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Unformatted text preview: Stat 3011, Fall 2010 Midterm 2 Stat3011zhdkhemnjExan12
November 22, 2010 Name. Ki; Instructions Do not open the exam until you have been instructed to do so. You have until 3:20 pm to complete the exam. The exam has 11 pages (including this page) and is worth 100 total points.
Read each question carefully. The exam is closed book. You may use a calculator and one sheet of paper (size 8.5” X
11”) with hand—written formulas or other notes on both sides. No sharing of calculators
or formula sheets is allowed. Use of any communication device during the exam is prohibited. Violation of this
policy will result in a score of 0 for this exam. This exam must be your own work entirely. You may not share information with
anyone. Violation of this policy will result in a score of 0 for this exam. You must provide sufﬁcient details to receive full credit for problems in Section 2. 1/11 Stat 3011, Fall 2010 Midterm 2 Section 1: Multiple Choice (10 Questions, 3 points each) For the multiple choice
questions, circle one of the listed choices for each question (no explanation is needed). 1. Suppose we conduct a hypothesis test and compute a pwalue of 0.73. Which of the
following conclusions should we make?
(a) We believe H0 is true
We fail to reject H0 in favor of HCl
(0) We believe H, is true
((1) The probability that H0 is true is 0.73 2. Suppose We take a random sample of size n = 25 from a normally distributed
population with mean p. and standard deviation 0. We compute the sample mean E
and the sample standard deviation 5. What is the distribution of the statistic f‘*rt
3/5
(a) FV(0,1)
(bi N (a CT/ 5)
((1) its 3. Stacy the statistician works for a pharmaceutical company. Her supervisor wants her
to perform a hypothesis test at the a = .05 level to interpret the results of a clinical
trial for a new drug. Stacy wants to conduct the test at the or = .01 level instead.
What would be a potential negative consequence of conducting the test at the .01 level
instead of the .05 level? (a) Stacy would need to collect a larger sample to conduct the test at the .01 level
) (13 Tests at the .01 level are more likely to detect differences that are not practically
signiﬁcant (0 he probability that Stacy will make a Type 11 error will increase if she conducts
the test at the .01 level with the same data 2/11 Stat 3011, Fall 2010 Midterm 2 rI‘he following information applies to questions 4 and 5: According to the web site city—data. com, the mean home sale price in 2008 across the
Twin Cities was $287,000. Suppose we also know that the standard deviation of home
sale prices in 2008 across the Twin Cities was $95,000, and that the distribution was
skewed to the right. 4. Suppose we randomly sample one Twin Cities home that was sold in 2008. What is
the probability that this home was sold for more than $382,000?
(a) 0.16
(b)(l32
(c) 0.84
@We do not have enough information to answer this question 5. Now suppose we take a random sample of 47 homes that were sold in the Twin Cities in
2008 and compute the average of their sale prices. What is the approximate probability
that the average sale price of this sample is higher than $300,000? I — 15.2..»
17 xﬁNQﬂlw’; a, a: %= ofH.
(b)(l45 1ﬁ$q
(c) 0.83 Pawﬁll; 13321“! (d) We do not have enough information to answer this question 6. Suppose we conduct a test of H0 : p = .4 vs. Ha : p 75 .4. We compute 13 = 0.47, and
our p—value is 0.07. What would the p—value be for testing Hg : p z: .4 vs. Ha : p > .4? 7. Which of the following is not true about conﬁdence intervals? (a) Conﬁdence intervals become wider as the conﬁdence level increases Conﬁdence intervals will vary from sample to sample he general form of a conﬁdence interval is (point estimate) :1: (standard error) 3/11 Stat 3011, Fall 2010 Midterm 2 The following information applies to questions 8 and 9: Suppose Fred draws a random sample of size n = 40 from a population with mean a
and standard deviation 0, and Sally draws a different random sample of size n : 400
from the same population. 8. Before looking at the data, which sample average will have the higher mean (expected
value)? (a) Fred’s sample average (b Sally’s sample average
oth sample averages will have the same mean (expected value) 9. Before looking at the data, which sample average will have the higher standard error? Fred’s sample average (b) Sally’s sample average (0) Both sample averages will have the same standard error A t—test analysis was done in R, but the test statistic and the p—value have been crossed
out: > t . test (x=Ferti1ity, conf . level=0 . 95 , alternative="two . sided" ,mu=1 . 5)
One Sample ttest data: Fertility
t = ******, df = 1?, p—value = ******
alternative hypothesis: true mean is not equal to 1.5
95 percent confidence interval:
1.502037 1.797963
sample estimates: mean of X
1.65 10. What is the p—value that was crossed out?
(a) 0.003 wt Wm [main “jar 'Hu, twosiclul +e5'i' di'
'0‘” at .05 level, Luause +t¢ 792 (1: (m5) (0) 0.053
(d) 0.950 do“ M“"‘l“°" ‘50 =/”° 4/11 Stat 3011, Fall 2010 Midterm 2 Section 2: Completely answer each question. For full credit, show all your work and
explain your reasoning clearly. Problem 11 (30 points total) A survey on the website BuzzDash.com asked whether the recent selection of Nancy
Pelosi as the House Minority Leader was a good choice or a bad choice. Of the 102
responses, 46 said it was a good choice while 56 said it was a bad choice. For the purposes
of this problem, assume the responders to this survey represent a random sample of all
Americans. (a) (8 points) Construct a 95% conﬁdence interval for the proportion of all Americans who
believe Pelosi’s selection as the House Minority Leader was a good choice. in‘ r lac. fmfor‘b'on a"; all AMENCans Wlw Lelia/e. Ptlasr was q
good choice. g: % “~.‘{Sl is «m zshmai'e oi: ns'fiﬂlEZIS'
inn—6.) =5BZ5 . 5/11 Stat 3011, Fall 2010 Midterm 2 (b) (10 points) Test the null hypothesis that half of all Americans believe Pelosi’s selection
was a good choice, versus the alternative that less than half of all Americans believe
her selection was a good choice. Perform your test at the a = .05 level, and interpret
your conclusion in the context of the problem. A'SSmPleS : R‘MJOM Sample, V’
n (p) = wars) 4! 215‘ ./
no?) :107—(5) =5! 2 IS I/ I “‘15”: H9: 2‘5 u u
+3?» Hg: P<.S (Mi! Hut. flnmse (855 Hum ) 42,2» Hist:5“ _
TM slalw'an‘c.‘ 2= P ~ '0‘” i~NCoﬂ :5} H» is ’mu.
hm : Paw.411) = .mn ,
Conclusion: P'Valw, 7 d, 50 I‘D «J‘cc'l’ H? in ‘FlVor o‘F‘ PM. interfrcl'a‘l‘ioni TlM's fa“ clues noi' Fmvt‘ll “ml{Ml Wilma ﬁr us to amid; ‘ku‘l‘ [(55 #mn hall: of a“ Hwﬂ‘cans Leliae Pains? was
a j"; CAM.“ a} H0161 Makerlb Leaclar. (c) (5 points) If, by chance, the conclusion you made in part (b) is in error, would it be a
Type I error or a Type II error? Brieﬂy explain your answer. This Woulcl lot. a Taft IT— crror. Wt arc. Fac'lihj i‘b reJ'e c'l’ H5
50 {51: wt make an (war. ii" won mun 'Hm'l‘ H4 is W Lui' we
have, gun'ch +0 red‘u’l' a H‘ :4 PW” bl; H31" 6/11 Stat 3011, Fall 2010 Midterm 2 (d) (7 points) Suppose you wish to design your own survey that will estimate the proportion
of Americans who believe Pelosi’s selection as House Minority Leader was a good choice.
You want your survey to estimate this proportion to within 5 percentage points with
99% conﬁdence. How many Americans do you need to sample? 2.1+ z.
n?— F*(l~]0*) , use a; lo) quwvln emf/3 ‘0”.5’ wall 96 M acceth answvaﬁw of» {ca . ‘ 2575‘)?
AZ .‘lSl(5Lt7)( .95 = ,me (2652.25) 3 (953.7 so: we. would Mal ‘lv Samﬂe 41" least éS? Prm‘cMs. 7/11 Stat 3011, Fall 2010 Midterm 2 Problem 12 (24 points total) One of the items in the ﬁrst class survey asked you to rate your interest level in statistics
before taking this course. The responses to this question were read into R, and some
preliminary analysis was done below. For the purposes of this problem, assume this class represents a random sample of
all incoming STAT 3011 students at the University of Minnesota. > stat.interest<*c(2,9,4,5,1,4,9,1,6,5,3,8,4,3,7,4,5,7,3,4,7,5,10,
5,7,8,6,6,3,10,8,6,8,6,6,3,4,5,8,7,8,4,5,7,4,7,2,4,5,6,5,7,5,9,
6,2,2,5,1,1,6,4,1,4,4,9,7,5,10,7,2,3,3,9,5,3,7) > summarsttat.interest)
Min. lst Qu. Median Mean 3rd Qu. Max. 1.000 4.000 5.000 5.273 7.000 10.000
> sd(stat.interest) [1] 2.365522 > lengtthtat . interest)
[1] 77 > qt(0 .975,df=76,lower=T)
[1] 1.991673 > qt (0 . 95 , df=76 , lower=T)
[1] 1.665151 > qt(0.90,df=76,lower=T)
[1] 1.292790 (a) (8 points) Construct a 90% conﬁdence interval for the mean interest level in statistics
for the population of all incoming STAT 3011 students. Wt have. a random Samfk 0F size. 77, SD assumptions are s 2.3%
CI: Kilt" W = 9.2.73tl.665’( = 51733: 0.4%
44.3w, 5.721? 8/11 Stat 3011, Fall 2010 Midterm 2 (b) (8 points) Perform a test of H0 : p = 5 vs. Ha : ,u 79 5 at the a = .05 level. ASSMF‘I’KMS 2 K4510!" 54mph /
n= 77 350 I/ o .. 1+0: :5” “1595 (33van . .  u"  9333's = I ~ "(211. wina Ha Psime
TU: sfuhant—i t“ ' ('0
5 [m mes/m hm : menu) 2. 2.00:1 43".“ Table 6 Hal“, .2 at, So HI large.ch Ho. Conclust'm 1 (c) (8 points) Explain your results from parts (a) and (b) in a way that would make sense
and be meaningful to someone who has not studied statistics. Th9. rcsuHs Sujjcd' ﬂnf' on a scale 19m I #10, Ha. aucmje.
M‘l‘crcs'l' Icvcl {4 sﬁ‘ffsﬁts ﬁr fnefar‘i'rlj {9 +312 SWIFT" 30” "5 “mu”! 5" 77w averaje 'Fw Halls farmlfmlar class was 5’13ka
Al'chr 'Ham 9 bu‘f m are ‘fS‘Z, 0.4M“? 'I‘Lc were): ﬁrst” claims
1‘5 Lehman ‘13 «ml 5.7 'Frm Fart (ca)J ml m. Jo and” have. {MKJA
evilmtc Jr” (Whit ﬁt verve 1t" all classes is Ja'rrfc'mt 'Fhm 5— i4
furt (lo). 04 «Lu/mac ) Stulm‘l's are maltmlcb fateres teal F1
slut‘str'cs Ln‘érc Mm} 57m 201/. 9/11' Stat 3011, Fall 2010 Midterm 2 Problem 13 (8 points total) Suppose that 20% of the subscribers of a cable television company watch the shopping
channel at least once a week. However, the cable'company does not know this. The company
is trying to decide whether to replace the shopping channel with a new local station. A survey
of 100 subscribers will be undertaken. r{he cable company has decided to keep the shOpping
channel if and only if the sample proportion of subscribers in their survey who watch the
shopping channel at least once a week is at least 0.25. What is the approximate probability
that the cable company will keep the shopping channel? P: .w is known. 3" 20 3/5—
I'la'f): 80 215 , Si by “Q; (Cu/Lth L‘Mt'l' Therm Fir owarﬁms) ﬁx NC r: 1e37= m, = New) S Hi? 2.23% l’CB '21:“) = Mia/.13) : b.3944 3: .1056 111% is «heal a l0“ll% (,Lama ﬁml‘ ﬂu, out
COM‘MMj will {duff ﬁe strp'fj diaan . 10/11 Stat 3011, Fall 2010 ' Midterm 2 Problem 14 (8 points total) Suppose that the mean amount of money spent by all individual early morning Black Friday
shoppers at a store is $110, and suppose the standard deviation is $50. Suppose the store
lets in 50 shoppers at random when it Opens on Black Friday. What is the probability that
these 50 shoppers will spend over $5000 combined? (Hint: The total will be more than $5000 when the sample average exceeds what value?) FC Wt.“ 5‘3cm 2 $5009) ‘3’ lilac) 'FV' 'Hnt. $0 7135—0 330) 50 L3 #1, (mini Und 7740mm ‘er Means) §Z x N91, ﬁt) =N(ue % ’33“) 2N0“), 7.07) too Jlo So f0? 2100): P(z .2 7.07 ) = PCZZ 4.4!!) = l— Pas—UH): lao7?3= .9207. 77‘!!! i5 ‘Joou'i' 0. OLAALC #156 511:7?ch an.”
growl more. Hum $5000 fumbch . 11/11 ...
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