Download STAT303 Sec 508-510 Fall 2008 Exam #2 Form A

STAT303 Sec 508-510
Fall 2008
Exam #2
Form A
Instructor: Julie Hagen Carroll
Name:
1. Don’t even open this until you are told to do so.
2. All graphs are on the last page which you may remove.
3. There are 20 multiple-choice questions on this exam, each worth 5 points. There is partial credit. Please mark your
answers clearly. Multiple marks will be counted wrong.
4. You will have 60 minutes to finish this exam.
5. If you have questions, please write out what you are thinking on the back of the page so that we can discuss it after
I return it to you.
6. If you are caught cheating or helping someone to cheat on this exam, you both will receive a grade of zero on the
exam. You must work alone.
7. This exam is worth the 15% of your course grade.
8. When you are finished please make sure you have marked your CORRECT section (Tuesday 12:45 is 508, 2:20 is 509,
and 3:55 is 510) and FORM and 20 answers, then turn in JUST your scantron to the correct pile for your section.
9. Good luck!
1
STAT303 sec 508-510
Exam #2, Form A
Fall 2008
1. Which of the following is (are) appropriate statements
5. In the fall of 2003, a magazine article reported that
about randomness and/or probability?
about 87% of adults drink milk. A local dairy farmers’ association is planning a new marketing campaign
A. A phenomenon is called random if individual outfor the tri-county area they represent. They randomly
comes are uncertain but in a large number of repepolled 800 people in the area. In this sample, 654 peotitions there is a regular distribution of outcomes.
ple said that they drink milk. If 87% is the correct
B. The word random in statistics is a description of
percentage of adults who drink milk, what is the sama kind of order that emerges in the long run.
pling distribution of the sample proportion, p800 ?
C. Probability describes only what happens in the
A. We can’t tell if the shape is normal since we don’t
long run.
have a plot of the data.
D. In a small or moderate number of repetitions, the
B. N (0.87, 0.000142 )
observed proportion of an outcome can be far from
C. N (0.87, 0.011892 )
the probability of the outcome.
D. N (0.00109, 0.000142 )
E. All of the above are appropriate statements.
E. N (0.87, 0.001092 )
2. Suppose we want to test H0 : µ = 30 vs. HA : µ 6= 30
but all we have is a 95% confidence interval for the true
6. I want to test H0 : µ = 100 vs. HA : µ > 100. I take a
mean, µ, (27.8,31.2). Which of the following is true?
sample of 15 from a normal population with standard
deviation, σ = 4.2 and get a sample mean x = 102.3.
A. We would reject the null since the sample mean is
What is my test statistic value and the correct p-value?
29.5 not 30.
A. 0.55 and 0.7088
B. We would fail to reject the null at the 5 and 1%
levels.
B. 0.55 and 0.2912
C. We would reject the null at the 5 and 1% levels.
C. 2.12 and 0.035
D. We would reject the null at 10% level.
D. 2.12 and 0.983
E. Two of the above are true.
E. 2.12 and 0.017
3. Suppose a simple random sample is selected from a population with mean, µ and variance, σ 2 . The central
limit theorem tells us that
7. Which of the following would be a Type II error for the
set of hypotheses above?
A. failing to prove the true mean is not 100
B. failing to prove the true mean is not 100 when it’s
actually more than 100
C. failing to prove the true mean is 100 when it’s
actually more than 100
D. failing to prove the true mean is more than 100
when it’s actually not 100
E. failing to prove the true mean is more than 100
when it’s actually more than 100
A. the sample mean, x, gets closer to the population
mean, µ, as the sample size increases.
B. if the sample size n is sufficiently large, the sample
will be approximately normal.
C. the mean of X will be µ if the sample size n is
sufficiently large.
D. if the sample size is sufficiently large, the distribution of X will be approximately normal with mean
µ and standard deviation, √σn .
8. The American Veterinary Association claims that the
annual cost of medical care for dogs averages $100 with
a standard deviation of $30. The cost for cats averages
$120 with a standard deviation of $35. So the average
of the difference in the cost of medical care for dogs
and cats is then $100 - $120 = - $20. The standard
deviation of that same difference equals $46. If the
difference in costs follows a normal distribution, what
is the probability that the cost for someone’s dog is
higher than for the cat?
E. the distribution X of will be normal only if the
population from which the sample is selected is
also normal.
4. What are the z critical values, the zα/2 , for a 39% confidence interval?
A.
B.
C.
D.
E.
±0.28
±0.51
±0.65
±0.61
±0.73
A.
B.
C.
D.
E.
2
0.2843
0.7157
0.3336
0.6664
0
STAT303 sec 508-510
Exam #2, Form A
Fall 2008
9. Suppose a simple random sample of 100 observations is 14. Suppose I use computer software to do the following.
to be selected from a population that is highly skewed
I generate ten random numbers from a N (500, 102 )
2
with mean µ = 4 and variance σ = 8. Which of the
distribution. From these ten numbers I compute a
following statements about the sampling distribution of
95% confidence interval for the mean using the formula
100
x̄ ± √
X is FALSE?
, where x̄ is the mean of the sample of ten ran10
dom numbers. I then repeat this process (generating a
A. The distribution of X will have mean µ = 4.
new set of 10 random numbers from a N (500, 102 ) disB. The distribution will be approximately normal.
tribution each time) until I have produced 1000 such
C. Because the distribution is highly skewed, the
intervals. Which of the following will be true?
shape of the distribution of X will also show skewness.
A. Approximately 95% of the intervals will contain
D. Even though the distribution of the population
the value 500.
variable is skewed, the distribution of X will be
B. Approximately 95% of the intervals will contain
approximately symmetric around µ = 4.
the value 100.
E. The standard deviation of the distribution of X
C. Approximately 97.5% of the intervals will contain
will be σ = 0.283.
the true mean because the probability that a standard normal random variable is less than 1.96 is
10. Which of the following is NOT a question you need to
0.975.
ask when reviewing a test of hypothesis conclusion?
D. All of the intervals will contain 500 since it is the
true mean of the data.
A. Who paid for the study?
E.
You must look at each interval to see whether it
B. What sample size did they use?
contains
500 or not. There is no way of knowing
C. What population does the sample represent?
how
many
really will.
D. All of the above are important questions to ask.
E. Two of the three questions above do not need to
90% (40.552, 48.448)
be asked.
95% (39.796, 49.204)
99% (38.318, 50.682)
11. What is the missing probability and then the mean of
the distribution below?
15. Using the information above, what is the correct range
of the p-value if I wanted to test H0 : µ = 50 vs. HA :
µ 6= 50?
X |
-1 |
0
|
2
|
4
|
----------------------------------------prob | 0.4 | ??? | 0.3 | 0.1 |
A.
B.
C.
D.
E.
A.
B.
C.
D.
E.
0 and µ = 0.6
0.2 and X = 0.6
0.2 and µ = 0.6
0 and µ = 1.4
0.2 and µ = 1.4
16. What is the P (p25 < 0.5) if p25 ∼ N (0.4, 0.0982 )? Note:
n = 25.
12. Using the previous distribution, how likely are you to
draw a 2 and a 4?
A.
B.
C.
D.
E.
A. It can’t be determined since p25 is not normal since
n < 30.
B. 1.02
C. 1
D. 0.1539
E. 0.8461
0.03
0.4
0.3
0.2
0.5
13. If the population variable is known to be normally distributed with mean µ50 and variance σ 2 = 225 and the
sample size used is to be n = 16, what is the probability
that the sample mean will be between 48.35 and 55.74,
i.e., P (48.35 < x̄ < 55.74)?
A.
B.
C.
D.
E.
p-value > 0.10
0.10 > p-value > 0.05
0.05 > p-value > 0.01
p-value< 0.01
You need a test statistic value to determine the
p-value
0.393
0.607
0.937
0.330
Not within ± 0.010 of any of the above.
3
STAT303 sec 508-510
Exam #2, Form A
17. Suppose you tested H0 : µ = 10 vs. HA : µ 6= 10 and
got a p-value = 0.016. Which of the following conclusions is/are correct?
A. Reject the null at the 5 and 10% levels and conclude that the true mean is not 10.
B. Reject the null at the 1% level and conclude that
the true mean is not 10.
C. Fail to reject at the 1% level and conclude that
the true mean is 10.
D. Two of the above are correct.
E. None of the above are correct.
18. When figure skaters need to find a partner for “pair
figure skating,” it is important to find a partner who is
compatible in weight. The weight of figure skaters can
be modeled by a normal distribution. For male skaters,
the mean is 170 lbs with a standard deviation of 10 lbs.
For female skaters, the mean is 110 lbs with a standard
deviation of 5 lbs. Let the random variable X = the
weight of female skaters and the random variable Y =
the weight of male skaters. What is the distribution of
the SUM of the weights, X + Y , if the weights of the
partners are independent?
A. We do not know if the distribution of the sum is
normal.
B. N (280, 112 )
C. N (280, 152 )
D. N (60, 52 )
E. N (60, 152 )
19. How likely would the AVERAGE of 9 male skaters be
more than 180 lbs, P (Y > 180)?
A.
B.
C.
D.
E.
0.0013
0.1587
0.8413
0.9987
0
20. How much would a female skater weigh if she was in
the 80th percentile?
A.
B.
C.
D.
E.
80% of 110, 88 lbs.
114.2
114
112.9
110.8
1E,2B,3D,4B,5C,6E,7E,8C,9C,10D,11C
12A,13B,14A,15C,16E,17A,18B,19A,20B
4
Fall 2008