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Unformatted text preview: Page I of“) Econ 15B: Probability and Statistics Final, 10:30 — 12:30, Thursday, June 14, 2007 Test Form: A Words of Wisdom: "In tfieﬁtture, I pfan on tanng more qfan active rot}: in the decisions I maﬁa ”
— Paris Hilton (6/07) FIGURE A (Critical bound indicated by the vertical line) 1. We commonly set on = .05:
a. For mathematical reasons that go beyond the scope of this course.
13. For a (somewhat complicated) reason that we proved in class.
c. For no mathematical reason at all.
d, Because that guarantees that we won’t need an overly large sample size. I None of the above. Page 3 ofIO 7. Up to the rounding error of the computing software, the function (1 —— @qnorm(.4))
calculates the following (assume that X has the appropriate distribution):
a. The p such that pr(X S p) = .4
b. The p such that pr (X S .4) = p
c. The p such that ﬂp) ; .4, where f is the pdf of X.
(1. The p such that ﬁ.4) = p, where f is the pdf of X.
None of the above. 8. As the size n of a random sample gets very large, the difference between the normal
distribution N(0, 1) and the T distribution T(n — 1) becomes:
a. Small.
b. Very small.
c. Insaner small.
d. Freakin’ tiny!
All of the above. [Hint e is the correct answer.] 9. @qtdist(4, 4) = a. 4
b. 8
@00 hard to be determined (by an intro statistics student) without a computer
. Undeﬁned e. None of the above 10. Suppose qN is the .975 quantile of N(0, l), and qT is the .975 quantile of the T distribution
with n — 1 degrees of freedom. Suppose also that [a, b] is the conﬁdence interval derived
from the Normal distribution, and [c, d] is the conﬁdence interval derived from the T
distribution. (Both conﬁdence intervals were derived from the same data set, and let us b—a_q_N d—c qr also assume 0 = 5.) Now consider the claim: a. This claim is true.
b. This claim is false, because the ﬁrst ratio also depends on the size of )7 . g @This claim is false, because (b — a) depends on X]— , but (d — c) depends on
H
y .
n —l d. b and c
e. None of the above. 11. If you calculate that a 95% conﬁdence interval for the mean u of a population is given by
(231.7, 278.5), then y0u can conclude: @ pr(231.7 s a s 278.5) = .95
b. pr(23l.7 < p < 278.5) = .95
c. pr(231.7 2 [J 2 278.5) = .95
d. a and b
6. None of the above. Page5 of“) 17. You are planning a clinical trial to see if you should reject the null hypothesis that a given
drug has no effect (against the alternative that it does some good). During this planning
phase, you discover that the drug is much more harmless than you had originally thought,
and potentially has quite helpful effects. Using the methods discussed in this course, you
might reﬂect this new information by deliberately doing what (even though this choice
might v o .n  effects): .” ecreasmg B 6. None of the above. 18. Ultimately, T distributions are: a. Probability distributions that reﬂect the deep underlying distribution of many
phenomena of actual interest. b. Very special forms of the normal distributions 0. Mostly just a statistical ‘trick’ for addressing certain questions by ‘reorganizing’
or summarizing data so that its new form comes from approximately a T
distribution. d. a and b @ None of the above. 19. To use Eviews to calculate a 95% conﬁdence interval using the student’s T distribution
and a set of 17 observations, your formula would include the subformula:
a. @qtdist(.025, 17)
b. @ctdist(.05, 17) ‘I c. @qtdist(.025, .975, 17) d. Cannot tell; need information about the (estimated) standard deviation of the data
set ® None of the above. 20. If the random variable X (using a random sample of size n) is an unbiased estimator of the ulation parameter 6, then:
@(X) = e
. lim n ——> 00 X = 6 c. For the given data set, X = (a, b), the conﬁdence interval at the given conﬁdence
' level which is always built into X (1. For the given data set, X = (a, b), the conﬁdence interval at the some selected
conﬁdence level, which is not necessarily built into X.
e. None of the above. 21. 13(82) = 02
a. True.
1). False; E(SZ) > 0'2.
<95 @ False; E(SZ) < 02.
d. False; sometimes E(Sz) < 02, and sometimes E(Sz) > 0'2.
e. None of the above. Page 7 ofIO 27. The distinctive structure of a twosided hypothesis test (as opposed to a onesided test) re ects the fact that:
® You want to show that 2? lies in one of the extreme ‘tails’ of It; ’3 distribution.
vb. You want to show that 35 lies in one of the extreme ‘tails’ of X’s distribution.
0. You cannot assume that u S pm) or that tt 2 um. (1. You do in fact know either that u 5 pm or that u .2 pm.
e. None of the above. 28. If c is your critical bound for a onesided hypothesis test, then if you switch to a two
sided test (leaving everything else the same), your critical bound will always be:
a. c . ,1, _. ,. _ ..._. .cW‘ “' " ""' "‘“hw—L. e. None of the above. 29. Compared to two—sided hypothesis tests, one—sided tests:
a Are more powerful
. Are less powerful Q. Have larger alevels t
d. a and c 6. None of the above. 30. Let ﬁx) be the pdf of the distribution NQJ, e2). 1 = dx
1 (x — m2
a. ex ——Hw——
J27rcr p[ 20'2 _(x#) “(xﬂy
I b. M03 exp1: 202 ] c_ (it—m mpg] VZEJZ 20' x (x  m2
d. — ex — —
J27r0'2 p[ 202 ] None of the above. 31. If it is extremely important that you don’t incorrectly reject the null hypothesis using a
give; data set, you would do the following: Make a very small, although that will mean increasing the size of [3.
. _ Make ct very small, although that will mean decreasing the size of B.
0. Make B very large, although that will mean increasing the size of 0L. (1. Make B very small, although that will mean increasing the size of 0L.
e. None of the above. Page 9 of“) Scratch Paper Page 60f10 22. Suppose the variance of a population is known to be 3, and for your desired conﬁdence
level 0, @qnorrn(c) = b. Then if your sample is of size 75, the length of your conﬁdence intervalis:
a. 1b
75
b. ——6—b
75 c. 2b 5 d. 3b 5 © None of the above. 23. Using the deﬁnitions of 0t and [3 as we have done in this course, the probability of a Type Ierroris:
@ or
b. (lea)
c. B
d. (1—5)
6. (17B 24. When T = t, the p—value for a onesided test using a Tdistribution with k degrees of
freedom is always given by (up to rounding error of the software): @gtdistﬁ, k.)____ ' tdist(t, k)
c. ist(t, k—l)
d. @qtdist(t,k1) ,le. None of the above 25. Using the deﬁnitions of a and B as we have done in this course, the probability of a Type Herroris:
a. on
b. (1—00
® B
i (1l3)
e. a—B 26. Using the deﬁnitions of on and [3 as we have done in this course, the probability of a
correct rejection of the null hypothesis is: a. Ci.
b. (1—41)
B C.
(6 (1 4%) e. on—B Page 4 of“) 12. The primary advantage to empirical science of a T—distribution over a normal distribution
in hypothesis testing is that Tdistributions
a. Account for the unknown variance of the underlying population.
b. Account for the degrees of freedom in the sample.
a and'b.
d. Tdistributions are easier to work with.
6. None of the above. 13. To shorten the length of a conﬁdence interval, you could
@ Increase the size of your sample.
. Decrease the size of your conﬁdence level.
c. Decrease the size of the Tdistribution
d. a and b
e. a, b, and c 14. Let it) be the pdf of N(0, l) and let f be the pdf of a Tdistribution with k degrees of
ﬁeedom. Then for all numbers x: a. ¢(x) (ﬁx).
b. ¢(x) =ﬂx).
y. C W) >ﬂxl @Cannot be determined: it depends on the size of k.
e. None of the above. 15. Using the distribution T(5), the quantile of O is:
a. 0
e 5
c. 1
d, — 00
e. Too hard to be determined (by an intro statistics student) without a computer 16. Suppose that on a two—tailed Ttest, your value I exceeds only the rightmost critical
boundary appropriate to this test. Since the test is twotailed, you lmow that your pValue
will be larger than it would be if you had merely performed the corresponding onetailed
Ttest. Nonetheless, it is absolutely gyaranteed that your resulting p—value will be below
your oclevel for the test that you did in fact perform. a. True.
b. False; it depends on how close t is to the (Jrlevel. c. False; it depends on how many degrees of freedom you have.
cl. b and c. ’ "N.
Islone of the above. “5134‘s?
I 2. 9 l7 4. l5 5. EE 6. % Page20f10 Referring to Figure A at the beginning of this exam, (1 — B) is given by the area of the
region(s) (regions are the shapes bounded by black lines):
a. m and n
(9 j and k
c. l, m, and n
d. 1 and m
6. None of the above Referring to Figure A at the beginning of this exam, on is given by the area of the
region(s) (regions are the shapes bounded by black lines): a. m b. l and m c. l, m, and n @k e. None of the above If r (t < 0) is your “studentized” score from your data, and H0 is the null hypothesis, then when you (appropriately) reject H0 in a onetailed composite hypothesis test at the on level
of significance, this means that: a. pr(Ho l t) = or
pr(TSt]H0)<oc
c. pr(HoTsr)<ot
d. Cannot tell ﬁom the information given.
e. None of the above. Let f be the pdf of the distribution NQJ, 0'2). ﬂu) = 74 when (and only when):
0’2 = V 27:  74 74 d. 0' = w
4221'
{L e.) None of the above. a a. Roughly speaking, the Central Limit Theorem is crucial to hypothesis testing because:
a. It ensures that X is (approximately) normally distributed.
Q It ensures that X is (approximately) normally distributed.
c. It enables us to form the null and alternative hypotheses.
d. It is what makes the (statistical) power relevant to the hypothesis test.
6. It is what makes 0t relevant to the hypothesis test. ...
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