EF 507
PS-Chapter 11
FALL 2008
1. A company produces electric devices operated by a thermostatic control. The
standard deviation of the temperature at which these controls actually operate should
not exceed 2.0 degrees Fahrenheit. For a random sample of 25 of these controls the
sample standard deviation of operating temperatures was 2.36 degrees Fahrenheit.
Stating any assumptions you need to make, test at the 5% level the null hypothesis
that the population standard deviation is at most 2.0 against the alternative that is
bigger.
ANSWER:
The hypothesis test assumes that the population values are normally distributed
H 0 :   2.0 and H1 :   2.0.
2
Reject H 0 if  2 >  24,0.05
 36.42
Test statistic:  2 
(n  1) s 2

(24)(2.36) 2
 33.418 , therefore, we fail to reject H 0 . We
(2)2
 02
conclude that the population standard deviation is at most 2 degrees Fahrenheit.
QUESTIONS 2 THROUGH 4 ARE BASED ON THE FOLLOWING
INFORMATION:
A new safety device is installed in an assembly-line operation. After the installation of
the device, a random sample of ten days output gave the following results for numbers
of finished components produced: 577, 642, 615, 655, 600, 570, 640, 620, 595, and
575. Management is concerned about the variability of daily output and views as
undesirable any variance above 525.
2. Calculate the sample variance
ANSWER: s 2  922.322
3. Test at the 10% significance level the null hypothesis that the population variance
for daily output does not exceed 525.
ANSWER: H0 :  2  525 and H1 :  2  525
2
Reject H 0 if  2 > 9,0.10
 14.684
2 
(n  1) s 2

(9)(922.322)
 15.811 , therefore, we reject H 0 and conclude that the
525

population variance for daily output exceeds 525.
2
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4. Of a random sample of 150 music industry management majors, 75 rated a sense of
humor as very important trait to their career performance. The same view was held by
81 of an independent random sample of 180 professional gulf management majors.
Test at the 5% level against a two-sided alternative the null hypothesis that the
population proportions of music industry management and professional gulf
management majors who rate a sense of humor as very important are the same.
ANSWER: M = Music industry management, and G = professional gulf management
H 0 : PM  PG  0 and H1 : PM  PG  0
Reject H 0 if Z   z0.025  1.96 or Z  z0.025  1.96
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pˆ M = 75 / 150 = 0.50 and pˆ G  81 / 180 = 0.45
An estimate of the common proportion is given by
nF pˆ F  nU pˆU (150)(0.5)  (180)(0.45)

 0.4727
nF  nU
150  180
pˆ F  pˆ u
0.5  0.45

 0.91 .
Test statistic: Z 
pˆ 0 (1  pˆ 0 )(1 nF  1/ nU )
(0.4727)(0.5273)(1/150  1/180)
pˆ 0 
Therefore, do not reject H 0 . We conclude that the population proportions of music
industry management and professional gulf management majors who rate a sense of
humor as very important are the same.
5. You are comparing the precision of two brands of stamping machines. From a
random sample of 12 units of output from the Brand A machine, you find that it
produces with a standard deviation of 15.2. For the Brand B machine, in a sample of
20 units of output, you find a standard deviation of 10.1. Is this sufficient evidence at
the 5% significance level to conclude that Brand B machines produce with lower
variance? Assume that the output of both machines follows a normal distribution.
ANSWER: H 0 : H 0 :  A2   B2 vs. H1 :  A2   B2
Reject H 0 if Fcalc  Fcrit  F11,19,0.05  F10,19,0.05  2.38
Since Fcalc  s A2 / sB2  (15.2)2 /(10.1)2  2.26 , we fail to reject H 0 at  = 0.05. There is
insufficient evidence to conclude that Brand B machines produce with lower variance.
QUESTIONS 6 THROUGH 8 ARE BASED ON THE FOLLOWING
INFORMATION:
The medical school of Wisconsin conducted an investigation to determine whether
women and men complete medical school in significantly different amounts of time,
on the average. Two independent random samples were selected and the following
summary information concerning times to completion of medical school computed:
Women
80
4.5 years
0.5 years
Sample Size
Sample mean
Population Standard
Deviation
Men
100
4.4 years
0.6 years
6. State the appropriate null and alternative hypotheses
ANSWER: H 0 :W   M  0 vs. H1:W   M  0
7. Perform the appropriate test of hypothesis to determine whether there is a
significant difference in time to completion of medical school between women and
men. Test using  = 0.05.
ANSWER:
Z
xW  xM

2
W
nW


2
M
nM

4.5  4.4
(0.5)2 (0.6) 2

80
100
 1.22
Reject H 0 if Z <  z.025  -1.96 or Z > z.025 = 1.96. Since Z = 1.22, do not reject H 0 . Thus
one cannot conclude there is a significant difference in mean time to completion of
medical school between women and men.
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8. Find the p-value associated with the test in Question 77 and explain how to use it
for testing the hypothesis in Question 76.
ANSWER: p-value =2  P(Z  1.22) = 2 (0.50 – 0.3888) = 0.2224. Since p-value >
 = 0.05, we fail to reject H 0 .
QUESTIONS 9 THROUGH 11 ARE BASED ON THE FOLLOWING
INFORMATION:
Before agreeing to purchase a large order of polyethylene sheaths for a particular type
of high pressure oil-filled submarine power cable, a company wants to see conclusive
evidence that the true standard deviation of sheath thickness is less than .05mm.
9. What hypotheses should be tested, and why?
ANSWER: Let  denote the population standard deviation. The appropriate
hypotheses are H 0 :   .05 vs. H1 :   .05 . With this formulation, the burden of proof
is on the data to show that the requirement has been met (the sheaths will not be used
unless H 0 can be rejected in favor of H 1 .)
10. In this context, what is Type I error?
ANSWER: Type I error: Conclude that the standard deviation is < .05 mm when it is
really equal to 0.05 mm.
11. In this context, what is Type I error?
ANSWER: Type II error: Conclude that the standard deviation is .05 mm when it is
really < 0.05
QUESTIONS 12 THROUGH 16 ARE BASED ON THE FOLLOWING
INFORMATION:
The Ford Motor Company, as part of its quality control program, began returning to
the supplier all shipments of steel that had defects or faulty chemistry. When Ford
began this program, the defective rate in 100 shipments was 9%. A recent survey
indicated that 2.2% in 136 shipments was defective. The quality control department
wants to determine at the 0.05 level of significance if this represents a significant
improvement in the quality of the steel.
12. State the null and alternative hypotheses.
ANSWER: Let P1 = proportion of defects before the quality control program, and
P2 = proportion of defects after the quality control program,
Then, the null and alternative hypotheses are H 0 : P1  P2  0 and H1 : P1  P2  0.
13. Calculate an estimate of the common proportion
ANSWER: p̂1 = 0.09 and p̂2  0.022. Then, an estimate of the common proportion is
given by pˆ 0 
n1 pˆ1  n2 pˆ 2 (100)(0.09)  (140)(0.022)

 0.0503
n1  n2
100  140
14. What is the test statistics?
ANSWER:
Z
pˆ F  pˆ u
0.09  0.022

 2.38
pˆ 0 (1  pˆ 0 )(1 nF  1/ nU )
(0.0503)(0.9497)(1/100  1/140)
15. What is the p-value?
ANSWER: p-value = P(Z > 2.38) = 0.50 – 0.4913 = 0.0087
16. What is the conclusion?
ANSWER: Since p-value = 0.0087 < 0.05, reject H . There is enough statistical
evidence to conclude that there is a significant improvement in the quality of the steel
0
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