2523   2511   1  P z 

14 .142 

 1  Pz  0.84   1  .5  .2995   .2005
At 1  2523 (the critical value) the power is 50%.
At 1  2535 the power is .7995.
2523  2546 

At 1  2546 the power is 1  Px  2523   2546   1  P z 

14 .142 

 1  Pz  1.63   1  .5  4484   .9484
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e)
f)
H 0 : m  9 H1 : m  9 . From the Poisson table for a mean of 9, p  value  2Px  15 
 21  Px  14   21  .95853   .08824 . Since this is above .05, accept H 0 .
Since the number in e) is pretty close to .05, try 16 (Look for probabilities above 97.5%!). Also try
numbers close to zero (Look for probabilities below 2.5%!).
For 16, p  value  2Px  16   21  Px  15   21  .97786   .02214 below .05
For 3, p  value  2Px  3  2.02123  .04246 below .05.
These tell us our reject region, so our accept region is 4 to 16. To find the power, take one minus the
probability of x being between 4 and 16 when the mean is 12.5.
P4  x  16   Px  16   Px  3  .86931  .00155  .86776 . So the power is 1 - .86776=13224.
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Computer Question
1. Turn in your first computer assignment. (2)
2. The data below was read into the computer with the marked data tagged ‘West’ and the rest tagged
‘East.’ A test was run resulting in the printout shown below the data. Answer the following questions:
a. What was H 0 ? H 1 ? (1)
b.
c.
If your confidence level is 99%, do you accept H 0 ? Why? (1)
How do you think that the computer found only 22 degrees of freedom? (1)
The output read:
Two sample T for Hprice
Location
N
Mean
East
23
122498
West
14
149327
StDev SE Mean
37196
7756
46418
12512
90% CI for mu(East) - mu(West) : ( -52106, -1552)
T-Test mu(East) = mu(West) (vs not =) : T = -1.82 P = 0.083
DF = 22.
Solution: a) If East is 1 and West is 2, H 0 : 1   2
H 1 : 1   2
b) Accept H 0 because the p-value (8.3%) is above the significance level (1%).
c) It probably used
DF 
 s12 s22 



n

 1 n2 
2
.
   
s12
2
n1
n1  1
s 22
2
n2
n2  1
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Not for release!
5. A consumer advocacy group compares the proportions of automobiles producer by two different
manufacturers that need major repairs in the first three years of ownership. The results are as below:
Manufacturer 1 Manufacturer 2
x1  45
x2  27
n1  1000
n2  1000
a. Test the hypothesis that the proportions are equal at the 90% level. (5)
b. Do a confidence interval for the difference at the 95% level. (4)
c. Test the hypothesis that the proportion that need repairs is greater for the first manufacturer than the
second at the 90% level. (2)
d. Find p-values for your results in a) and c). (2)
Solution: From page 10 of the Syllabus Supplement:
Interval for
Confidence
Hypotheses
Interval
Difference
p  p  z 2 sp H 0 :p  p0
between
p  p1  p2
H 1 : p  p0
proportions
q  1 p
sp 
Test Ratio
z
p0  p01  p02
or p 0  0
p1q1 p2 q 2

n1
n2
p  p0
 p
If p  0
 p 
p01q 01 p02 q 02

n1
n2
Or use
H 0 : p  p 0
p1 
x1
45

 .045
n1 1000
p2 
x2
27

 .027
n1 1000
n1 p1  n2 p2
45  27

 .036
n1  n2
1000  1000
p0 
s p 
a)
H1 : p  p 0 or H 0 : p1  p 2
p1q1 p2 q2


n1
n2
(i) z 
0
pcv  p0  z 2  p
If p0  0
 p 
p0 q 0  1 n1 
1
n2

n p  n2 p2
p0  1 1
n1  n2
s p
  .10 z 2  1.645
H1 : p1  p 2
p  p1  p 2  .045  .027  .018
1
1
p0 q0  
 n1 n2

 


.036 .964 
1
 1000

1 
  .0833
1000 
.045 .955   .027 .973   .0832
p  p 0
p
p 
Critical Value
1000

1000
.018
 2.161 . This is outside the interval 1.645 so reject H 0 .
.00833
or (ii) pcv  p0  z  p  0  1.645.00833  .0137 . This interval does not include .018 so reject
2
H0 .
or (ii) p  p  z s p  .018  1.645.00832  .018  .0137 This interval does not include 0 so reject
2
H0 .
b)
c)
p  p  z 2 s p  .018  1.96.00832  .018  0163 .
H 0 : p1  p 2 H1 : p1  p 2 . this is the same as H 0 : p  0 H1 : p  0   .10
the material in a) (i) z  2.161 This is above 1.282 so reject H 0 . Or (ii)
z  1.282 Use
pcv  p0  z  p  0  1.282.00833  .011 . Since this is below .018, reject H 0 . Or (iii)
p  p  z 2 s p  .018  1.282.00832  .018  .011  .007. this is above 0, so reject H 0 .
d) For a), 2Pz  2.16   2.5  4846   .0308 . For b), Pz  2.16   .5  4846   .0308
6