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Advanced Placement Statistics Practice Test
This is a practice test for the multiplechoice
section of the Advanced Placement Statistics Exam.
Like the actual AP Statistics test,
this test includes 40 questions covering a variety
of topics.
Test Instructions
Each multiplechoice question has five response options.
Choose the best answer. Then, click the "Next" button or the
"Back" button to see another question.
The multiplechoice portion of the Advanced Placement (AP)
Statistics Exam is a timed test, lasting 90 minutes.
Use the Timer (top left) to simulate an actual test, or pause
the Timer to work at your own pace.
Use the Test Performance monitor (bottom left) to check your test
results. Test questions are colorcoded.
 Blue. A question that has not been answered.
 Green. A question that has been answered correctly.
 Red. A question that has been answered incorrectly.
Each test question in the Test Performance monitor is a link. To
jump to a particular question, click its link.
Choose the best answer to the question. To see another question,
click the "Next" button or the "Back" button.
To see the solution, click the "Show solution" button.
Problem 2
A coin is tossed three times. What is the probability that
it lands on heads exactly one time?
(A) 0.125
(B) 0.250
(C) 0.333
(D) 0.375
(E) 0.500
Solution
The correct answer is (D). If you toss a coin three times, there
are a total of eight possible outcomes. They are: HHH, HHT,
HTH, THH, HTT, THT, TTH, and TTT. Of the eight possible outcomes,
three have exactly one head. They are: HTT, THT, and TTH.
Therefore, the probability that three flips of a coin will
produce exactly one head is 3/8 or 0.375.
Problem 3
An auto analyst is conducting a satisfaction survey, sampling
from a list of 10,000 new car buyers. The list includes 2,500 Ford
buyers, 2,500 GM buyers, 2,500 Honda buyers, and 2,500 Toyota buyers.
The analyst selects a sample of 400 car buyers, by randomly
sampling 100 buyers of each brand.
Is this an example of a simple random sample?
(A) Yes, because each buyer in the sample was randomly sampled.
(B) Yes, because each buyer in the sample had an equal chance of being sampled.
(C) Yes, because car buyers of every brand were equally represented in the sample.
(D) No, because every possible 400buyer sample did not have an equal chance of being chosen.
(E) No, because the population consisted of purchasers of four different brands of car.
Solution
The correct answer is (D). A
simple random sample requires that
every
sample
of size n (in this problem, n
is equal to 400) have an equal chance of being selected. In this
problem, there was a 100 percent chance that the sample would
include 100 purchasers of each brand of car. There was
zero percent chance that the sample would include, for example,
99 Ford buyers, 101 Honda buyers, 100 Toyota buyers, and 100
GM buyers. Thus, all possible samples of size 400 did not have
an equal chance of being selected; so this cannot be a simple
random sample.
The fact that each buyer in the sample was randomly sampled is
a necessary condition for a simple random sample, but it is not
sufficient.
Similarly, the fact that each buyer in the sample had an equal
chance of being selected is characteristic of a simple
random sample, but it is not sufficient. The sampling method in
this problem used random sampling and gave each buyer an equal
chance of being selected;
but the sampling method was actually
stratified random sampling.
The fact that car buyers of every brand were equally
represented in the sample is irrelevant to whether the sampling
method was simple random sampling. Similarly, the fact that
population consisted of buyers of different car brands is
irrelevant.
Problem 8
Nine hundred (900) high school freshmen were randomly selected for
a national survey. Among survey participants, the mean gradepoint
average (GPA) was 2.7, and the standard deviation was 0.4. What
is the margin of error, assuming a 95% confidence level?
(A) 0.013
(B) 0.025
(C) 0.500
(D) 1.960
(E) None of the above.
Solution
The correct answer is (B). To compute the margin of error, we
need to find the
critical value and the standard error of the mean.
To find the critical value, we take the following steps.
 Compute alpha (α): α = 1  (confidence level / 100)
= 1  0.95 = 0.05
 Find the critical probability (p*): p* = 1  α/2
= 1  0.05/2 = 0.975
 Find the critical z score.
Since the sample size is large, the sampling distribution will
be roughly normal in shape. Therefore, we can express the
critical value as a z score. For this problem, it will
be the z score having
a cumulative probability equal to 0.975.
Then, using an online calculator (e.g., Stat Trek's free
normal distribution calculator),
a handheld
graphing calculator, or
the standard normal distribution table, we find the cumulative
probability associated with the zscore.
Using the
Normal Distribution Calculator,
we find that the critical value is 1.96.
(On the actual AP Statistics exam, you may need to use a
graphing calculator or a normal table, since Stat Trek's
analytical tools will not be available.)
Next, we find the standard error of the mean, using the following
equation:
SE_{x} = s / sqrt( n )
= 0.4 / sqrt( 900 ) = 0.4 / 30 = 0.013
And finally, we compute the margin of error (ME).
ME = Critical value x Standard error
= 1.96 * 0.013 = 0.025
Problem 12
Suppose we want to estimate the average weight of an adult male in
Dekalb County, Georgia. We draw a random sample of 1,000 men from a
population of 1,000,000 men and weigh them. We find that the average
man in our sample weighs 180 pounds, and the standard deviation of
the sample is 30 pounds. What is the 95% confidence interval.
(A) 180 + 1.86
(B) 180 + 3.0
(C) 180 + 5.88
(D) 180 + 30
(E) None of the above.
Solution
The answer is (A). To specify the confidence interval, we work
through the four steps below.
 Identify a sample statistic. Since we are trying to estimate
the mean weight in the population, we choose the mean weight
in our sample (180) as the sample statistic.
 Select a confidence level. In this case, the confidence level
is defined for us in the problem. We are working with a 95%
confidence level.
 Find the margin of error. Previously, we described
how to compute the margin of error.
The key steps are shown below.
 Find standard error. The standard error (SE) of the
mean is:
SE = s / sqrt( n ) = 30 / sqrt(1000) = 30/31.62 = 0.95
 Find critical value. The critical value is a factor used to
compute the margin of error. To express the critical value
as a
t score
(t*), follow these steps.
 Compute alpha (α): α = 1  (confidence level / 100) = 0.05
 Find the critical probability (p*): p* = 1  α/2 = 1  0.05/2 = 0.975
 Find the
degrees of freedom (df): df = n  1 = 1000  1 = 999
 The critical value is
the t score having 999 degrees of freedom and a
cumulative probability
equal to 0.975.
Using an online calculator (e.g., Stat Trek's free
t Distribution Calculator),
a handheld
graphing calculator, or
a t distribution table, we find that the t score associated
with a cumulative probability of 0.975 is 1.96.
(On the actual AP Statistics exam, you may need to use a
graphing calculator or a t table, since Stat Trek's
analytical tools will not be available.)
Note: We might also have expressed the critical value as a
z score.
Because the sample size is large, a z score analysis produces
the same result  a critical value equal to 1.96.
 Compute margin of error (ME): ME = critical value * standard error
= 1.96 * 0.95 = 1.86
 Specify the confidence interval. The range of the confidence
interval is defined by the sample statistic +
margin of error.
And the uncertainty is denoted by the confidence level.
Therefore, we can be 95% confident that the population means falls
within the interval 180 + 1.86.
Problem 15
A major metropolitan newspaper selected a simple random sample of
1,600 readers from their list of
100,000 subscribers. They asked whether the paper should increase its
coverage of local news. Forty percent of the sample wanted more local
news. What is the 99% confidence interval for the proportion of
readers who would like more coverage of local news?
(A) 0.30 to 0.50
(B) 0.32 to 0.48
(C) 0.35 to 0.45
(D) 0.37 to 0.43
(E) 0.39 to 0.41
Solution
The answer is (D). The approach that we used to solve this
problem is valid when the following conditions are met.
 The sampling method must be
simple random sampling. This condition is satisfied;
the problem statement says that we used simple random sampling.
 The sample should include at least 10 successes and 10 failures.
Suppose we classify a "more local news" response as a success,
and any other response as a failure. Then, we have 0.40 * 1600 = 640
successes, and 0.60 * 1600 = 960 failures  plenty of successes
and failures.
 If the population size is much larger than the sample
size, we can use an "approximate" formula for the standard
deviation or the standard error. This condition is satisfied,
so we will use a simple "approximate" formula for the standard
error.
Since the above requirements are satisfied, we can use the following
fourstep approach to construct a confidence interval.
 Identify a sample statistic. Since we are trying to estimate
a population proportion, we choose the sample proportion
(0.40) as the sample statistic.
 Select a confidence level. The confidence level is defined
for us in the problem statement. We are working with a 99%
confidence level.
 Find the margin of error. Elsewhere on this site, we show
how to compute the margin of error when the sampling
distribution is approximately normal. The key steps are
shown below.
 Specify the confidence interval. The range of the confidence
interval is defined by the sample statistic +
margin of error. And the uncertainty is denoted
by the confidence level.
Therefore, the 99% confidence interval is 0.37 to 0.43. That is, we are 99%
confident that the true population proportion is in the range
defined by 0.4 + 0.03.
Problem 16
Suppose a simple random sample of 150 students is drawn
from a population of 3000
college students. Among sampled students, the average IQ score is
115 with a standard deviation of 10. What is the 99%
confidence interval for the students' IQ score?
(A) 115 + 0.01
(B) 115 + 0.82
(C) 115 + 2.1
(D) 115 + 2.6
(E) None of the above.
Solution
The correct answer is (C). The approach that we used to solve this
problem is valid when the following conditions are met.
Since the above requirements are satisfied, we can use the following
fourstep approach to construct a confidence interval.
 Identify a sample statistic. Since we are trying to estimate
a population mean, we choose the sample mean
(115) as the sample statistic.
 Select a confidence level. In this analysis, the confidence level
is defined for us in the problem. We are working with a 99%
confidence level.
 Find the margin of error. Elsewhere on this site, we show
how to compute the margin of error when the sampling
distribution is approximately normal. The key steps are
shown below.
 Find standard deviation or standard error. Since we do not
know the standard deviation of the population, we cannot compute the
standard deviation of the sample mean; instead, we compute the standard
error (SE). Because the sample size is much smaller than the
population size, we can use the "approximate" formula for the
standard error.
SE =
s / sqrt( n ) = 10 / sqrt(150) = 10 / 12.25 = 0.82
 Find critical value. The critical value is a factor used to
compute the margin of error. Because the standard deviation
of the population is unknown, we express the critical
value as a
t score
rather than a
z score.
To find the critical value, we take these steps.
 Compute alpha (α): α = 1  (confidence level / 100) = 1  99/100 = 0.01
 Find the critical probability (p*): p* = 1  α/2 = 1  0.01/2 = 0.995
 Find the
degrees of freedom (df): df = n  1 = 150  1 = 149
 The critical value is
the t score having 149 degrees of freedom and a
cumulative probability
equal to 0.995.
Using an online calculator (e.g., Stat Trek's free
t Distribution Calculator),
a handheld
graphing calculator, or
a t distribution table, we find that the t score associated
with a cumulative probability of 0.995 is 2.61.
(On the actual AP Statistics exam, you may need to use a
graphing calculator or a t table, since Stat Trek's
analytical tools will not be available.)
Note: We might also have expressed the critical value as a
z score.
Because the sample size is fairly large, a z score analysis produces
a similar result  a critical value equal to 2.58.
 Compute margin of error (ME): ME = critical value * standard error
= 2.61 * 0.82 = 2.1
 Specify the confidence interval. The range of the confidence
interval is defined by the sample statistic +
margin of error. And the uncertainty is denoted
by the confidence level.
Therefore, the 99% confidence interval is 112.9 to 117.1. That is, we are 99%
confident that the true population mean is in the range
defined by 115 + 2.1.
Problem 31
An archer claims that 25% of her shots will be in the center
of the target (i.e., a bullseye). A sports writer plans to test
this claim by sampling 300 shots. If the 300 shots result in
60 or fewer bullseyes (i.e., 20% bullseyes),
the writer will reject the archer's claim.
What is the probability that the sports writer will reject
the archer's claim, when it is actually true?
(A) 0.01
(B) 0.02
(C) 0.04
(D) 0.08
(E) 0.16
Solution
The answer is (B). The solution to this problem involves finding
the Pvalue,
the probability of making 60 or fewer bullseyes, assuming
that 25% of the archer's shots are normally bullseyes.
In other words, the
null hypothesis is P = 0.25.
To find the Pvalue, take the following steps.
 Calculate the standard deviation (σ), assuming that the
null hypothesis is true.
σ = sqrt[ P * ( 1  P ) / n ]
= sqrt [(0.25 * 0.75) / 300] = sqrt(0.000625) = 0.025
where P is the hypothesized value of population proportion in
the null hypothesis, and n is the sample size.
 Compute the zscore test statistic (z).
z = (p  P) / σ = (0.20  0.25)/0.025 = 2
where P is the hypothesized value of population proportion in
the null hypothesis, and p is the proportion of bullseyes observed in
the sample.
 The Pvalue is the probability that a zscore will be
less than 2.0.
Using an online calculator (e.g., Stat Trek's free
Normal Distribution Calculator),
a handheld
graphing calculator, or
a normal distribution table, we find that the z score of 2.0
has a cumulative probability of 0.023.
(On the actual AP Statistics exam, you may need to use a
graphing calculator or a normal distribution table, since Stat
Trek's analytical tools will not be available.)
Note: This problem can also be treated as a
binomial experiment. Previously, we showed
how to analyze a binomial experiment. The binomial experiment
is actually the more exact analysis. It produces a Pvalue
with a cumulative probability of 0.0245. Without a computer,
the binomial approach is computationally demanding. Therefore,
many statistics texts emphasize the approach presented above,
which is a normal approximation of the binomial.
Problem 32
The Acme Car Company claims that at most 8% of its new cars have
a manufacturing defect. A quality control inspector randomly
selects 300 new cars and finds that 33 have a defect.
Should she reject the 8% claim? Assume that the significance
level is 0.05.
(A) Yes, because the Pvalue is 0.016.
(B) Yes, because the Pvalue is 0.028.
(C) No, because the Pvalue is 0.16.
(D) No, because the Pvalue is 0.28.
(E) There is not enough information to reach a conclusion.
Solution
The answer is (B). The solution to this problem takes four steps:
(1) state the hypotheses, (2) formulate an analysis plan,
(3) analyze sample data, and (4) interpret results.
We work through those steps below:
State the hypotheses. The first step is to
state the null hypothesis and an alternative hypothesis.
Null hypothesis: P <= 0.08
Alternative hypothesis: P > 0.08
Note that these hypotheses constitute a onetailed test.
The null hypothesis will be rejected only if the sample
proportion is too big.
Formulate an analysis plan. For this analysis,
the significance level is 0.05. The test method,
shown in the next section, is a
onesample ztest.
Analyze sample data. Using sample data, we
calculate the standard deviation (σ) and compute the zscore
test statistic (z).
σ = sqrt[ P * ( 1  P ) / n ]
= sqrt [(0.08 * 0.92) / 300] = sqrt(0.0002453) = 0.0157
z = (p  P) / σ = (.11  .08)/0.0157 = 1.91
where P is the hypothesized value of population proportion in
the null hypothesis, p is the sample proportion,
and n is the sample size.
Since we have a
onetailed test, the Pvalue is the probability that the
zscore is greater than 1.91.
We use the
Normal Distribution Calculator
to find P(z > 1.91) = 0.028. Thus, the Pvalue = 0.028.
Interpret results. Since the Pvalue (0.028) is
less than the significance level (0.05), we cannot accept the
null hypothesis.
Note: If you use this approach on an exam, you may also want to mention
why this approach is appropriate. Specifically, the approach is
appropriate because the sampling method was simple random sampling,
the sample included at least 10 successes and 10 failures, and
the population size was at least 10 times the sample size.
Problem 36
A sports writer hypothesized that Tiger Woods plays better on
par 3 holes than on par 4 holes. He reviewed Woods' performance
in a random sample
of golf tournaments. On the par 3 holes, Woods made a birdie in
20 out of 80 attempts. On the par 4 holes, he made a birdie in
40 out of 200 attempts. How would you interpret this result?
(A) The Pvalue is < 0.001,
very strong evidence that Woods plays better on par 3 holes.
(B) The Pvalue is between 0.001 and 0.01,
strong evidence that Woods plays better on par 3 holes.
(C) The Pvalue is between 0.01 and 0.05,
moderate evidence that Woods plays better on par 3 holes.
(D) The Pvalue is between 0.05 and 0.10,
some evidence that Woods plays better on par 3 holes.
(E) The Pvalue is > 0.10,
little or no support for the notion that Woods plays better on par 3 holes.
Solution
The answer is (E). The solution to this problem takes four steps:
(1) state the hypotheses, (2) formulate an analysis plan,
(3) analyze sample data, and (4) interpret results.
We work through those steps below:
State the hypotheses. The first step is to
state the null hypothesis and an alternative hypothesis.
Null hypothesis: P_{3} <= P_{4}
Alternative hypothesis: P_{3} > P_{4}
Note that these hypotheses constitute a onetailed test.
The null hypothesis will be rejected if the proportion
of birdies on par 3 holes (p_{3}) is sufficiently
greater than the proportion of birdies on par 4 holes
(p_{4}).
Formulate an analysis plan. For this analysis,
the test method is a twoproportion ztest, which is
shown below.
Analyze sample data. Using sample data, we
calculate the pooled sample proportion (p) and the standard error
(SE). Using those measures, we compute the zscore
test statistic (z).
p = (p_{3} * n_{3} + p_{4} * n_{4})
/ (n_{3} + n_{4})
= [(0.25 * 80) + (0.20 * 200)] / (80 + 200) = 50/280 = 0.214
SE =
sqrt{ p * ( 1  p ) * [ (1/n_{3}) + (1/n_{4}) ] }
SE = sqrt [ 0.214 * 0.786 * ( 1/80 + 1/200 ) ]
= sqrt[ 0.214 * 0.786 * 0.0175 }= sqrt [0.0029548] = 0.0544
z = (p_{3}  p_{4}) / SE
= (0.25  0.20)/0.0544
= 0.92
where p_{3} is the sample proportion of birdies on par 3,
where p_{4} is the sample proportion of birdies on par 4,
n_{3} is the number of par 3 holes,
and n_{4} is the number of par 4 holes.
Since we have a
onetailed test, the Pvalue is the probability that the
zscore is greater than 0.92.
We use the
Normal Distribution Calculator
to find P(z > 0.92) = 0.18. Thus, the Pvalue = 0.18.
Interpret results. Since the Pvalue (0.18) is
greater than 0.10, we have little support for the notion that
Woods plays better on par 3 holes. In short,
we cannot reject the null hypothesis.
Note: If you use this approach on an exam, you may also want to mention
why this approach is appropriate.
Specifically, the approach is appropriate because
the sampling method was simple random sampling, the
samples were independent, each population was at least 10 times
larger than its sample, and
each sample included at least 10 successes and 10 failures.
Problem 38
Acme Corporation manufactures light bulbs. The CEO claims that an average Acme
light bulb lasts 300 days. A researcher randomly selects 15 bulbs for testing.
The sampled bulbs last an average of 290 days, with a standard deviation of 50 days. If
the CEO's claim were true, what is the probability that 15 randomly selected
bulbs would have an average life of no more than 290 days?
(A) 0.100
(B) 0.226
(C) 0.334
(D) 0.443
(E) .775
Solution
The answer is (B). The first thing we need to do is compute the t score,
based on the following equation:
t = [ x  μ ]
/ [ s / sqrt( n ) ]
t = ( 290  300 ) / [ 50 / sqrt( 15) ] = 10 / 12.909945 =  0.7745966
where x is the sample mean, μ
is the population mean, s is the standard deviation of the sample, and n is the
sample size.
Then, using an online calculator (e.g., Stat Trek's free
T Distribution Calculator),
a handheld
graphing calculator, or
the t distribution table, we find the cumulative
probability associated with the t score.
For this practice test, we can use the
T Distribution Calculator; but on
the actual AP Statistics Exam, you may need to use a graphing
calculator or a t distribution table.
Since we know the t score, we select "T score" from the Random Variable
dropdown box of the T Distribution Calculator. Then, we enter
the following data:

The degrees of freedom are equal to 15  1 = 14.

The t score is equal to  0.7745966.
The calculator displays the cumulative probability: 0.226. Hence, if the true
bulb life were 300 days, there is a
22.6% chance that the average bulb life for 15 randomly selected bulbs would
be less than or equal to 290 days.
Problem 40
A public opinion poll surveyed a simple random sample of voters.
Respondents were classified by gender (male or female) and by
voting preference (Republican, Democrat, or Independent).
Results are shown below.

Voting Preferences 

Republican 
Democrat 
Independent 
Row total 
Male 
200 
150 
50 
400 
Female 
250 
300 
50 
600 
Column total 
450 
450 
100 
1000 
If you conduct a chisquare test of independence, what is the
expected frequency count of male Independents?
(A) 40
(B) 50
(C) 60
(D) 180
(E) 270
Solution
The answer is (A). To apply the chisquare
test for independence, we compute the expected frequency counts
for each cell of the table, using the following equation.
The computation for males who are classified as Independents
is shown below.
E_{r,c} = (n_{r} * n_{c}) / n
E_{1,3} = (400 * 100) / 1000 = 40000/1000 = 40
where
r is the number of levels of gender,
c is the number of levels of the voting preference,
n_{r} is the number of observations from level r
of gender,
n_{c} is the number of observations from level c
of voting preference,
n is the number of observations in the sample,
E_{r,c} is the expected frequency count when gender is level
r and voting preference is level c, and
O_{r,c} is the observed frequency count when gender is level
r voting preference is level c.