EF 507
PS-Chapter 10
1. QUESTIONS 1
INFORMATION:
AND
2
ARE
BASED
FALL 2008
ON
THE
FOLLOWING
The supervisor of a production line believes that the average time to assemble an
electronic component is 14 minutes. Assume that assembly time is normally
distributed with a standard deviation of 3.4 minutes. The supervisor times the
assembly of 14 components, and finds that the average time for completion was 11.6
minutes.
1. What are the appropriate null and alternative hypotheses?
A) H 0 :  
B) H 0 :  
C) H 0 :  =
D) H 0 :  
ANSWER:
14
14
14
14
C
and
and
and
and
H 1 :  < 14
H 1 :  > 14
H 1 :   14
H 1 :  = 14
2. Which of the following statements is most accurate?
A)
Unable to reject the null hypothesis at   0.10.
B)
Reject the null hypothesis at  = 0.025, but not at  = 0.05.
C)
Reject the null hypothesis at  = 0.05, but not at  = 0.01.
D)
Reject the null hypothesis at  = 0.008.
ANSWER: D
QUESTIONS 3 THROUGH 6 ARE BASED ON THE FOLLOWING
INFORMATION:
The manufacturer of a certain chewing gum claims that at least 80% of dentists
surveyed prefer their type of gum. You decide to test their claim. You find that in a
sample of 200 dentists, 74.1% do actually prefer their gum.
3. What are the null and alternative hypotheses for your test?
A) H 0 : P  0.80 and H 1 : P < 0.80
B) H 0 : P = 0.80 and H 1 : P  0.80
C) H 0 : P  0.80 and H 1 : P > 0.80
D) H 0 : P > 0.80 and H 1 : P  0.80
ANSWER: A
4. What would your decision rule be?
A)
Reject H 0 if ( pˆ  P0 ) / P0 1  P0  / n  z / 2 .
B)
Reject H 0 if ( pˆ  P0 ) / p 1  p  / n  z .
C)
Reject H 0 if ( pˆ  P0 ) / P0 1  P0  / n   z / 2 .
D)
Reject H 0 if ( pˆ  P0 ) / P0 1  P0  / n < - z .
ANSWER: D
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5. The value of the test statistic is
A)
2.086.
B)
1.444.
C)
-2.086.
D)
-1.444.
ANSWER: C
6. Which of the following statements is most accurate?
A)
Unable to reject the null hypothesis at   0.10.
B)
Reject the null hypothesis at  = 0.05.
C)
Reject the null hypothesis at  = 0.10, but not 0.05.
D)
Reject the null hypothesis at  = 0.01.
ANSWER: B
7. A local transportation planning group is concerned about the lack of car-pooling on
the part of commuters. They are afraid that the proportion of local drivers car-pooling
is below the national average of 20%. A survey of 356 local drivers reveals that
18.7% of them car pool. What is your conclusion?
A)
There is evidence that the proportion of local people car-pooling is not
below the national average.
B)
There is evidence that the proportion of local people car-pooling is below
the national average.
C)
There is no evidence that the proportion of local people car-pooling is
below the national average.
D)
There is no evidence that the proportion of local people car-pooling is not
below the national average.
ANSWER: C
8. Suppose that you want to test H 0 : P = 0.53 vs. H 0 : P  0.53. The appropriate
critical value would be:
A)
B)
C)
tn1,  / 2
D)
tn 1, 
z
z / 2
ANSWER:
B
9. Suppose that you want to test H 0 : P = 0.53 vs. H 0 : P  0.53. The test statistic
would be:
A)
( p̂ - P0 ) / P0 (1  P0 ) / n
B)
(p - P0 ) / P(1  P) / n
C)
( p̂ - P0 ) / P0 (1  P0 ) / n
D)
(p - P0 ) / [( P0 )(1 - P0 ) /n ]
ANSWER: A
2
QUESTIONS 10 THROUGH 12 ARE BASED ON THE FOLLOWING
INFORMATION:
The state lottery office claims that the average household income of those people
playing the lottery is greater than $37,000. Assume that the distribution of household
income of those people playing the lottery is normally distributed with a standard
deviation of $5,756. Suppose that for a sample of 25 households, it is found that the
average income was $36,243.
12. What are the appropriate null and alternative hypotheses?
A) H 0 :  = $37,000
B) H 0 :  = $37,000
C) H 0 :  = $37,000
D) H 0 :  > $37,000
ANSWER: C
and
and
and
and
H1 :
H1 :
H1 :
H1 :




> $37,000
 $37,000
< $37,000
 $37,000
13. What is the test statistic for this test?
A)
t = 0.66.
B)
Z = 0.66.
C)
Z = 1.92.
D)
t = 1.92.
ANSWER: B
14. What is the most accurate estimate of the p-value?
A)
p-value > 0.10
B)
p-value < 0.05
C)
p-value < 0.10
D)
p-value < 0.01
ANSWER: A
15. Suppose you have the following null and alternative hypotheses: H 0 :  = 277 and
H 1 :   277, and you know that  = 13.5. Take a random sample of 20 observations
and let  = 0.05. For what values of sample mean will you reject the null hypothesis?
A)
Sample mean > 282.9.
B)
Sample mean < 271.1.
C)
Sample mean > 282.9.
D)
Sample mean > 271.1.
ANSWER: B
16. Which of the following statements is not true?
A.
A test statistic is a function of the sample data on which the decision to
reject or not reject the null hypothesis is to be based.
B.
A rejection region consists of the set of all test statistic values for which
the null hypothesis will be rejected.
C.
A rejection region consists of the set of all test statistic values for which
the alternative hypothesis will be rejected.
D.
A good hypothesis-testing procedure is one for which the probability of
making either type I or type II error is small.
ANSWER: C
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17. The supervisor of a production line believes that the average time to assemble an
electronic component is 14 minutes. Assume that assembly time is normally
distributed with a standard deviation of 3.4 minutes. The supervisor times the
assembly of 14 components, and finds that the average time for completion was 11.6
minutes. Is there evidence that the average amount of time required to assemble a
component is something other than 14 minutes? Use  = 0.01
ANSWER: H 0 :  =14 vs. H1 :   14
Test statistic Z= (11.6 - 14) / (3.4 / 6.32) = -4.46. Since p-value = 2P(Z < -4.46) 
0.0041, we reject H 0 at  = 0.01. We conclude that there is sufficient evidence to
disprove the claim that the average time to assemble an electronic component is 14
minutes (The population mean appears to be less than 14 minutes.)
18. The manufacturer of a certain chewing gum claims that four out of five dentists
surveyed prefer their type of gum. You decide to test their claim. You find that in a
sample of 200 doctors, 74.1% do actually prefer their gum. Is this evidence sufficient
to doubt the manufacturer’s claim? Use  = 0.025
ANSWER: H 0 : P = 0.80 vs. H1 : P < 0.80
Test statistics Z =  pˆ  P0  / P0 (1  P0 ) / n  2.09 . Since p-value = P(Z < -2.09) =
0.0183, we reject H 0 at  = 0.025. There is sufficient evidence to doubt the
manufacturer’s claim.
19. The manufacturer of bags of cement claims that they fill each bag with at least
50.2 pounds of cement. Assume that the standard deviation for the amount in each
bag is 1.2 pounds. The decision rule is adopted to shut down the filling machine if the
sample mean weight for a sample of 40 bags is below 49.8. Suppose that the true
mean weight of bags is 50 pounds. Using this decision rule, what is the probability of
a Type II error?
ANSWER:
  P  X  X crit |       P  X  49.8 |   50  = P(Z > -1.05) = 0.8531
20. An assembly line will be shut down for maintenance if the defect rate exceeds
2.3%. Suppose you adopt a 5% significance level and take a random sample 200
items off the assembly line and compute the proportions that are defective. For what
values of the sample proportion will the assembly line be shut down?
ANSWER:
H 0 : P = 0.023 vs. H 1 : P > 0.023
Reject H 0 if  pˆ  P0  /
 P0 1  P0  / n 
> Z.05 = 1.645, that is, assembly line will be shut
down for maintenance if ( p̂ -0.023) / 0.0106 > 1.645, or equivalently if sample
proportion exceeds 0.0404
QUESTIONS 21 THROUGH 24 ARE BASED ON THE FOLLOWING
INFORMATION:
A pharmaceutical manufacturer is concerned that the mean impurity concentration in
pills should not exceed 2%. It is known that from a particular production run impurity
concentrations follow a normal distribution with standard deviation 0.32%. A random
sample of 64 pills from a production run was checked, and the sample mean impurity
concentration was found to be 2.05%.
21. Test at the 5% level the null hypothesis that the population mean impurity
concentration is 2% or less against the alternative that it is more than 2%.
4
ANSWER: H 0 :   2 and H1 :   2 . Reject H 0 if Z  Z.05  1.645
Z
2.05  2
 1.25 , therefore, we fail to reject H 0 at the 5% level. We conclude that
0.32 / 64
the population mean impurity concentration is 2% or less.
22. Calculate the p-value for this test.
ANSWER: p -value = P( Z > 1.25) = 0.50 – 0.3944 = 0.1056
23. Suppose that the alternative hypothesis in Question 103 had been two-sided rather
than one-sided. State, without doing the calculations, whether the p-value of the test
would be higher than, lower than, or the same as that found in Question 104. Explain
your reasoning.
ANSWER: The p-value would be higher (namely; 0.2112) – since the p-value now
corresponds to the area in both the lower and upper tails of the distribution whereas
before it was the area in the upper tail only.
24. In the context of this problem, explain why a one-sided alternative hypothesis is
more appropriate than a two-sided alternative.
ANSWER: A one-sided alternative is more appropriate since we are not interested in
detecting possible low levels of impurity, only high levels of impurity.
QUESTIONS 25 THROUGH 29 ARE BASED ON THE FOLLOWING
INFORMATION:
Consider a problem with the hypothesis test H 0 :   4 vs. H1 :   4 , and the following
decision rule: Reject H 0 if
x 4
 1.645 .
0.08 / 25
25.
Rewrite the decision rule explicitly in terms of x
ANSWER:
x 4
 1.645  x  4  1.645 0.08/ 25  4.026  Reject H 0 if x > 4.026
0.08 / 25


26. Compute the probability of Type II error and the power of the test for  = 4.06
ANSWER:
4.026  4.06 

0.08/ 25 

= P( Z  2.13)  0.50  0.4834  0.0166

  P( X  4.026 |   4.06)  P  Z 
Power = 1 -  = 1 – 0.0166 = 0.9834
27. Compute the probability of Type II error and the power of the test for  = 4.04
ANSWER:
4.026  4.04 

0.08/ 25 

= P( Z  0.88)  0.50  0.3106  0.1894

  P( X  4.026 |   4.04)  P  Z 
Power = 1 -  = 1 – 0.1894 = 0.8106
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28. Compute the probability of Type II error and the power of the test for  = 4.01
ANSWER:
4.026  4.01 

0.08/ 25 

= P( Z  1.0)  0.50  0.3413  0.8413

  P( X  4.026 |   4.01)  P  Z 
Power = 1 -  = 1 – .8413 = 0.1587
29. Compute the probability of Type II error and the power of the test for  = 4.0
ANSWER:
4.026  4.0 

0.08/ 25 

= P( Z  1.63)  0.50  0.0.4484  0.9484

  P( X  4.026 |   4.0)  P  Z 
Power = 1 -  = 1 – .9484 = 0.0516
QUESTIONS 30 THROUGH 34 ARE BASED ON THE FOLLOWING
INFORMATION:
A wine producer claims that the proportion of its customers who cannot distinguish its
product from frozen grape juice is at most 0.12. The producer decides to test this null
hypothesis against the alternative that the true proportion is more than 0.12. The
decision rule adopted is to reject the null hypothesis if the sample proportion who
cannot distinguish between these two flavors exceeds 0.15.
30. If a random sample of 100 customers is chosen, what is the probability of a Type I
error, using this decision rule?
ANSWER:
H 0 : P  0.12 and H1 : P  0.12 .
The decision rule is: Reject H 0 if p̂ > 0.15


0.15  0.12


(0.12)(0.88) /100 

 P( Z  0.92)  0.50  0.0.3212  0.1788
  P( pˆ  0.15 | P  0.12)  P  Z 
31. If a random sample of 400 customers is selected, what is the probability of a Type
I error, using this decision rule?
ANSWER:


0.15  0.12


(0.12)(0.88) / 400 

 P( Z  1.85)  0.50  0.4678  0.0322
  P( pˆ  0.15 | P  0.12)  P  Z 
32. Suppose that the true proportion of customers who cannot distinguish between
these flavors is 0.10. If a random sample of 100 customers is selected, what is the
probability of a Type II error?
ANSWER:


0.15  0.20


(0.20)(0.80) /100 

= P( Z  1.25)  0.5  0.3944  0.1056
  P( pˆ  0.15 | P  0.20)  P  Z 
6