STAT202, Safi
Fall 2003
Practice Final Exam
Solutions
SECTION 1: MULTIPLE-CHOICE
1. You have measured the systolic blood pressure of a random sample of 25 employees of a company. A
95% confidence interval for the mean systolic blood pressure for the employees is computed to be
(122,138). Which of the following statements gives a valid interpretation of this interval?
(a) About 95% of the samples of employees have a systolic blood pressure between 122 and 138.
(b) About 95% of the employees in the company have a systolic blood pressure between 122 and
138.
(c) If the sampling procedure were repeated many times, then approximately 95% of the
resulting confidence intervals would contain the mean systolic blood pressure for employees
in the company.
(d) If the sampling procedure were repeated many times, then approximately 95% of the sample
means would be between 122 and 138.
2. In a hypothesis test of a null H0: = 0 versus the alternative Ha:  0, which of the following
statements is NOT equivalent to the others?
(a)
(b)
(c)
(d)
We reject the null hypothesis at the 10% level.
The 95% confidence interval for does not cover 0.
The p-value is less than 0.1.
The 90% confidence interval for does not cover 0.
3. A student wishes to determine the difference in the amount of caffeine in regular and decaffeinated
coffee served at a dinner. She obtains 45 servings of regular and 45 servings of decaf coffee and
measures the caffeine level in each. What analysis should she use to test a difference in the mean of
caffeine level between regular and decaf coffee?
(a) One sample z-test.
(b) Chi-square test.
(c) Two sample t-test.
(d) Matched samples t-test.
4. In a study of leg strength 20 soccer players submit to strength tests. A left-leg strength and a right-leg
strength measurement are made on each player. What analysis should we use to see if there is a
difference between the right and left leg strengths on average?
(a) One sample z-test.
(b) One sample t-test.
(c) Two sample t-test.
(d) Matched pairs t-test.
5. A statistician selects a random sample of 200 seeds from a large shipment of a certain variety of
tomato seeds and tests the sample for percentage germination. If 155 of the 200 seeds germinate, then
a 95% confidence interval for p, the population proportion of seeds that germinate is:
(a)
(b)
(c)
(d)
(.726, .824)
(.717, .833)
(.706, .844)
(.712, .827)
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6. Which of the following is not a property of the t distribution?
(a) The t curve is centered at 0.
(b) The t curve is bell-shaped.
(c) The t curve is more spread out than a z curve.
(d) The t curve tends to spread out as the degrees of freedom increases.
7. Based on a sample, we obtain a 95% confidence interval for the proportion of females as (0.42, 0.51).
What is the sample size?
(a) 468
(b) 472
(c) 118
(d) 452
8. A city planner wants to estimate the proportion of residents in the city who use the public
transportation system regularly. She selects a simple random sample (SRS) of 200 residents and finds
that 32 of the residents regularly use the public transportation system. The Margin of Error for the
90% Confidence Interval for the population proportion is
(a) 0.0426
(b) 0.0011
(c) 0.0508
(d) 0.0429
9. Suppose we are testing the hypothesis Ho: p=0.3 vs. Ha: p0.3 and the only information we are given
from a sample of size n=30 is the 90% confidence interval (0.28, 0.42). Then
(a) Since 0.30 is in the interval, we reject H0.
(b) Since 0.30 is in the interval, we do not reject H0.
(c) Since 0.30 is not in the interval, we do not reject H0.
(d) Since 0.30 is not in the interval, we reject H0.
10. What would be the type I error when testing the hypotheses above?
(a) You claim the true proportion is not 0.30 when it really is.
(b) You claim the true proportion is 0.30 when it really is 0.30
(c) You claim the true proportion is not 0.30 when it is not 0.30.
(d) You claim the true proportion is 0.30 when it is not 0.30
11. A paired difference experiment is conducted to compare the starting salaries of male and female
college graduates who find jobs. Pairs are formed by choosing a male and a female with same major
and similar grade-point averages. Suppose a random sample of 5 pairs and the starting salaries (in
thousands) are as follows:
Pair
1
2
3
4
5
Male
25.9
20.0
28.7
13.5
18.8
Female
24.9
18.5
27.7
13.0
17.8
To test whether the mean starting salary for females is less than that of males with = 0.05, the
absolute value of the test statistic is:
(a) 0.2657
(b) 0.3535
(c) 5.6580
(d) 6.3246
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12. A marketing study was planned for the purpose of determining p, the proportion of consumers who
prefer Brand A soft drink over its main competitor, Brand B. The manager of the project wants to be
99% confident that the point estimate will be no more than 0.04 units from the true value of p.
Determine the sample size needed if no previous information about p is available.
(a) 4148
(b) 1037
(c) 1033
(d) 65
13. A corporation needs to estimate p, the proportion of its employees that commute to work by car. A
pilot study of 100 employees indicated that 70% of them commuted to work by car. Use this
information to determine the number of employees that should be included in the sample if the
company wants to be 90% confident that the estimate of p will be no more than 0.03 units from the
true value of p?
(a) 628
(b) 241
(c) 632
(d) 752
14. A manufacturer receives parts from two suppliers. A SRS of 400 parts from the first supplier
contains 20 defectives. A SRS of 100 parts from the second supplier contains 10 defectives. Let
p1 and p 2 be the proportion of defective parts from the first and the second supplier
respectively. A 95% confidence interval for p1  p2 is
(a) (-0.1126, 0.01256)
(b) (-0.1196, 0.0084)
(c) (-0.01256, 0.1126)
(d) (-0.0084, 0.1196)
15. Popular wisdom is that eating pre-sweetened cereal tends to increase the number of dental caries
(cavities) in children. A sample of children was (with parental consent) entered into a study and
followed for several years. Each child was classified as a sweetened-cereal lover or a non-sweetened
cereal lover. At the end of the study, the amount of tooth damage was measured. Here is the
summary data:
Sugar Bombed No sugar
Sample size
10
15
Sample mean
6.41
5.20
Sample standard deviation 5.0
15.0
A 95% confidence interval for the means difference assuming unequal variances is
(a)
6.41  5.20  2.262
(b)
6.41  5.20   2.262
(c)
6.41  5.20  1.96
(d)
6.41  5.20  2.06912.11
5 15

10 15
25 225

10 15
25 225

10 15
1
1

10 15
3
16. In a restaurant, 90% of people order coffee with their dinner. A simple random sample of 144 patrons
of the restaurant is taken. The probability that at least 94.5% of those patrons will order coffee with
their meal is
(a) 0.4641
(b) 0.5359
(c) 0.0359
(d) 0.0520
17. Which of the following is an example of a matched pairs design?
(a) A teacher compares the pre-test and the post-test scores of students.
(b) A teacher compares the scores of students using a computer-based method of instruction with the
scores of other students using a traditional method of instruction.
(c) A teacher compares the scores of students in her class on a standardized test with national
average score.
(d) A teacher calculates the average of scores of students on a pair of tests and wishes to see if this
average is larger than 80%.
18. Wages for a particular industry are normally distributed with an average of $11.90 an hour and a
standard deviation of $0.40. What percentage of workers receive between $10.90 and $12.90
(a) 0.9938
(b) 0.0062
(c) 0.9876
(d) 0.0124
19. The scores of a reference population on the Wechsler Intelligence Scale for Children (WISC) are
normally distributed with mean, µ = 100, and standard deviation, =15. What score must a child
achieve on the WISC in order to fall in the top 6.3% of the population?
(a) 77.05
(b) -77.05
(c) 129.4
(d) 122.95
For problems 20-21, use the following information:
Parents in Lake Wobegone think their children are smarter than average. To try and prove this they
take a random sample of 2500 children from Lake Wobegone and find their average IQ is 100.8 with a
standard deviation of 15. The average IQ outside of Lake Wobegone is 100.
20. The P-value of the test is
(a) 0.051
(b) 0.012
(c) 0.008
(d) 0.004
21. A 90% confidence interval for the average IQ of Lake Wobegone children is
(a) (100.3,101.3)
(b) (100,101)
(c) (99.1,101.9)
(d) (93.4,108.2)
4
22. A nationwide test has a mean of 75 and a standard deviation of 10. A random sample of 64 examinees
was drawn and their scores were recorded. The probability that the sample mean is greater than 78 is
(a) 0.0082
(b) 0.4918
(c) 0.1179
(d) 0.3821
23. An airplane is only allowed a gross passenger weight of 8000 kg. If the weights of passengers
traveling by air between Toronto and Vancouver have a mean of 78 kg and a standard deviation of 7
kg, the approximate probability that the combined weight of 100 passengers will exceed 8,000 kg is:
(a) 0.4978
(b) 0.3987
(c) 0.0044
(d) 0.0022
24. Salary information for a random sample of male and female employees of a large company is shown
below
Male Female
Sample size
64
36
Sample mean (in $1,000)
44
41
Sample variance
128
72
If we are interested in testing whether or not the average salary of males is significantly greater than
that of females, the test statistic is
(a) 2.0
(b) 1.5
(c) 1.96
(d) 1.645
25. In order to estimate the average time spent on the computer terminals per student at a local university,
data were collected from a sample of 81 business students over a one-week period. Assume the
population standard deviation is 1.2 hours. For a 95% confidence interval, the margin of error for
estimating the population mean is
(a) 0.26
(b) 1.96
(c) 0.03
(d) 0.21
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
For problems 26-27, use the following information:
A researcher wants to estimate the difference between the average response times (in minutes) of
emergency 911 telephone calls for two cities. Based on a simple random sample of 75 calls from each
city last month, the sample averages were 13 minutes for City 1 and 9 minutes for City 2. The sample
standard deviations were 2.1 minutes for City 1 and 1.7 minutes for City 2.
26. A 90% confidence interval for the difference between the two population mean response times is
(a) (3.38, 4.62)
(b) (3.55, 4.45)
(c) (3.49, 4.51)
(d) (3.62, 4.60)
27. We want to test using =0.1 the hypothesis that the two population mean response times are different.
Based on the confidence interval above we
(a) Reject Ho.
(b) Do not reject Ho.
(c) Reject Ha.
(d) We can not tell.

For problems 28-29, use the following information:
The shelf life of a carbonated beverage is of interest. Ten bottles are randomly selected and tested.
The following results are obtained in days:
108
124
124
106
115
138
163
159
We want to test if the true mean shelf life is greater than 125 days.
134
139
28. The alternative hypothesis is
(a) H a : X  125
(b) H o :   125
(c) H a : p  125
(d) H a :   125
29. Which of the following is the test statistic?
(a) z=0.31
(b) t=0.97
(c) z=0.97
(d) t=3.07
30. Which of the following statements about confidence intervals is INCORRECT?
(a) If we keep the sample size fixed, the confidence interval gets wider as we increase the confidence
coefficient.
(b) A confidence interval for a mean always contains the sample mean.
(c) If we keep the confidence coefficient fixed, the confidence interval gets narrower as we increase
the sample size.
(d) If the population standard deviation increases, the confidence interval decreases in width.
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31. Which of the calculated values of a test statistic would have the smallest P-value?
(a) z = 3.05
(b) t = 3.05 with 10 degrees of freedom.
(c) t = 3.05 with 15 degrees of freedom.
(d) t = 3.05 with 30 degrees of freedom.
32. When taking a sample from a population, we use the ___________ to estimate the _____________.
(a) statistic, sample
(b) statistic, parameter
(c) parameter, statistic
(d) parameter, sample
33. The weight of potato chips in packaged bags is normally distributed with mean µ=15 ounces and
variance 2= 4. A simple random sample of 16 bags was selected, the probability that the mean of the
selected sample weighs 14 ounces or less
(a) 0.3085
(b) 0.4013
(c) 0.1587
(d) 0.0228
34. Which of the following does not affect the P-value when testing statistical hypothesis?
(a)
(b)
(c)
(d)
Sample size.
Null hypothesis.
Significance level.
Alternative hypothesis.
35. In testing hypotheses, which of the following would be strong evidence against the null hypothesis
(a) a small significance level.
(b) a small p-value
(c) a large p-value
(d) the selection of α of .025
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SECTION TWO: FREE-RESPONSE PROBLEMS
Question (1)
Bastien, Inc. has been manufacturing small automobiles that have averaged 50 miles per gallon of
gasoline in highway driving. The company has developed a more efficient engine for its small cars and
now advertises that its new small cars average more than 50 miles per gallon in highway driving. An
independent testing service road-tested 25 of the automobiles. The sample showed an average of 51.5
miles per gallon with a standard deviation of 6 miles per gallon. At 0.05 level of significance, test to
determine whether or not the manufacturer's advertising campaign is legitimate.


Hypotheses
H0: =50 H0:  50
Test statistic:
n  25, x  51.5, s  6
x   0 51.5  50
t

 1.25, df  25  1  24
s n
6 / 25
 P-value=P(T24>1.25)=tcdf(1.25, 1E99,24)=0.1117
Conclusion: We do not reject H0 and conclude that the advertising campaign is not legitimate
Question (2)
The Excellent Drug Company claims its aspirin tablets will relieve headaches faster than any other
aspirin on the market. To determine whether Excellent's claim is valid, random samples of size 15 are
chosen from aspirins made by Excellent and the Simple Drug Company. An aspirin is given to each of
the 30 randomly selected persons suffering from headaches and the number of minutes required for each
to recover from the headache is recorded. The sample results are:
Excellent
Simple
Sample mean
8.4
8.9
Sample standard deviation
2.05
2.14
A 5% significance level test is performed to determine whether Excellent's aspirin cures headaches
significantly faster than Simple's aspirin.
Let mean time required for Excellent’s aspirin and mean time required for Simple’s aspirin
 Hypotheses
H0:  H0: 
 Test statistic:
t


x1  x 2
s12 s 22

n1 n 2

8.4  8.9
2.042 2.152

15
15
 0.65
df=min(15-1, 15-1) =14
P-value=P(T14<-0.65)=tcdf(-1E99,-0.65,14)=0.26
Conclusion: We do not reject H0 and conclude that Excellent’s aspirin does not cure headaches
faster than Simple’s aspirin.
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Question (3)
The Montana Highway Patrol is interested in determining whether Montana residents or nonresidents
drive faster on a particular stretch of Interstate 90. Independent random samples of the speeds of cars
having Montana license plates and cars licensed in other states results in the summary data listed below.
Group
Montana
Others
Sample size
14
17
Sample Mean
73.2
76.6
Sample standard deviation
3.8
4.7
Assume the population variances are the same. At 0.05 level of significance, is there sufficient evidence
to conclude that nonresidents drive faster on this stretch of Interstate 90 than residents of Montana?
 Ho:
Ha:

Test Statistic:

P-value or critical region

Conclusion:
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Question (4)
The Montana Highway Patrol is interested in determining whether Montana residents or nonresidents
drive faster on a particular stretch of Interstate 90. Independent random samples of the speeds of cars
having Montana license plates and cars licensed in other states results in the summary data listed below.
Nonresidents
Montana
n1  35
x 1  76.6
s 1  4 .7
n2  40
x 2  73.2
s 2  3. 8
Assume the population variances are the same. At 0.05 level of significance, is there sufficient evidence
to conclude that nonresidents drive faster on this stretch of Interstate 90 than residents of Montana?

Ho:
Ha:

Test Statistic:

P-value or critical region

Conclusion:
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Question (5)
A poll of 500 randomly selected U.S. adults in 1997 found 185 did not own a credit card. Test if more
than 60% of U.S. adults have credit cards. Use =0.05.

Ho:
Ha:

Test Statistic:

P-value:

Conclusion:
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Question (6)
Thousands of high quality lambs are imported from Australia and New Zealand. It is claimed by New
Zealand that their lambs on average are heavier. The weights of lambs from each of these countries can
be assumed to normally distributed with unknown means and with unknown true standard deviations
which can be assumed to be equal. The USDA sampled 13 lambs from the New Zealand shipment and 17
from the Australian shipment and obtained the weights for each lamb in the sample. The Table below
summarizes their findings.
New Zealand
13
11.1
0.7
Sample size
Sample mean
Sample standard deviation
Australia
17
10.5
0.8
a) State H0 and Ha.

Ho:
Ha:
b) Calculate the test statistic.
c) Perform the test and state bounds on the “P-value” or give the rejection region.
d) State your conclusion about what this data indicates about New Zealand’s claim. Use =0.05.
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Question (7)
A paired difference experiment is conducted to compare the starting salaries of male and female college
graduates who find jobs. Pairs are formed by choosing a male and a female with same major and similar
grade-point averages. Suppose a random sample of 5 pairs and the starting salaries (in thousands) are as
follows:
Pair
Male
Female
1
25.9
24.9
2
20.0
18.5
3
28.7
27.7
4
13.5
13.0
5
18.8
17.8
At 0.01 level of significance, test whether the mean starting salary for males is higher than that of
females.
(a) State H0 and Ha.

Ho:
Ha:
(b) Calculate the test statistic.
(c) Find the “P-value” or give the rejection region.
(d) State your conclusion.
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Question (8)
In 1990, 5.8% of job applicants who were tested for drugs failed the test. At the 0.01 level, test the claim
that the failure rate is now lower if a random sample of 1520 current job applicants results in 58 failures.
(a) State H0 and Ha.
H0: p=0.058
Ha: p<0.058
(b) Calculate the test statistic.
58
 0.0382,
1520
p  p0
0.0382  0.058
ˆ
z 

 3.3
SE ˆp
0.006
n  1520, x  58,
p
ˆ
p0 (1  p0 )

n
SE ˆp 
(0.058)(0.942)
 0.006
1520
(c) Find the “P-value” or give the rejection region.
P-value=P(Z<-3.3)=normalcdf(-1E99,-3.3)=0.00048
(d) State your conclusion.
Conclusion: We reject H0 and conclude that the failure rate is now lower than 5.8%.
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