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Hypothesis Tests – 1 sample Means
1. Marriage. In 1960, census results indicated that the age
at which American men first married had a mean of
23.3 years. It is widely suspected that young people
today are waiting longer to get married. We want to
find out if the mean age of first marriage has increased
during the past 40 years.
a) Write appropriate hypotheses.
b) We plan to test our hypothesis by selecting a
random sample of 40 men who married for the first
time last year. Do you think the necessary
assumptions for inference are satisfied? Explain.
c) Describe the approximate sampling distribution
model for the mean age in such samples.
d) The men in our sample married at an average age
of 24.2 years with a standard deviation of 5.3
years. What's the P-value for this result?
e) Explain (in context) what this P-value means.
f) What's your conclusion?
2. Fuel economy. A company with a large fleet of cars
hopes to keep gasoline costs down, and sets a goal of
attaining a fleet average of at least 26 miles per gallon.
To see if the goal is being met, they check the gasoline
usage for 50 company trips chosen at random, finding
a mean of 25.02 mpg and a standard deviation of 4.83
mpg. Is this strong evidence that they have failed to
attain their fuel economy goal?
a) Write appropriate hypotheses.
b) Are the necessary assumptions to perform
inference satisfied?
c) Describe the sampling distribution model of mean
fuel economy for samples like this.
d) Find the P-value.
e) Explain what the P-value means in this context.
f) State an appropriate conclusion.
3. Cars. One of the important factors in auto safety is the
weight of the vehicle. Insurance companies are
interested in knowing the average weight of cars
currently licensed in the United States; they believe it
is 3000 pounds. To see if that estimate is correct, they
checked a random sample of 91 cars. For that group the
mean weight was 2919 pounds, with a standard
deviation of 531.5 pounds. Is this strong evidence that
the mean weight of all cars is not 3000 pounds?
4. Portable phones. A manufacturer claims that a new
design for a portable phone has increased the range to
150 feet, allowing many customers to use the phone
throughout their homes and yards. An independent
testing laboratory found that a random sample of 44 of
these phones worked over an average distance of 142
feet, with a standard deviation of 12 feet. Is there
evidence that the manufacturer's claim is false?
5. Chips ahoy. In 1998, as an advertising campaign, the
Nabisco Company announced a "1000 Chips
Challenge," claiming that every 18-ounce bag of their
Chips Ahoy cookies contained 1000 chocolate chips.
Dedicated Statistics students at the Air Force Academy
(no kidding) purchased some randomly selected bags
of cookies, and counted the chocolate chips. Some of
their data are given below. (Chance, 12, no. 1 [1999])
1219
1244
1214
1258
1087
1356
1200
1132
1419
1191
1121
1270
1325
1295
1345
1135
a) Check the assumptions and conditions for
inference. Comment on any concerns you have.
b) What does this evidence say about Nabisco's
claim?
6. Yogurt. Consumer Reports tested 14 brands of vanilla
yogurt and found the following numbers of calories per
serving:
160 130 200 170 220 190 230
80 120 120 180 100 140 170
a) Check the assumptions and conditions for
inference.
b) A diet guide claims that you will get 120 calories
from a serving of vanilla yogurt. What does this
evidence indicate?
7. Maze. Psychology experiments sometimes involve
testing the ability of rats to navigate mazes. The mazes
are classified according to difficulty, as measured by
the mean length of time it takes rats to find the food at
the end. One researcher needs a maze that will take rats
an average of about one minute to solve. He tests one
maze on several rats, collecting the data at the right.
a) Plot the data. Do you think
Time (sec)
the conditions for inference
38.4
57.6
are satisfied? Explain.
46.2
55.5
b) Test the hypothesis that the
62.5
49.5
mean completion time for
38.0
40.9
this maze is 60 seconds.
62.8
44.3
What is your conclusion?
33.9
93.8
c) Eliminate the outlier, and
50.4
47.9
test the hypothesis again.
35.0
69.2
What is your conclusion?
52.8
46.2
d) Do you think this maze
60.1
56.3
meets the "one-minute
55.1
average" requirement?
Explain.
8. Braking. A tire manufacturer is considering a newly
designed tread pattern for its all-weather tires. Tests
have indicated that these tires will provide better gas
mileage and longer treadlife. The last remaining test is
for braking effectiveness. The company hopes the tire
will allow a car traveling at 60 mph to come to a
complete stop within an average of 125 feet after the
brakes are applied. They will adopt the new tread
pattern unless there is strong evidence that the tires do
not meet this objective. The distances (in feet) for 10
stops on a test track were 129, 128, 130, 132, 135, 123,
102, 125, 128, and 130. Should the company adopt the
new tread pattern? Test an appropriate hypothesis and
state your conclusion. Explain how you dealt with the
outlier, and why you made the recommendation you
did.
Answers:
1. a) H0:  = 23.3; HA:  > 23.3
b) We have a random sample that comprises less than 10%
of the population. Population may not be normally
distributed, as it would be easier to have a few much older
men at their first marriage than some very young men.
However, with a sample size of 40, y should be
approximately Normal. We should check the histogram for
severity of skewness and possible outliers.
c) t39(23.3, s/ 40 )
d) 0.1414
e) If the average age at first marriage is still 23.3 years,
there is a 14.5% chance of getting a sample mean of 24.2
years or older simply from natural sampling variation.
f) We have not shown that the average age at first marriage
has increased from the mean of 23.3 years.
2. a) H0:  = 26; HA:  < 26
b) We have a representative sample, less than 10% of all
trips, and a large enough sample that skewness should not
be a problem.
c) t49(26, 4.83/ 50 )
d) 0.0789
e) If the average fuel economy is 26 mpg, the chance of
obtaining a sample mean of 25.02 or less by natural
sampling variation is 8%.
f) Since 0.05 < P < 0.10, there is some evidence that the
company may not be achieving the fuel economy goal.
3. No. Based on this large random sample, t = -1.45; P-value =
0.1494. Because the P-value is high, we fail to reject H0.
These data do not provide evidence to support the claim
that the average weight of all cars is not 3000 pounds. We
retain the null hypothesis that the mean is 3000 pounds.
4. Yes. Based on this large random sample, t = -4.42, P-value =
3.29 X 10-5. Because the P-value is low, we reject H0. These
data show that the range of these phones is not 150 feet; in
fact, it is less.
5. a) Random sample; less than 10% of all Chips Ahoy bags; the
nearly Normal condition seems reasonable from a Normal
probability plot. The histogram is roughly unimodal and
symmetric with no outliers.
b) t = 10.105, P-value = 4.35 x 10-8. Since p-value < α, we
reject that the population mean amount of chocolate chips
in Chips Ahoy is 1000.
6. a) Random sample; this is less than 10% of all vanilla yogurt
brands. Nearly Normal condition is reasonable by
examining a Normal probability plot. The histogram is
roughly unimodal (although somewhat uniform) and
symmetric with no outliers.
b) t = 3.165, P-value = 0.0075. Since p-value < α, we reject
that the population mean amount of calories in vanilla
yogurt is 120.
7. a) The Normal probability plot is relatively straight, with one
outlier at 93.8 sec. Without the outlier, the conditions seem
to be met. The histogram is roughly unimodal and
symmetric with no other outliers.
b) t = -2.63, P-value = 0.0160. With the outlier included, we
might conclude that the mean completion time for the maze
is not 60 seconds; in fact, it is less.
c) t = -4.46, P-value = 0.0003. Because the P-value is so
small, we reject H0. Without the outlier, we conclude that
the average completion time for the maze is not 60 seconds;
in fact, it is less. The outlier here did not change the
conclusion.
d) The maze does not meet the "one-minute average"
requirement. Both tests rejected a null hypothesis of a mean
of 60 seconds.
8. The data value of 102 feet is an outlier. When this is
removed, the Normal probability plot is relatively straight.
The Nearly Normal Condition seems satisfied. With the
outlier removed, the histogram is roughly unimodal and
symmetric with no other outliers. H0:  = 125; HA:  > 125
feet. With the outlier eliminated, x = 128.89, t = 3.29, Pvalue = 0.01. With a P-value this low, we reject H0. There is
strong evidence to suggest that the tires will not bring the
car to a complete stop within 125 feet. On the basis of these
data, the company should not adopt the new tread pattern.
Only 2 out of the 10 data values were less than the desired
125 feet, and 1 of these was an outlier.