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CHAPTER 10 REVIEW
1) Each of the following situations requires a significance test about a population mean μ. State the
appropriate null hypothesis H0 and alternative hypothesis Ha in each case.
a) Larry’s car averages 32 miles per gallon on the highway. He now switches to a new motor oil that
is advertised as increasing gas mileage. After driving 3000 highway miles with the new oil, he wants
to determine if his gas mileage actually has increased.
b) A university gives credit in French language courses to students who pass a placement test. The
language department wants to see if students who get credit in this way differ in their understanding of
spoken French from students who actually take the French courses. Some faculty think the students
who test out of the courses are better, but others argue that they are weaker in oral comprehension.
Experience has shown that the mean score of students in the courses on a standard listening test is 24.
The language department gives the same listening test to a sample of 40 students who passed the credit
examination to see if their performance is different.
ANS:
a) H0: μ = 32 mpg; Ha: μ > 32 mpg
b) H0: μ = 24; Ha: μ ≠ 24
2) Sulfur compounds cause “off-odors” in wine, so winemakers want to know the odor threshold, the
lowest concentration of a compound that the human nose can detect. The odor threshold for dimethyl
sulfide (DMS) in trained wine tasters is about 25 micrograms per liter of wine (μg/l). The untrained
noses of consumers may be less sensitive, however. Here are the DMS odor thresholds for 10
untrained students:
31 31 43 36 23 34 32 30 20 24
Assume that the standard deviation of the odor threshold for untrained noses is known to be σ = 7 μg/l.
a) Make a stemplot to verify that the distribution is roughly symmetric with no outliers.
b) Give a 95% confidence interval for the mean DMS odor threshold among all students.
c) Are you convinced that the mean odor threshold for students is higher than the published threshold,
25 μg/l? Carry out a significance test to justify your answer.
ANS:
a) 2 | 0 3 4
3|011246
4|3
b) Confidence interval = 30.4  1.960( 7
)  (26.06,34.74)
10
c) H0: μ = 25
Ha: μ > 25
Method 1: Since 25 lies outside the confidence interval, we reject the null hypothesis at the
5% significant level.
Method 2:
30.4  25
z
 2.44
7 10
Since 2.44 > 1.960 (z > z*), we reject the null hypothesis.
Method 3:
z = 2.44  P-value = 1 – 0.9927 = 0.0073
Since 0.0073 < 0.05, we reject the null hypothesis.
3) Researchers studying iron deficiency in infants examined infants who were following different
feeding patterns. One group of 26 infants was being breast-fed. At 6 months of age, these children
had mean hemoglobin level x-bar = 12.9 grams per 100 milliliters of blood. Assume that the
population standard deviation is σ = 1.6. Give a 90% confidence interval for the mean hemoglobin
level of breast-fed infants.
ANS:
  (12.38,13.42)
Confidence interval = 12.9  1.6451.6

26


4) Here are the Degree of Reading Power (DRP) scores for an SRS of 44 third-grade students from a
suburban school district:
40 26 39 14 42 18 25 43 46 27 19
47 19 26 35 34 15 44 40 38 31 46
52 25 35 35 33 29 34 41 49 28 52
47 35 48 22 33 41 51 27 14 54 45
DRP scores are approximately normal. Suppose that the standard deviation of scores in this school
district is known to be σ = 11. The researcher believes that the mean score μ of all third graders in this
district is higher than the national mean, which is 32.
a) State H0 and Ha to test this suspicion.
b) Carry out the test.
ANS:
a) H0: μ = 32 vs. Ha: μ > 32
35.09  32
b) z 
 1.86  P  value  0.031
11
44
Since 0.031 < 0.05, there is a strong evidence against H0.
5) The survey of Study Habits and Attitudes (SSHA) is a psychological test that measures the
motivation, attitude toward school, and study habits of students. Scores range from 0 to 200. The
mean score for U.S. college students is about 115, and the standard deviation is about 30. A teacher
who suspects that older students have better attitudes toward school gives the SSHA to 20 students
who are at least 20 years of age. Their mean score is x-bar = 135.2.
Assuming that σ = 30 for the population of older students, carry out a test of
H0: μ = 115
Ha: μ > 115
Report the P-value of your test, and state your conclusion clearly.
ANS:
135.2  115
z
 3.01  P  value  0.0013
30
20
Since 0.0013 < 0.01, there is an overwhelming evidence against the H0.
6) The mean yield of corn in the United States is about 120 bushels per acre. A survey of 40 farmers
this year gives a sample mean yield of x-bar = 123.8 bushels per acre. We want to know whether this
is good evidence that the national mean this year is not 120 bushels per acre. Assume that the farmers
surveyed are an SRS from the population of all commercial corn growers and that the standard
deviation of the yield in this population is σ = 10 bushels per acre. State the hypotheses and give the
P-value. Are you convinced that the population mean is not 120 bushels per acre?
ANS:
H0: μ = 120
Ha: μ ≠ 120
123.8  120
z
 2.40  P  value  0.016
10
40
Since 0.016 < 0.05, there is a strong evidence against the H0.
7) A company that manufacturers classroom chairs for high school students claims that the mean
breaking strength of the chairs they make is 300 pounds. From year of production, they have seen that
σ = 15 pounds. One of the chairs collapsed beneath a 220-pound student last week. You wonder
whether the manufacturer is exaggerating about the breaking strength of their chairs.
a) State the null and alternative hypotheses in words and symbols.
b) Describe a Type I error and a Type II error in this situation. Which is more serious?
c) There are 30 chairs in your classroom. You decide to determine the breaking strength of each
chair, and then to find the mean of those values. What values of x-bar would cause you to reject H0 at
the 5% significance level?
d) If the truth is that μ = 270 pounds, find the probability that you will make a Type II error.
ANS:
a) “Exaggeration” in this case would mean that the mean breaking strength of the company’s chairs
was actually less than the claimed value of 300 pounds. Let μ = the mean breaking strength of all
chairs produced by the company, we let H0: μ = 300, Ha: μ < 300.
b) Type I error: concluding that the company’s claim is exaggerated (μ < 300) when in fact it is
legitimate (μ = 300).
Type II error: concluding that the company’s claim is legitimate when in fact it is invalid.
Type II error would be more serious in this case, because allowing the company to continue the
“false advertising” of its chairs’ strength could lead to injuries, lawsuits, and other serious
consequences.
c) In order to reject H0 at level α = 0.05, we must have z ≤ -1.645, or
  295.495. All values of x-bar at or below 295.495 pounds would cause
x  300  (1.645)15

30 

us to reject H0.
d) P(Type II error) = P(fail to reject H0 when μ = 270) = P(x-bar > 295.495 when μ = 270)
295.495  270
z
 9.31  p  value  0
15
30
8) An airplane is interested in determining the average number of unoccupied seats for all of its
flights. It selects an SRSV of 100 flights and determines that the average number of unoccupied seats
for the sample is 12.5 seats with a sample standard deviation of 3.9 seats. Construct a 95% confidence
interval for the true number of unoccupied seats for all flights. Interpret the confidence interval in the
context of the problem.
ANS:
For a 95% confidence interval, z* = 1.960
  (11.7,13.3)
12.5  1.96 3.9

100


We are 95% confident that the true mean number of unoccupied seats is between 11.7 and 13.3 seats.
9) A machine for inflating tires, when properly calibrated, inflates tires to 32 pounds, but it is known
that the machine varies with a standard deviation of about 0.8 lbs. How large a sample is needed in
order to be 99% confident that the mean inflation pressure is within ±0.10 lbs.?
ANS:
2
 2.576(0.8) 
n
  424.49
0.1


Because the sample size needed must be an integer, 424 might not be enough. You would need a
sample of 425 tires. In general, you should round up to the next integer when choosing a sample size
by this method.
10) Suppose it turns out that Todd’s 50 throws are approximately normally distributed with mean 47.5
yards and standard deviation 8 yards. His claim is that he can average 50 yards per throw. What is the
probability of getting a finding this far below expected by chance alone if his true average is 50 yards?
Is this finding significant at α = 0.05? At α = 0.01?
ANS:
47.5  50
P( x  47.5)  P( x  47.5)  P( z 
 2.21)  0.014
8
50
Two-tailed test
Since 47.5 < 50
It is significant at the 0.05 level because the P-value = 0.014 < 0.05.
It is not significant at the 0.01 level because the P-value = 0.014 > 0.01.
11) The hypothesis that the Giants would win the World Series in 2002 was held by many of their
fans. What type of error has been made by a fan that refuses to accept the fact that the Giants actually
lost the series to the Angels?
ANS:
The hypothesis is false but the fan has failed to reject it. That is a type-II error.
12) A hotel chain claims that the average stay for its business clients is 5 days with a standard
deviation of 2.3 days. The hotel believes that the true stat may actually be fewer than 5 days. A study
conducted the hotel of 100 randomly selected clients yield a mean of 4.55 days with a standard
deviation of 3.1 days. What is the probability of getting a finding as extreme, or more extreme, as
4.55 if the true mean is really 5 days? That is, what is the P-value of this finding?
ANS:
In this problem, H0: μ = 5 and Ha: μ < 5.
4.55  5
P( x  4.55)  P( z 
 1.96)  0.0250
2.3
100
Thus, the P-value = 0.0250
13) A university is worried that they might not have sufficient housing for their students for the next
academic year. It’s very expensive to build additional housing, so they are operating under the
assumption (hypothesis) that the housing they have is sufficient, and they will spend the money to
build additional housing only if they are convinced it is necessary (that is, they reject their hypothesis.)
a) For the university’s assumption, what is the risk involved in making a type-I error?
b) For the university’s assumption, what is the risk involved in making a type-II error?
ANS:
a) A type-I error is made when we mistakenly reject a true null hypothesis. In this situation, that the
means that we would mistakenly reject the true hypothesis that the available housing is sufficient. The
risk would be that a lot of money would be spent on building additional housing when it wasn’t
necessary.
b) A type-II error is made when we mistakenly failed to reject a false hypothesis. In this situation that
means we would fail to reject the false hypothesis that the available housing is sufficient. The risk is
that the university would have insufficient housing for its students.
14) Professor Olsen has taught statistics for 41 years and has kept the scores of every test he has ever
given. Every test has been worth 100 points. He is interested in the average test score over the years.
He doesn’t want to put all of the scores (there are thousands of them) into a computer to figure out the
exact average so he asks his daughter, Anna, to randomly select 50 of the tests and use those to come
up with an estimate of the population average. Anna has been studying statistics at college and
decides to create a 98% confidence interval for the true average test score. The mean test score for the
50 random selected tests she selects is 73.5 with a standard deviation of 7.1. What does she tell her
father?
ANS:
  (71.85,75.15)
Confidence interval = 73.5  2.326 7.1

50 

Anna tells her father that he can be 98% confident that the true average test score for his 41 years of
teaching is between 71.85 and 75.15.
15) Ball bearings must be manufactured to strict tolerances or they will not function properly. If a
quality control inspector randomly samples ball bearings from a production process, what is the
minimum sample size required to be 99% confident that the mean diameter is within 0.04 mm? Past
data indicate that the standard deviation of bearing diameters is 0.10 mm.
ANS:
Margin of error = 0.04 mm
2
 2.576(0.1) 
n
  41.4736
 0.04 
Round up to next integer, we need to sample 42 ball bearings for the required margin of error.
16) In the year 1976, the mean beef consumption of U.S. residents was 89 pounds per person. A
health studies researcher wishes to know if this has changed and will do a study. Write the appropriate
null and alternative hypotheses to test whether mean beef consumption has changed.
ANS:
The null hypothesis is the statement of no difference and is H0: μ = 89 pounds. The researcher seeks
evidence of a change in consumption, so the alternative hypothesis is Ha: μ ≠ 89 pounds.
17) A pharmaceutical company claimed on television ads that its new medicine could prevent the
common cold. Soon afterward, a consumer protection group started receiving complaints from
consumers that the medicine did not perform as advertised. The consumer protection group decided to
investigate the company’s claim.
a) Which decision by consumer protection group might lead to the type-I error?
b) Which decision by consumer protection group might lead to the type-II error?
ANS:
The consumer protection group decided to test the following set of hypotheses:
H0: The medicine prevents the common cold, as advertised.
Ha: The medicine does not prevent the common cold.
a) Concluding that the medicine does not prevent the common cold when it does will lead to the typeI error.
b) Concluding that the medicine does prevent the common cold when it does not will lead to the typeII error.
18) Most students enroll in courses with the intention of completing them. But after taking into
account their midterm grades, some students will consider dropping a course. In other words, they
want to decide whether to
H0: Complete the course
Ha: Drop the course
a) Which decision by a student may lead to the type-I error?
b) Which decision by a student may lead to the type-II error?
ANS:
a) Dropping a course when the student should have completed it.
b) Staying in a course when the student should have dropped it.
19) A factory cans fresh pineapple in sugar syrup. The manager in charge is interested in estimating
the average amount of sugar per can to within 2 mg of the true mean. From previous experiments, he
knows that the standard deviation of the sugar content is approximately 15 mg. Each test of measuring
the amount of sugar in a can costs $5.00. One thousand dollars have been budgeted for this
experiment. Does the manager have enough funds to estimate the average amount of sugar per can
with a 95% confidence?
ANS:
The manager is interested in determining the sample size needed to estimate μ = the mean amount of
sugar per can.
Confidence level = 0.95  1.960
The desired margin of error = 2 mg and standard deviation = 15 mg.
2
 1.960(15) 
So the required sample size is n  
  216.09
2


The manager should sample at least 217 cans. It will cost him at least 217(5) = $1,085. Because he
has a budget of $1,000, he does not have enough money.
20) A company that manufactured bicycles received steel rods in large shipments from a supplier.
From past experience, he knows that σ = 0.12 inches (approximately) and that the lengths are
approximately normally distributed. The manager takes a random sample of 10 rods. The lengths of
the sample rods are as follows:
11.90, 11.94, 12.05, 12.07, 11.97, 12.01, 12.08, 12.05, 12.12, 12.14
a) The manager wants to estimate the mean length of the latest shipment of rods using a 90%
confidence interval. Help this manager to construct a confidence interval and then to interpret it.
b) The company needs rods with a length of 12 inches in order to assemble the bikes properly.
Because too-large and too-small rods are unsuitable, shipments not meeting the requirement are sent
back to the supplier. Help this manager to determine whether the shipment should be accepted or
rejected.
ANS:
The manager is interested in estimating μ = true mean length of rods received in the latest shipment.
a) To construct a 90% confidence interval, we need z* = 1.645.
  (11.97,12.1)
Confidence interval = 12.033  1.645 0.12

10 

Find the mean
from the above data
We are 90% confident that the true mean length of rods from this shipment is between 11.97 and 12.10
inches.
b) H0: μ = 12 inches vs. Ha: μ ≠ 12 inches
Let’s use α = 0.10.
Method 1: Since 12 lies within (11.97, 12.1), we accept the null hypothesis.
Method 2: z* = 1.645
12.033  12
z
 0.87
0.12
10
Since |0.87| < 1.645, we reject the null hypothesis.
Method 3: z = 0.87  P-value = 1 – 0.8078 = 0.1922
Since 0.1922 > 0.1, we reject the null hypothesis.
21) A bottling machine is operating with a standard deviation of 0.12 ounce. Suppose that in an SRS
of 36 bottles the machine inserted an average of 16.1 ounces into each bottle.
a) Estimate the mean number of ounces in all the bottles this machine fills. More specifically, given
an interval within which we are 95% certain that the mean lies.
b) How about a 99% confidence interval estimate?
ANS:
a) 95%  z* = 1.960
 0.12 
Confidence interval = 16.1  1.960
  (16.06, 16.14)
 36 
b) 99%  z* = 2.576
 0.12 
Confidence interval = 16.1  2.576
  (16.05, 16.15)
 36 
22) At a certain plant, batteries are being produced with a life expectancy that has a variance of 5.76
months squared. Suppose the mean life expectancy in an SRS of 64 batteries is 12.35 months.
a) Find a 90% confidence interval estimate of life expectancy for all the batteries produced at this
plant.
b) What would the 90% confidence interval estimate be if the sample mean of 12.35 had come from a
sample of 100 batteries?
ANS:
a) The standard deviation of the population is   5.76  2.4 and the standard deviation of the
2.4
sample means is  x 
 0.3.
64
90%  z* = 1.645
 2.4 
Confidence interval = 12.35  1.645
  (11.86, 12.84)
 64 
b) 90%  z* = 1.645
 2.4 
Confidence interval = 12.35  1.645
  (11.96, 12.75)
 100 
23) A manufacturer claims that a new brand of air-conditioning unit uses only 6.5 kilowatts of
electricity per day. A consumer agency believes the true figure is higher and runs a test on a sample of
size 50. If the sample mean is 7.0 kilowatts with a standard deviation of 1.4, should the
manufacturer’s claim be rejected at a significance level of 5%? Of 1%?
ANS:
H0: μ = 6.5 vs. Ha: μ > 6.5
α = 0.05 or 0.01
7  6.5
P ( x  7)  P ( z 
 2.53)  0.9943
1.4
50
P-value = 1 – 0.9943 = 0.0057
At both the 0.05 and 0.01 level, we reject the null hypothesis because p-value = 0.0057 < 0.05 and
0.01.
24) A coffee dispensing machine is supposed to deliver 8 ounces of liquid into each paper cup, but a
consumer believes that the actual amount is less. As a test he plans to obtain a sample of 36 cups of
the dispensed liquid and, if the mean content is less than 7.75 ounces, to reject the 8-ounce claim. If
he machine operates with a standard deviation of 0.9 ounces, what is the probability that the consumer
will mistakenly reject the 8-ounce claim even though the claim is true?
ANS:
H0: μ = 8 and Ha: μ < 8.
0.9
The standard deviation of sample means is  x 
 0.15
36
7.75  8
 1.67  0.0475
The z-score for 7.75 is z 
0.15
Thus, if the 8-ounce claim is correct, there is a 0.0475 probability that the consumer will still obtain a
sample mean less than 7.75 and will mistakenly reject the claim.
25) A factory manager claims that the plant’s smokestacks spew forth only 350 pounds of pollution
per day. A government investigator suspects that the true value is higher and plans hypothesis test
with critical value of 375 pounds. Suppose the standard deviation in daily pollution poundage is 150
and the true mean is 385 pounds. If the sample size is 100 days, what is the probability that the
investigator will mistakenly fail to reject the factory manager’s false claim?
ANS:
x 
150
100
 15.0
With a true mean of 385, the z-score for 375 is
375  385
 0.67 and the probability to the left of 375
15
is 0.2514.
26) Crop researchers plant 15 plots with a new variety of corn. The yields in bushels per acre are
183.0 193.1 131.0 152.3 174.0 179.2 181.9 143.8 169.0
173.2 156.1 143.0 132.5 171.1 155.0
Assume that σ = 20 bushels per acre.
a) Find the 90% confidence interval for the mean yield μ for this variety of corn.
b) Find the 95% confidence interval.
c) Find the 99% confidence interval.
ANS:
a) 90%  z* = 1.645
  (154.05, 171.04)
Confidence interval = 162.55  1.645 20

15 

b) 95%  1.960
  (152.43, 172.67)
Confidence interval = 162.55  1.960 20

15 

c) 99%  2.576
  (149.25, 175.85)
Confidence interval = 162.55  2.576 20

15 

27) To access the accuracy of a laboratory scale, a standard weight known to weigh 15 grams is
weighed repeatedly. The scale readings are normally distributed with unknown mean (this mean is 15
grams if the scale has no bias.) The standard deviation of the scale readings is known to be 0.002
gram.
a) The weight is weighed ten times. The mean result is 15.0023 grams. Give a 96% confidence
interval for the mean of repeated measurements of the weight.
b) How many measurements must be averaged to get a margin of error of ±0.001 with 96%
confidence?
ANS:
a) 96%  z* = 2.054
  (15.001, 15.004)
Confidence interval = 15.0023  2.054 0.002

10


2
 0.002(2.054) 
b) n  
  16.88  17
0.001


28) The National Center for Health Statistics reports that the mean systolic blood pressure for males
35 to 44 years of age is 128 and the standard deviation in this population is 15. The medical director
of a large company looks at the medical records of 72 executives in this age group and finds that the
mean systolic blood pressure n this sample is x-bar = 126.07. Is this evidence that the company’s
executives have a different mean blood pressure from the general population? Assume that executives
have the same σ = 15 as the general population of middle-aged males.
ANS:
H0: μ = 128
Ha: μ ≠ 128
α = 0.05
126.07  128
z
= -1.09  0.1379
15
72
Since 0.1379 > 0.05, x-bar = 126.07 is not a good evidence that executives differ from other men. We
fail to reject our null hypothesis (H0).
29) A company wants to estimate the mean net weight of all 32-ounce packages of its Yummy Taste
cookies at 95% confidence. It is know that the standard deviation of net weights is 0.1 ounce. Find
the sample size that will yield the margin of error within 0.02 ounces of the population mean.
ANS:
2
 1.960(0.1) 
n
  96.04  97
 0.02 
30) An ecologist would like to estimate the mean carbon monoxide level of the air in a particular city.
The carbon monoxide levels are measured on 14 days during a month and recorded. A histogram of
the 25 readings is roughly symmetrical, with no outlying values. The mean and standard deviation of
these values are 5.4 and 2.2, respectively. Assume the 25 days can be considered a simple random
sample of all days. Compute the 95% confidence interval.
ANS:
95%  z* = 1.960
  (4.5376, 6.2624)
Confidence interval = 5.4  1.960 2.2

25 

31) A random sample of 50 adults identified by their primary care physicians as being overweight
were fed a special high-protein diet for a month and had an average weight loss of 8.5 pounds. If the
standard deviation of the sample was 3.7, what is the probability that the average weight change for
the population of adults identified by their primary care physicians as being overweight would be less
than 10 pounds? Use a 0.01 level of significance.
ANS:
H0: μ = 10
Ha: μ < 10
α = 0.01
8.5  10
z
 2.87  0.0021
3.7
50
Since 0.0021 < 0.01, we reject the null hypothesis.
32) A study was done to determine if 12 to 15 year old girls who want to be engineers differ in IQ
from the average. The mean IQ of all girls in this age range is assumed to be about 100, based on
prior research. A random sample of 49 girls is selected who state that they want to be engineers and
their IQ is measured. The mean IQ of the girls in the sample is 104.5 with a sample standard deviation
of 15. Does this finding provide evidence, at the 0.05 level of significance, that the mean IQ of 12- to
15-year-old girls who want to be engineers differs from the average? Use all 3 methods to answer this
question.
ANS:
Let μ = the true mean IQ of girls who want to be engineers.
H0: μ = 100
Ha: μ ≠ 100
Method 1:
95%  z* = 1.960
  (100.3, 108.7)
Confidence interval = 104.5  1.96015

49 

Since 100 lies outside this interval, we reject the null hypothesis.
Method 2:
104.5  100
z
 2 .1
15
49
Since |2.1| > 1.960, we reject the null hypothesis.
Method 3:
z-score = 2.1  P-value = 1 – 0.9821 = 0.0179
Since 0.0179 > 0.01, we reject the null hypothesis.