Stat 2470, Practice Exam #2, Spring 2014
Name __________________________________
Instructions: You may use your calculator for any functions the TI-84/83 model calculator is capable of
using, such as probability distributions and obtaining graphs of any data. To show work on these
problems, report the functions and their syntax as entered. Other things, such as integrating, must be
done by hand unless specifically directed otherwise. Round means to one more place than the original
data, variances and standard deviations to two more than the original data. Round probabilities to
three significant figures or use exact values. In order to receive partial credit on any problem, you must
show some work or I will have nothing to award partial credit on. Be sure to complete all the requested
parts of each problem.
1.
An instructor has given a short test consisting of two parts. For a randomly selected student, let X = the
number of points earned on the first part and Y = the number of points earned on the second part. Suppose
that the joint pmf of X and Y is given in the accompanying table.
p(x,y)
0
5
10
2.
0
.02
.04
.01
5
.06
.15
.15
10
.02
.20
.14
15
.10
.10
.01
a.
If the score recorded in the grade book is the total number of points earned on the two parts, what is the
expected recorded score E(X + Y)?
b.
If the maximum of the two scores is recorded, what is the expected recorded score?
Abby and Bianca have agreed to meet for lunch between noon and 1:00 P.M. Denote Abby’s arrival time by X,
Bianca’s by Y, and suppose X and Y are independent with pdf’s.
What is the expected amount of time that the one who arrives first must wait for the other person? [Hint:
h(X, Y ) = |X – Y|.]
3. Explain in your own words why the Central Limit Theorem is important. What are the main features
of the theorem? (Give at least 2.)
4. Define a Minimum Variance Unbiased Estimator (MVUE).
5. What is the standard error of an estimator? How do you calculate it and why is it important?
6.
Which of the following statements are not true?
a. Maximizing the likelihood function gives the parameter values for which the observed sample is
most likely to have been generated---that is, the parameter values that “agree most likely” with
the observed data.
b. Different principles of estimation may yield different estimators of the unknown parameters.
c. The maximum likelihood estimator of the population standard deviation is the sample standard
deviation S.
d. None of the above statements are true.
7.
A random sample of bike helmets manufactured by a certain company is selected. Let
= the number
among the
that are flawed and let
= (flawed). Assume that only
is observed, rather than the
sequence of
a.
b.
Derive the maximum likelihood estimator of
Is the estimator of part (a) unbiased?
. If
= 25 and
c. If = 25 and =5, what is the mle of the probability
examined is flawed?
8.
=5, what is the estimate?
that none of the next five helmets
Let
denote the proportion of allotted time that a randomly selected student spends working on a certain
aptitude test. Suppose the pdf of
is
where
a.
b.
> -1. A random sample of ten students yields data
Use the method of moments to obtain an estimator of and then compute the estimate for this data.
Obtain the maximum likelihood estimator of and then compute the estimate for the given data.
9. A random sample of 50 observations produced a mean value of 55 and standard deviation of 6.25. The 95%
confidence interval for the population mean is between __________ and __________. (two decimal places)
10. The 5th percentile of a chi-squared distribution with 10 degrees of freedom is equal to __________.
11. The 90th percentile of a chi-squared distribution with 15 degrees of freedom is equal to __________.
12. Which of the following statements are true?
a. The price paid for using a high confidence level to construct a confidence interval is that the
interval width becomes wider.
b. The only 100% confidence interval for the mean
is
.
c. If we wish to estimate the mean
deviation
of a normal population when the value of the standard
is known, and be within an amount B with
determining the necessary sample size n is
confidence, the formula for
.
d. All of the above statements are true.
e. None of the above statements are true.
13. In developing a confidence interval for the population mean
, a sample of 50 observations was used, and the
confidence interval was 15.24
1.20. Had the sample size been 200 instead of 50, the confidence interval
would have been
a. 7.62 1.20
b. 15.24 .30
c. 15.24 .60
d. 3.81 1.20
e. None of the above answers are correct.
14. A random sample of 10 observations was selected from a normal population distribution. The sample mean
and sample standard deviations were 20 and 3.2, respectively. A 95% prediction interval for a single
observation selected from the same population is
a. 20 6.152
b. 20 4.244
c. 20 7.962
d. 20 7.592
e. None of the above answers are correct.
15. A random sample of 100 lightning flashes in a certain region resulted in a sample average radar echo duration
of .81 sec and a sample standard deviation of .34 sec. Calculate a 99% (two-sided) confidence interval for
the true average echo duration , and interpret the resulting interval.
16. A sample of 14 joint specimens of a particular type gave a sample mean proportional limit stress of 8.50 MPa
and a sample standard deviation of .80 MPa.
a.
Calculate and interpret a 95% lower confidence bound for the true average proportional limit stress
of all such joints. What, if any, assumptions did you make about the distribution of proportional limit
stress?
b.
Calculate and interpret a 95% lower prediction bound for the proportional limit stress of a single joint
of this type.
17. A __________ error involves not rejecting the null hypothesis
18. A __________ error consists of rejecting the null hypothesis
is false.
is true.
19. A __________ error is usually more serious than a __________ error.
20. Suppose a test procedure about the population mean
is performed, when the population is normal and the
sample size n is small, then if the alternative hypothesis is
test is __________.
the rejection region for a level
21. The smallest level of significance
for which the null hypothesis
would be rejected is the tail area
captured by the computed value of the test statistic. This smallest is referred to as the __________.
22. If the calculated test statistic for an upper-tailed z test is 2.15, then the P-value is __________.
23. If the calculated test statistic for two-tailed z test is -1.84, then the P-value is __________.
24. Which of the following statements are not correctly stated?
a. The two possible conclusions from a hypothesis-testing analysis are rejecting the null hypothesis
or accepting
.
is referred to as the “research hypothesis” since it
is the claim that the researcher would really like to validate.
c. In our treatment of hypothesis testing, the null hypothesis will always be stated as an equality
claim.
d. A test statistic is a rule, based on sample data, for deciding whether to reject the null hypothesis.
e. All of the above statements are correctly stated.
b. In many situations, the alternative hypothesis
25. A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCi/L of radon.
The resulting readings were as follows:
104.3
a.
b.
89.6
89.9
95.6
95.2
90.0
98.8 103.7
98.3
106.4
102.0
91.1
Does this data suggest that the population mean reading under these conditions differs from 100?
State and test the appropriate hypotheses using =.05
Suppose that prior to the experiment, a value of
=7.5 had been assumed. How many
determinations would then have been appropriate to obtain
for the alternative
26. A certain pen has been designed so that true average writing lifetime under controlled conditions (involving
the use of a writing machine) is at least 12 hours. A random sample of 18 pens is selected, the writing
lifetime of each is determined, and a normal probability plot of the resulting data supports the use of a
one-sample t test.
a.
b.
c.
d.
What hypotheses should be tested if the investigator believe a priori that the design specification has
been satisfied?
What conclusion is appropriate if the hypotheses of part (a) are tested, t = -2.5, and
?
What conclusion is appropriate if the hypotheses of part (a) are tested, t = -2, and
?
What should be concluded if the hypotheses of part (a) are tested and t = -3.25?
Answers
1.
+ (0 + 5)(.06) + … + (10 + 15)(.01) = 14.10
a.
= (0)(.02) + (5)(.06) +…+(15)(.01) = 9.60
b.
2.
=
11.
6. C
7. We wish to take the derivative of
set it equal to zero and solve for
setting this equal to zero and solving for
yields
. For n = 25 and x = 5,
b.
is an unbiased estimator of
.
c.
8. a.
so the moment estimator
Since
b.
so the log likelihood is
Taking
Taking
9. (53.27, 56.73)
10. 3.94
11. 22.307
12. D
13. C
14. D
15.
16.
and equating to 0 yields
is the solution to
a.
A 95% lower confidence bound:
With 95% confidence, the
value of the true mean proportional limit stress of all such joints lies in the interval (8.11, ). If this
interval is calculated for sample after sample, in the long run 955 of these intervals will include the
true mean proportional limit stress of all such joints. We must assume that the sample observations
were taken from a normally distributed population.
b.
A 95% lower prediction bound:
If this bound is calculated for sample after sample, in the long run 95% of these bounds will provide a
lower bound for the corresponding future values of the proportional limit stress of a single joint of this
type.
17. Type II
18. Type I
19. type I, type II
20.
21. P-value
22. 0.0158
23. 0.0658
24. A
25. n = 12,
a.
Parameter of Interest:
pCi/L of radon.
Null Hypotheses:
= true average reading of this type of radon detector when exposed to 100
, and Alternative Hypothesis:
The test statistic value is
The critical region is either
Fail to reject
. The data do not indicate that these readings differ significantly from 100.
b.
26. a.
From the chart of “
30.
The appropriate hypotheses are
curves for t-tests” available in your text, df
29, thus n
b.
-2.5 < so we would reject
The data indicates that the pens do not meet the
design specifications.
c.
-2.0 is not
, so we would not reject
. There is not enough evidence to say that
the pens don’t satisfy the design specifications.
d.
Since
the p-value is between .001 and .005, which gives strong
evidence to support the alternative hypothesis.