Download Chapter 8 Practice Key

Directions: SHOW YOUR WORK. Include the items discussed in class (PHATC steps;
include original claim, the null hypothesis, the alternative hypothesis, a graph, test
statistic, critical values, etc.) and write a conclusive statement regarding the claim.
1. A researcher wants to check the claim that convicted burglars spend an average of
18.7 months in jail. She takes a random sample of 11 such cases from court files
and finds that X = 21.2 months and s = 7.1 months. Assume the population is
normal.
a) Test the null hypothesis that  = 18.7 at the 0.05 significance level.
 = average amount of time spent in jail
Original Claim:   18.7
H 0 :   18.7
H a :   18.7
 SRS
 Population normal (given)
  is unknown = t distribution
21.2  18.7
= 1.17; p-value = tcdf (1.17, 1000,10) = .27 which is not less than .05.
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Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that
the mean is 18.7 months.
t
b) What is a type I error in this scenario?
Rejecting   18.7 when in fact   18.7 is true.
2. A manufacturer makes steel rods that are supposed to have a mean length of 50
cm. A retailer claims that the bars are running long. A sample of 40 bars is taken
and their mean length is determined to be X = 51 cm with a standard deviation of
s = 3.6.
a) Test the retailer’s claim that  <50 cm at the 1 percent significance level.
 = average mean length of steel rods in cm.
Original Claim:   50
H 0 :   50
H a :   50
 SRS
 n>30
  is unknown = t distribution
51  50
=1.76; p-value = .043 is not less than .01 so Fail to Reject H 0 . There is not
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40
sufficient evidence to support the claim that the mean length is greater than 50 cm.
t
b) What is a type II error in this scenario?
Failing to reject an untrue H 0 . That is, we do not reject   50 when in fact   50 .
3. In a sample of 164 children selected randomly from one town, it is found that 36
of them suffer from asthma.
a) At the 0.05 significance level, test the claim that the proportion of all
children in the town who suffer from asthma is 11%.
Original claim: p = .11
p = proportion of children who suffer from asthma.
H 0 : p = .11
H a : p  .11
36
pˆ 
=.2195
164
 SRS
 2NIP
 Np = (164) (.2195)=35.998>5 and Nq = (164)(.7805)=128.002>5
.2195  .11
=4.48; pvalue (find by looking up 4.48 in Table A-2 and subtracting
z
(.11)(.7805)
164
from 1) = .0000739<.05. Reject H 0 and conclude there is sufficient evidence to warrant
rejection of the claim that the proportion of all children who suffer is 11%.
b) What is Type I error in this scenario?
Rejection a true H 0 ; that is, rejecting p = .11 in favor of p  .11 when in fact
p = .11 is true.
c) What is the probability of a Type I error in this scenario?
P(Type I Error) = .05.
4. A researcher claims that the amounts of acetaminophen in a certain brand of cold
tablets have a mean different from the 600 mg claimed by the manufacturer. The
mean acetaminophen content for a random sample of n = 49 tablets is 604.2mg
with a population standard deviation of 4.2 mg.
a) Test this claim at the 0.02 level of significance.
 =the mean amount of acetaminophen in a tablet.
Original Claim:   50
H 0 :   600
H a :   600
 SRS
 n>30
  is known = z distribution
604.2  600
=7.00; pvalue is .0001, multiply this by 2 gives you .0002. Critical
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Values are 2.33 . Reject the H 0 ; There is sufficient evidence to support the claim
that the mean acetaminophen content is different than 600 mg.
z
b) What is a Type II error in this scenario?
Failing to Reject a bad (untrue) H 0 . That is, not rejecting   600 when in
fact   600 is true.
5. A manufacturer makes ball bearings that are normally distributed and are
supposed to have a mean weight of 30 g. A retailer suspects that the mean weight
is actually less than 30g. The mean weight for a random sample of 16 ball
bearings is 29.5 g with a standard deviation of 4.3 g. At the 0.05 significance
level, test the claim that the mean is less than 30g.
 = mean weight of ball bearings
Original Claim:   50
H 0 :   30
H a :   30
 SRS
 N is not greater than 30, but the population is normally distributed, so the sample
is, too.
  is unknown = t distribution
29.5  30
= -.47; p-value = .3226; Fail to Reject H 0 . There is not sufficient evidence
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to support the claim that the mean length is less than 30g.
t
6. A light-bulb manufacturer advertises that the average life for its light bulbs is 900
hours. A random sample of 15 of its light bulbs resulted in the following lives in
hours:
995
590
510
539
739
917
571
555
916
728
664
693
708
887
849
At the 10% significance level, do the data provide evidence that the mean life for
the company’s light bulbs differs from the advertised mean?
 = mean life in hours of light bulbs
Original Claim:   900
H 0 :   900
H a :   900
 SRS
 N is not greater than 30, and we don’t know if the population is normal; proceed
with caution.
  is unknown = t distribution
724.06  900
= -4.342; p-value = .000676; Reject H 0 . There is sufficient evidence to
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reject the claim that the mean life for light bulbs is equal to 900 hours.
t
7. According to a recent poll, 53% of Americans would vote for the incumbent
president. If a random sample of 100 people results in 45% who would vote for
the incumbent, test the claim that the actual percentage is 53%. Use a 0.10
significance level.
Original claim: p = .53
p = proportion of people who vote for an incumbent president.
H 0 : p = .53
H a : p  .53
p̂  .45
 SRS
 2NIP
 Np = (100) (.45)=45 >5 and Nq = (100)(.55)=55 >5
.45  .53
= -1.60; Critical Values: 1.645 ; 2 * pvalue = 2(.0548)=.1096 which is
z
(.53)(.55)
100
not less than .05. Fail to reject H 0 and conclude there is not sufficient evidence to
warrant rejection of the claim that the proportion of people who vote for an incumbent
president is 53%.
8. In test of a computer component, it is found that the mean time between failures is
937 hours. A modification is made which is supposed to increase reliability by
increasing the time between failures. Tests on a sample of 36 modified
components produce a mean time between failures of 983 hours, with a
population standard deviation of 52 hours. At the 0.01 level of significance, test
the claim that for the modified components, the mean time between failures is
greater than 937 hours.
 =the mean time between failures in computer components.
Original Claim:   937
H 0 :   537
H a :   537
 SRS
 n>30
  is known = z distribution
983  937
= 5.31; Critical Value is 2.33; pvalue is .0001, multiply this by 2 to get
52
36
.0002. Reject the H 0 ; There is sufficient evidence to support the claim that the mean
time between failure sin computer components is not 537 hours.
z
9. A large software company gives job applicants a test of programming ability and
the mean for that test has been 160 in the past. 25 job applicants are randomly
selected from one large university and they produce a mean score and standard
deviation of 183 and 12, respectively. Use a 0.05 level of significance to test the
claim that this sample comes from a normal population with a mean score greater
than 160.
 = mean score on test
Original Claim:   160
H 0 :   160
H a :   160
 SRS
 N is not greater than 30, but the population is normally distributed, so the sample
is, too.
  is unknown = t distribution
183  160
= 9.583; p-value = .0001; Reject H 0 . There is sufficient evidence to
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25
support the claim that the mean score on the test is greater than 160.
t
10. A brochure claims that the average maximum height a certain type of plant is
0.7m. A gardener suspects that this estimate is not accurate locally due to soil
conditions. A random sample of 42 mature plants is taken. The mean height of the
plants in the sample is 0.65 m with a population standard deviation of 0.2m. Test
the claim made in the brochure at the 1 percent level of significance.
 = the average height of a certain plant.
Original Claim:   .7
H 0 :   .7
H a :   .7
 SRS
 n>30
  is known = z distribution
.65  .7
= -1.62; Critical Values are 2.575 ; pvalue = 2(.0526) = .1052; Fail to
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reject the H 0 ; There is not sufficient evidence to reject the claim that the mean height
of a certain plant is .7.
z
11. To judge certain safety features of a car, an engineer must test the claim that the
reaction time of drivers to a given emergency situation has a standard deviation of
more than 0.010 seconds. What can she conclude at the 0.05 level of significance
if she gets s = 0.014 for a sample of size n = 15? Assume the population is
normally distributed.
 = standard deviation of reaction time to an emergency in seconds
Claim:   .01
H 0 :   .01
H a :   .01
 SRS
 Population is normal
14 (.014) 2
=27.44; pvalue =  2 cdf(27.44, 1000, 14) = .04999 < .05; Reject Ho and
 =
2
(.010)
conclude that   .01 . Critical Value is: 23.685 and it’s a right tailed test.
2
12. The specifications for the mass production of certain springs require that the
standard deviation of their compressed lengths be less than 0.040 cm. If a random
sample of size n = 35 from a certain production lot yields s = 0.053 and the
probability of a Type I error is not to exceed 0.01, does this constitute evidence
for the null or for the alternative hypothesis? Assume the population is normal.
 = standard deviation of compressed lengths
Claim:   .04
H 0 :   .04
H a :   .04
 SRS
 Population is normal
34 (.053) 2
=59.69; pvalue =  2 cdf(0, 59.69, 34) = .9958 is not less than .05; Fail to
2
(.04)
reject Ho and conclude there is insufficient evidence to conclude that   .04 . Critical
Value is: 14.954 and it’s a left tailed test.
2 =