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Name: ______________________ Class: _________________ Date: _________
ID: A
AP Statistics - Multiple Choice Sample (36 Minutes)
Directions: Solve each of the following problems, using the available space for scratchwork. Decide which is the
best of the choices given and fill in the corresponding oval on the answer sheet. No credit will be given for
anything written in the test book. Do not spend too much time on any one problem.
____
1. In the scatterplot of y versus x shown above, the least squares regression line is superimposed on the
plot. Which of the following points has the largest residual?
a.
b.
c.
d.
e.
A
B
C
D
E
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Name: ______________________
____
2. Under which of the following conditions is it preferable to use stratified random sampling rather than
simple random sampling?
a.
b.
c.
d.
e.
____
The population can be divided into a large number of strata so that each stratum
contains only a few individuals.
The population can be divided into a small number of strata so that each stratum
contains a large number of individuals.
The population can be divided into strata so that the individuals in each stratum are
as much alike as possible.
The population can be divided into strata so that the individuals in each stratum are
as different as possible.
The population can be divided into strata of equal sizes so that each individual in the
population still has the same chance of being selected.
3. All bags entering a research facility are screened. Ninety-seven percent of the bags that contain
forbidden material trigger an alarm. Fifteen percent of the bags that do not contain forbidden material
also trigger the alarm. If 1 out of every 1,000 bags entering the building contains forbidden material,
what is the probability that a bag that triggers the alarm will actually contain forbidden material?
a.
b.
c.
d.
e.
____
ID: A
0.00097
0.00640
0.03000
0.14550
0.97000
4. A candy company claims that 10 percent of its candies are blue. A random sample of 200 of these
candies is taken, and 16 are found to be blue. Which of the following tests would be most appropriate
for establishing whether the candy company needs to change its claim?
a.
b.
c.
d.
e.
Matched pairs t-test
One-sample proportion z-test
Two-sample t-test
Two-sample proportion z-test
Chi-square test of association
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Name: ______________________
____
5. Some descriptive statistics for a set of test scores are shown above. For this test, a certain student has
a standardized score of z = –1.2. What score did this student receive on the test?
a.
b.
c.
d.
e.
____
266.28
779.42
1008.02
1083.38
1311.98
6. In a test of H 0 : µ = 8 versus H a : µ ≠ 8, a sample of size 220 leads to a p-value of
0.034. Which of the following must be true?
a.
b.
c.
d.
e.
____
ID: A
A 95% confidence interval for µ calculated from these data will not include µ = 8.
At the 5% level if H 0 is rejected, the probability of a Type II error is 0.034.
The 95% confidence interval for µ calculated from these data will be centered at
µ = 8.
The null hypothesis should not be rejected at the 5% level.
The sample size is insufficient to draw a conclusion with 95% confidence.
7. A summer resort rents rowboats to customers but does not allow more than four people to a boat.
Each boat is designed to hold no more than 800 pounds. Suppose the distribution of adult males who
rent boats, including their clothes and gear, is normal with a mean of 190 pounds and standard
deviation of 10 pounds. If the weights of individual passengers are independent, what is the
probability that a group of four adult male passengers will exceed the acceptable weight limit of 800
pounds?
a.
b.
c.
d.
e.
0.023
0.046
0.159
0.317
0.977
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Name: ______________________
____
ID: A
8. Consider a data set of positive values, at least two of which are not equal. Which of the following
sample statistics will be changed when each value in this data set is multiplied by a constant whose
absolute value is greater than 1?
I. The mean
II. The median
III. The standard deviation
a.
b.
c.
d.
e.
____
9. Each person in a simple random sample of 2,000 received a survey, and 317 people returned their
survey. How could nonresponse cause the results of the survey to be biased?
a.
b.
c.
d.
e.
____
I only
II only
III only
I and II only
I, II, and III
Those who did not respond reduced the sample size, and small samples have more
bias than large samples.
Those who did not respond caused a violation of the assumption of independence.
Those who did not respond were indistinguishable from those who did not receive
the survey.
Those who did not respond represent a stratum, changing the simple random sample
into a stratified random sample.
Those who did respond may differ in some important way from those who did not
respond.
10. In a certain game, a fair die is rolled and a player gains 20 points if the die shows a “6.” If the die
does not show a “6,” the player loses 3 points. If the die were to be rolled 100 times, what would be
the expected total gain or loss for the player?
a.
b.
c.
d.
e.
A gain of about 1,700 points
A gain of about 583 points
A gain of about 83 points
A loss of about 250 points
A loss of about 300 points
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Name: ______________________
____
11. The Attila Barbell Company makes bars for weight lifting. The weights of the bars are independent
and are normally distributed with a mean of 720 ounces (45 pounds) and a standard deviation of 4
ounces. The bars are shipped 10 in a box to the retailers. The weights of the empty boxes are
normally distributed with a mean of 320 ounces and a standard deviation of 8 ounces. The weights of
the boxes filled with 10 bars are expected to be normally distributed with a mean of 7,520 ounces and
a standard deviation of
a.
b.
c.
d.
e.
____
ID: A
12 ounces
80 ounces
224 ounces
48 ounces
1664 ounces
12. Exercise physiologists are investigating the relationship between lean body mass (in kilograms) and
the resting metabolic rate (in calories per day) in sedentary males.
Based on the computer output above, which of the following is the best interpretation of the value of
the slope of the regression line?
a.
b.
c.
d.
e.
For each additional kilogram of lean body mass, the resting metabolic rate increases
on average by 22.563 calories per day.
For each additional kilogram of lean body mass, the resting metabolic rate increases
on average by 264.0 calories per day.
For each additional kilogram of lean body mass, the resting metabolic rate increases
on average by 144.9 calories per day.
For each additional calorie per day for the resting metabolic rate, the lean body mass
increases on average by 22.563 kilograms.
For each additional calorie per day for the resting metabolic rate, the lean body mass
increases on average by 264.0 kilograms.
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Name: ______________________
____
ID: A
13. An investigator was studying a territorial species of Central American termites, Nasutitermes corniger.
Forty-nine termite pairs were randomly selected; both members of each of these pairs were from the
same colony. Fifty-five additional termite pairs were randomly selected; the two members in each of
these pairs were from different colonies. The pairs were placed in petri dishes and observed to see
whether they exhibited aggressive behavior. The results are shown in the table below.
2
A Chi-square test for homogeneity was conducted, resulting in χ = 7.638. The expected counts are
shown in parentheses in the table. Which of the following sets of statements follows from these
results?
a.
b.
2
χ is not significant at the 0.05 level.
2
χ is significant, 0.01 < p < 0.05; the counts in the table suggest that termite pairs
from the same colony are less likely to be aggressive than termite pairs from
different colonies.
2
c. χ is significant, 0.01 < p < 0.05; the counts in the table suggest that termite pairs
from different colonies are less likely to be aggressive than termite pairs from the
same colony.
2
d. χ is significant, p < 0.01; the counts in the table suggest that termite pairs from the
same colony are less likely to be aggressive than termite pairs from different
colonies.
2
e. χ is significant, p < 0.01; the counts in the table suggest that termite pairs from
different colonies are less likely to be aggressive than termite pairs from the same
colony.
6
Name: ______________________
____
ID: A
14. Consider n pairs of numbers ÊÁË x 1 , y 1 ˆ˜ , ÊÁË x 2 , y 2 ˆ˜ , . . . , and ÊÁË x n , y n ˆ˜ . The mean and standard deviation
of the x-values are x = 5 and s x = 4, respectively. The mean and standard deviation of the y-values
are y = 10 and s y = 10, respectively. Of the following, which could be the least squares regression
line?
a.
b.
c.
d.
e.
____
8y = 8.5 + 0.3x
8y = 10.0 + 0.4x
15. The mayor of a large city will run for governor if he believes that more than 30 percent of the voters
in the state already support him. He will have a survey firm ask a random sample of n voters whether
or not they support him. He will use a large sample test for proportions to test the null hypothesis that
the proportion of all voters who support him is 30 percent or less against the alternative that the
percentage is higher than 30 percent. Suppose that 35 percent of all voters in the state actually support
him. In which of the following situations would the power for this test be highest?
a.
b.
c.
d.
e.
____
y8 = −5.0 + 3.0x
8y = 3.0x
8y = 5.0 + 2.5x
The mayor uses a significance level of 0.01 and n = 250 voters.
The mayor uses a significance level of 0.01 and n = 500 voters.
The mayor uses a significance level of 0.01 and n = 1,000 voters.
The mayor uses a significance level of 0.05 and n = 500 voters.
The mayor uses a significance level of 0.05 and n = 1,000 voters.
16. George and Michelle each claimed to have the better recipe for chocolate chip cookies. They decided
to conduct a study to determine whose cookies were really better. They each baked a batch of cookies
using their own recipe. George asked a random sample of his friends to taste his cookies and to
complete a questionnaire on their quality. Michelle asked a random sample of her friends to complete
the same questionnaire for her cookies. They then compared the results. Which of the following
statements about this study is false?
a.
b.
c.
d.
e.
Because George and Michelle have a different population of friends, their sampling
procedure makes it difficult to compare the recipes.
Because George and Michelle each used only their own respective recipes, their
cooking ability is confounded with the recipe quality.
Because George and Michelle each used only the ovens in their houses, the recipe
quality is confounded with the characteristics of the oven.
Because George and Michelle used the same questionnaire, their results will
generalize to the combined population of their friends.
Because George and Michelle each baked one batch, there is no replication of the
cookie recipes.
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Name: ______________________
____
17. A large company is considering opening a franchise in St. Louis and wants to estimate the mean
household income for the area using a simple random sample of households. Based on information
from a pilot study, the company assumes that the standard deviation of household incomes is σ =
$7,200. Of the following, which is the least number of households that should be surveyed to obtain
an estimate that is within $200 of the true mean household income with 95 percent confidence?
a.
b.
c.
d.
e.
____
ID: A
75
1,300
5,200
5,500
7,700
18. Courtney has constructed a cricket out of paper and rubber bands. According to the instructions for
making the cricket, when it jumps it will land on its feet half of the time and on its back the other half
of the time. In the first 50 jumps, Courtney’s cricket landed on its feet 35 times. In the next 10 jumps,
it landed on its feet only twice. Based on this experience, Courtney can conclude that
a.
b.
c.
d.
e.
the cricket was due to land on its feet less than half the time during the final 10
jumps, since it had landed too often on its feet during the first 50 jumps
a confidence interval for estimating the cricket’s true probability of landing on its
feet is wider after the final 10 jumps than it was before the final 10 jumps
a confidence interval for estimating the cricket’s true probability of landing on its
feet after the final 10 jumps is exactly the same as it was before the final 10 jumps
a confidence interval for estimating the cricket’s true probability of landing on its
feet is more narrow after the final 10 jumps than it was before the final 10 jumps
a confidence interval for estimating the cricket’s true probability of landing on its
feet based on the initial 50 jumps does not include 0.2, so there must be a defect in
the cricket’s construction to account for the poor showing in the final 10 jumps
8
ID: A
AP Statistics - Multiple Choice Sample (36 Minutes)
Answer Section
MULTIPLE CHOICE
1. ANS: A
PTS: 1
2. ANS: C
PTS: 1
3. ANS: B
Draw a conditional probability tree:
F=bag contains forbidden material
T=bag triggers alarm
P(F) = 1 = 0.001
1000
P(T| F) = 0.97--->P(F ∩ T) = (.001)(.97) =.00097
P(∼ T| F) = 0.03--->P(F ∩ ∼ T) = (.001)(.03) =.00003
P(∼ F) = 999 = 0.999
1000
P(T| ∼ F) = 0.15------>P(∼ F ∩ T) = (.999)(.15) =.14985
P(∼ T| ∼ F) = 0.85--->P(∼ F ∩ ∼ T) = (.999)(.85) =.84915
P(T) = P(T ∩ F) + P(T ∩ ∼ F) =.00097 +.14985 =.15082
∴ P(F| T) =
P(F ∩ T)
= .00097 =.0064315
P(T)
.15082
Alternately you could imagine 100,000 bags. 100 would contain forbidden material and 99,900 would
not. In this scenerio 97 of the bags with forbidden material would trigger the alarm (.97 * 100) and
14,985 of the bags without forbidden material would also trigger the alarm (.15 * 99,900). This
means that 15,082 bags trigger the alarm (97+14985). Of those, only 97 contain forbidden material.
Therefore the probability that the triggered bag will contain forbidden material is 97/15,082 or
.0644315.
PTS: 1
4. ANS: B
PTS:
1
1
ID: A
5. ANS: B
x−µ
z=
σ
therefore:
−1.2 = x − 1045.7 or x = 779.42
221.9
PTS: 1
6. ANS: A
7. ANS: A
Method A:
PTS:
1
Think of this as a sampling distribution of x with n=4 and therefore the mean of the x’s would be
190 lbs and the standard deviation would be σ = 10 = 5. The probability would be
n
4
P ( x > 200). Use calculator normcdf(200, 9999, 190, 5) = .023. *or you could use the empirical rule
since z=2 and you would get probability of .025.
Method B:
Consider 4 separate distributions with mean of 190 and variance of 100 (standard deviation squared).
Simply add the means and variances together creating a new distribution with mean of 760 and
variance of 400 and standard deviation of 20. Use your calculator to find the
P (X > 800)=normcdf(800, 9999, 760, 20) = .023.
PTS:
8. ANS:
9. ANS:
10. ANS:
1
E
E
C
PTS:
1
PTS:
PTS:
1
1
ÊÁ ˆ˜
ÊÁ ˆ˜
E(X) = Σx i p i = (20) ÁÁÁÁ 1 ˜˜˜˜ + (−3) ÁÁÁÁ 5 ˜˜˜˜ = 5 ≈ 0.833
ÁË 6 ˜
ÁË 6 ˜
6
If you were to play 100 times and expect to gain approximately 0.83 points per play then your total
gain would be approximately 83 points in total (0.83)(100).
2
ID: A
11. ANS: C
The variance of the weights is 16 (square standard deviation)
Ten weights would have a variance of 160 (16*10)
The variance of a box is 64
When adding the two distributions (one box with ten weights) add the means and variances.
mean = 7200 + 320 = 7520
variance = 160 + 64 = 224
therefore the standard deviation is
224
PTS: 1
12. ANS: A
explanatory variable is mass in kilograms
response variable is metabolic rate in calories
slope: for each increase of 1 in explanatory, we would expect an increase (positive slope) of slope
response variable.
slope is given as the coefficient of the explanatory variable.
PTS: 1
13. ANS: E
2
χ cdf(7.638,9999,1) = 0.006
therefore p is less than an alpha of .01 so we reject
answer is e, because termite pairs from different colonies are less likely to be agressive. We expect
37.5 to be agressive, but the actual count is 31.
2
Alternatively you can enter a 2x2 matrix [A] in your Calculator and complete a χ -test. You will get
the same p-value. The Expected Counts will automatically be put into matrix [B].
PTS:
1
3
ID: A
14. ANS: D
Use the following formulas: b 1 = r
sy
sx
and b 0 = y − b 1 x
Therefore b 0 = 10 − 5b 1 and b 1 = 2.5r or r =
b1
by manipulating the formulas.
2.5
Because −1 ≤ r ≤ 1, we can conclude that −2.5 ≤ b 1 ≤ 2.5. Therefore we eliminate answers (a) and
(b) which each have a slope out of these bounds.
Next, substitute the b 0 and the b 1 into the y-intercept equation for each response.
c. 5.0 ≠ 10 − 5(2.5) = −2.5
d. 8.5 = 10 − 5(0.3) = 8.5
e. 10.0 ≠ 10 − 5(0.4) = 8
only answer (d) gives a true statement.
PTS: 1
15. ANS: E
Two things that increase the power of a test (chance that you will reject the null hypothesis).
1) > significance level (increases chance of type I error)
2) >sample size (also reduces chance of an error)
answer (e) has the higher significance level and the highest sample size.
PTS: 1
16. ANS: D
(d) is false because even though they used the same questionnaire, we cannot combine the
populations because each population tasted a different recipe.
PTS:
1
4
ID: A
17. ANS: C
Margin of error = critical value * standard deviation
standard deviation of means = σ
n
therefore:
1.96 7200 ≤ 200
n
Since this is a multiple choice question, you can simply substitute each sample size starting with the
least until you find one that will give you a margin of error ≤ 200.
a. 1.96 7200 ≈ 1630
75
b. 1.96 7200 ≈ 391
1300
c. 1.96 7200 ≈ 196
5200
answers (d) and (e) will also give you margins less than 200, but you want to find the smallest sample
size.
PTS: 1
18. ANS: D
You do not need to calculate the confidence intervals, as this will waste time. simply remember that
larger sample sizes decrease the interval, therefore, (d) is the correct answer.
PTS:
1
5