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Bowdoin Math 1300, Spring 2016 - Midterm Exam #2 - S.Ol\,ATlOI\IS
Instructor: Mario Micheli Date/time: Tuesday, April 19, 2016, 2:30 p.m. - 3:55 p.m. Place: Searles 215, Bowdoin College
• You have 85 minutes to complete this midterm. This exam has 9 pages, including this cover sheet. The last pages include the Normal table as well as the table of Student's t critical values.
• There are 6 problems. I sugge~t that you start with the problems that you find the easiest. If you are stuck, move on to the next problem!
• No books, no printed material, and no solved homework assignments are allowed. Besides pencil, paper, and a calculator, only two one-sided (or one two-sided), hand-written, letter-sized sheet of paper with your own choice of definitions, results and formulas are allowed.
• Please do your work on these sheets; you may use the back of each sheet if necessary.
• Show your work. Ambiguous or otherwise unreadable answers will be marked incorrect. So: write clearly and provide only one answer to each question.
Your Name (LAST, First) & Bowdoin ID#: __________________________________________________________________________ _
I certify that the work appearing on this exam is completely my own. ___________________________________________________ _
Problem 1. (20 points total) Consider independent rolls of a fair, six-sided die. Let N be the number of
times that we roll a 6, out of n = 180 rolls. \ l' = ~ )
(a) (2 points) What is the probability distribution of N? (That is, what 'kind' of random variable is it?)
(b) (4 points) What is the exact probability that the number 6 appears between 29 and 31 times out of 180 rolls (with 29 and 31 included)? (You may leave the answer in terms of binomial coefficients and powers of fractions.)
(c) (4 points) What is the average number of times that a 6 appears in 180 rolls?
( d) (4 points) Roughly sketch the histogram for the number of times that the number 6 appears in 180 rolls.
( e) (6 points) Using the Normal Table on the second last page of this exam, find the approximate probability of rolling 35 or fewer sixes in 180 rolls.
(c) E (N) == ~ t = 180· ~ ::.1EJ (<1) 'N ~ \\.QN t.: V o.,~(N ) =: J\; p( \ - f) ::: l'Ro .l . i. =- 25 b {] )
Problem 2. (14 points) Suppose that a random sample of n = 25 observations is selected from a population that is Normally distributed with mean equal to 106 and standard deviation equal to 12.
(a) (4 pts.) Give the mean and standard deviation of the sampling distribution of the sample mean X.
(b) (4 pts.) Find the probability that X exceeds 110.
(c) (6 pts.) Find the probability that X deviates from the population mean /-L = 106 by no more than 4.
(b) 'r(X> (10) -\-?(X~\\o) 1- T(110 - lOG) Z.~
(c) f(lX- \061~4) = r(-L, ~ X-lOb ~ 4)
-r{\OZ~X~\IO)-
_ 1 ( 1\0 - \ 0 6 ) _ ~ ( 102.., - 100_) 2.~ Z.4
= p(l.Gl) - ~ (-\.61) = ~(I.~l) - [1-~(1.61)1
= 2 ~ ( l. 01) - 1 - Z· ( o. 3 S ~ S) - I
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Problem 3. (16 points total) A paint manufacturer wants to determine the average drying time of a new interior wall paint. For n = 12 test areas of equal size he obtained a sample mean drying time of X = 66.3 minutes and a sample standard deviation of s = 8.4 minutes.
(a) (8 points) Construct the 95% confidence interval for the true mean fl.
(b) (8 points) With the above statistics, with what confidence % does the the true mean f1 (of the drying times of the new interior wall paint) belong to the interval [62.995, 69.605]?
(0-) V1 = 12,) d-f :: II ? Ih~ Cr~t~ va1M.{.. jor Ov ke.. p"LobOvkl.\.-~
(b) T~~ ~J..t w~d+"" 0t L'he- CoY\t~d.lY\c,e. iVL~ex-vvt is
G3.605 - X
( ~lM-\lai~+ t.:) )
No\J.J s(.t c·
CNV\o. sotv t, ~Ol("'
G3~605 - 66.~ = 3.305
~.9. Go 5 - GZ . .9 3 ~ _ 3. 305 ) . t
s 3.3oS c. L • e_
{;: )
8.~
{Ii;
C, We., Oo.Q.;t : c= [Ii: (6. 30G) g.~
4
_ 1.363
Problem 4. (8 pts.) Suppose you wish to estimate a population mean /-l based on a random sample of n observations, and prior experience suggest that the population standard deviation is given by cr = 12.7. If you wish to estimate /-l correct to within 1.6, with probability equal to 0.95, what sample size n should you use?
Problem 5. (18 points total) A course can be taken for credit either by attending lecture sessions at fixed times and days, or by doing online session that can be done at the student's own pace and at those times the student chooses. The course coordinator wants to determine if these two ways of taking the course resulted in a significant difference in achievement as measured by the final exam for the course. The table below gives the scores on an examination with 45 possible points for one group of nl = 9 students and a second group of n2 = 9 students who took the course with conventional lectures. We want to determine, via a one-sided hypothesis test, whether these data present sufficient evidence to indicate that the average grade for students who take the course online is significantly higher than those who attend a conventional class.
Hint: I will save you some time-the sample standard deviations of the two groups are 81 = 4.9411 and S2 = 4.4752, respectively.
(a) (4 points) Compute the standard error for the difference of the two sample means.
(b) (6 points) Consider the hypotheses Ho : J.Ll = J.L2 (Le. J.Ll - J.L2 = 0) and HA : J.Ll > J.L2 (i.e. J.Ll - J.L2 > 0), and compute the appropriate test statistic (that is, the t-score).
(c) (4 points) What is the rejection region for the null hypothesis, with a significance level a = 0.05? With the above test statistic, do we reject or accept Ho?
(d) (4 points) Find the p-value of the test statistic (or rather: give a range of the type a < p-value < b).
( 0-) SE- _ x-y
= I S~ + S; ~ rz;:;;::;)-;-;( ~. ~ 152 y- ~ !,2 _ zu 1 ~ ~I ~2 V ~ s - -
CIRCLE one statement which is FALSE. (No justification is required.) for- n k~-L (a) Var(aX) = a 2Var(X) (b) If X and Yare independent, then Var(X - Y) = Var(X) + Var(Y).
({c))If X", N(2, 9), then Xg2 '" N(O, 1). ~ \N~t- is h -",,(. is: (a) If X '" Binomial(100, 1/2), then Var(X) = 25.
True or false? "A 90% confidence interval for the average number of children per household based on a simple random sample is found to be I 90% = [.7,2.1]. Under these assumptions, we conclude that 90% of households have between .7 and 2.1 children." (Justify your answer.)
FALSE I 'N t M..,\J-t gOo/I> c.o", ~~d.tV\u... ~ \:''''-e. W\<.a.V\., :\to 0+ ckUy--tVv
~'lv I \:;"'-Il ~\A.,s~ho(J~ \y\, ('IV tht \'vVt~-r"a.-t [4 7) • 2 I J, True or false ? "The eenter of a 95% eonfidence interval for the population mean is a random variable."
From two independent surveys of the same population, two 90% eonfidence intervals for the population mean are eonstructed. What is the probability that neither illterval eontains the respective population mean? What is the probability that both do?
Consider the two-sided hypotheses Ho : P = 0 and HA : fL i- 0, and suppose we are performing a i-test with signifieance level 0: = 0.05. If the sample size n inereases, does the rejection region R for Ho beeome bigger or smaller? (Justi(y!) " I' _ I
T ~ {, ..! !-J e,CAWV\ '(" ~ ~IM-. . \:7-<-CoV\<\ ~ S b ~% ~ ~
becoYv\~s ~W\a1\..Gr (~~ ~ loJc\<.; o~ S+u.d~t S t Mt'(~b\JJ-~'~'v\.j, Suppose: Ho : ~I- = P-o, HA : fL > fLo , n = 426, Z = 2.48 (z-score statistie). What is the p-value?
f-va-b=l- p(Z.48) = \-,g<BZ.=jO.0066! li~ t. 2,l'iR
6.8. Define the p-value of a test sta.tistic, as precisely as you can. It ts t~e, p1,o\..::.Q;b~ tJ~.J
Ll'v\.d.~ l~ v\'V\-Ll ~fo~{.~S ~o,.J to bvt, 0.., ~+Q.A~skc t:"' .... a./\-· is
o' .. A- ~$i- o..S ~Y\. fo..vov- 0t -t..~ Q.JeA~~~(. ~rcJ...k{,s\.S: o.s -l--k c..v...'("("~vvT s ~s kC. .
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Normallable . " Table shows values oft(:) forz fromO. [()3,59 by srepsof .ot Example; to find ~(1.23),IO(}k in rOw 1.2 and column .03 [ofindt(L2+ .03)-~(1.23) ==.8901. Use ~(zl=l- cJ(--t) fotnegative~.