Download Unit 5 Practice Test MULTIPLE CHOICE. Choose the

Unit 5 Practice Test
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
1) We have calculated a 95% confidence interval and would prefer for our next confidence interval to
have a smaller margin of error without losing any confidence. In order to do this, we can
I. change the z* value to a smaller number.
II. take a larger sample.
III. take a smaller sample.
A) I and II
B) II only
C) I only
D) III only
E) I and III
1)
2) Which is true about a 98% confidence interval for a population proportion based on a given
sample?
I. We are 98% confident that other sample proportions will be in our interval.
II. There is a 98% chance that our interval contains the population proportion.
III. The interval is wider than a 95% confidence interval would be.
A) III only
B) None
C) I and II
D) I only
E) II only
2)
3) We have calculated a confidence interval based on a sample of size n = 100. Now we want to get a
better estimate with a margin of error that is only one-fourth as large. How large does our new
sample need to be?
A) 50
B) 1600
C) 400
D) 200
E) 25
3)
4) A certain population is bimodal. We want to estimate its mean, so we will collect a sample. Which
should be true if we use a large sample rather than a small one?
I. The distribution of our sample data will be clearly bimodal.
II. The sampling distribution of the sample means will be approximately normal.
III. The variability of the sample means will be smaller.
A) II only
B) III only
C) I only
D) II and III
E) I, II, and III
4)
5) The manager of an orchard expects about 70% of his apples to exceed the weight requirement for
"Grade A" designation. At least how many apples must he sample to be 90% confident of
estimating the true proportion within ± 4%?
A) 356
B) 19
C) 23
D) 89
E) 505
5)
6) A P-value indicates
A) the probability of the observed statistic given that the alternative hypothesis is true.
B) the probability that the alternative hypothesis is true.
C) the probability the null is true given the observed statistic.
D) the probability of the observed statistic given that the null hypothesis is true.
E) the probability that the null hypothesis is true.
6)
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7) A statistics professor wants to see if more than 80% of her students enjoyed taking her class. At the
end of the term, she takes a random sample of students from her large class and asks, in an
anonymous survey, if the students enjoyed taking her class. Which set of hypotheses should she
test?
A) H0 : p = 0.80
7)
HA: p < 0.80
B) H0 : p > 0.80
HA: p = 0.80
C) H0 : p < 0.80
HA: p ≠ 0.80
D) H0 : p = 0.80
HA: p > 0.80
E) H0 : p < 0.80
HA: p > 0.80
8) Not wanting to risk poor sales for a new soda flavor, a company decides to run one more taste test
on potential customers, this time requiring a higher approval rating than they had for earlier tests.
This higher standard of proof will increase
I. the risk of Type I error
II. the risk of Type II error
III. power
A) III only
B) I and III
C) II only
D) I only
E) I and II
8)
9) Suppose that a manufacturer is testing one of its machines to make sure that the machine is
producing more than 97% good parts (H 0 : p = 0.97 and HA : p > 0.97) . The test results in a
9)
P-value of 0.122. Unknown to the manufacturer, the machine is actually producing 99% good
parts. What probably happens as a result of the testing?
A) They fail to reject H0 , making a Type I error.
B) They fail to reject H0 , making a Type II error.
C) They correctly fail to reject H 0.
D) They correctly reject H0 .
E) They reject H0 , making a Type I error.
10) Which of the following is true about Type I and Type II errors?
I. Type I errors are always worse than Type II errors.
II. The severity of Type I and Type II errors depends on the situation being tested.
III. In any given situation, the higher the risk of Type I error, the lower the risk of Type II error.
A) I only
B) I and III
C) III only
D) II and III
E) II only
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
11) Approval rating The President's job approval rating is always a hot topic. Your local
paper conducts a poll of 100 randomly selected adults to determine the President's job
approval rating. A CNN/USA Today/Gallup poll conducts a poll of 1010 randomly selected
adults. Which poll is more likely to report that the President's approval rating is below
50%, assuming that his actual approval rating is 54%? Explain.
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11)
10)
12) Cereal A box of Raspberry Crunch cereal contains a mean of 13 ounces with a standard
deviation of 0.5 ounce. The distribution of the contents of cereal boxes is approximately
Normal. What is the probability that a case of 12 cereal boxes contains a total of more than
160 ounces?
12)
13) Exercise A random sample of 150 men found that 88 of the men exercise regularly, while
a random sample of 200 women found that 130 of the women exercise regularly.
13)
a. Based on the results, construct and interpret a 95% confidence interval for the difference
in the proportions of women and men who exercise regularly.
b. A friend says that she believes that a higher proportion of women than men exercise
regularly. Does your confidence interval support this conclusion? Explain.
14) Internet access A recent Gallup poll found that 28% of U.S. teens aged 13-17 have a
computer with Internet access in their rooms. The poll was based on a random sample of
1028 teens and reported a margin of error of ±3%. What level of confidence did Gallup use
for this poll?
14)
15) Sleep Do more than 50% of U.S. adults feel they get enough sleep? According to Gallup's
December 2004 Lifestyle poll, 55% of U.S. adults said that that they get enough sleep. The
poll was based on a random sample of 1003 U.S. adults. Test an appropriate hypothesis
and state your conclusion in the context of the problem.
15)
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Answer Key
Testname: UNTITLED1
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
B
A
B
E
A
D
D
C
B
D
The smaller poll would have more variability and would thus be more likely to vary from the actual approval rating of
54%. We would expect the larger poll to be more consistent with the 54% rating. So, it is more likely that the smaller
poll would report that the President’s approval rating is below 50%.
12) Two methods can be used to solve this problem:
Method 1:
Let B = weight of one box of cereal and T = weight of 12 boxes of cereal. We are told that the contents of the boxes are
approximately Normal, and we can assume that the content amounts are independent from box to box.
E(T) = E(B1 + B2 +...+ B12 ) = E(B1 ) + E(B2) + ... + E(B12 ) = 156 ounces
Since the content amounts are independent,
Var(T) = Var(B1 + B2 +...+ B12 ) = Var(B1 ) + Var(B2 ) + ... + Var(B12 ) = 3
SD(T) = Var(T) = 3 = 1.73 oz.
We model T with N(156, 1.73).
160 - 156
z=
= 2.31 and P(T > 160) = P(z > 2.31) = 0.0104
1.73
There is a 1.04% chance that a case of 12 cereal boxes will weigh more than 160 ounces.
Method 2:
Using the Central Limit Theorem approach, let y = average content of boxes in the case. Since the contents are
0.5
Normally distributed, y is modeled by N 13,
.
12
Py >
160
13.33 - 13
= P(y > 13.33) = P z >
= P(z > 2.31) = 0.0104.
12
0.5
12
There is a 1.04% chance that a case of 12 cereal boxes will weigh more than 160 ounces.
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Answer Key
Testname: UNTITLED1
13) a. Conditions:
* Randomization Condition: We are told that we have random samples.
* 10% Condition: We have less than 10% of all men and less than 10% of all women.
* Independent samples condition: The two groups are clearly independent of each other.
* Success/Failure Condition: Of the men, 88 exercise regularly and 62 do not; of the women, 130 exercise regularly and
70 do not. The observed number of both successes and failures in both groups is at least 10.
With the conditions satisfied, the sampling distribution of the difference in proportions is approximately Normal with
a mean of pM - pW , the true difference between the population proportions. We can find a two-proportion
z-interval.
^
We know: nM = 150, p M =
^
^
88
130
= 0.587, nW = 200, p W =
= 0.650
150
200
^
We estimate SD(p M - p W) as
^
^
^
^
^
p M qM
p W qW
+
=
nM
nW
^
SE(p M - p W) =
^
(0.587)(0.413) (0.65)(0.35)
+
= 0.0525
150
200
^
ME = z* × SE(p M - p W) = 1.96(0.0525) = 0.1029
^
^
The observed difference in sample proportions = p M - p W = 0.587 - 0.650 = -0.063, so the 95%
confidence interval is -0.063 ± 0.1029, or -16.6% to 4.0%.
We are 95% confident that the proportion of women who exercise regularly is between 4.0% lower and 16.6% higher
than the proportion of men who exercise regularly.
b. Since zero is contained in my confidence interval, I cannot say that a higher proportion of women than men exercise
regularly. My confidence interval does not support my friend's claim.
^^
14) Since ME = z *
pq
, we have 0.03 = z*
n
(0.28)(0.72)
or z* =
1028
0.03
≈ 2.14.
(0.28)(0.72)
1028
Our confidence level is approximately P(-2.14 < z < 2.14) = 0.9676, or 97%.
15) Hypothesis: H0 : p = 0.50
HA : p > 0.50
Plan: Okay to use the Normal model because the trials are independent (random sample of U.S. adults), these 1003
U.S. adults are less than 10% of all U.S. adults, and np 0 = (1003)(0.50) = 501.5 ≥ 10 and nq0 = (1003)(0.50) = 501.5 ≥ 10.
We will do a one-proportion z-test.
Mechanics: SD(p 0 ) =
^
P(p > 0.55) = P(z >
p 0 q0
n
=
^
(0.50)(0.50)
= 0.0158; sample proportion: p = 0.55
1003
0.55 - 0.50
) = P(z > 3.16) = 0.0008
0.0158
With a P-value of 0.0008, I reject the null hypothesis. There is strong evidence that the proportion of U.S. adults who
feel they get enough sleep is more than 50%.
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