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Unformatted text preview: ph 141 summer 2010
midterm 1 NAME k F )1 Show all work on these sheets. Total 100 points. Working time is 2 hours. Part 1. 20 points Aﬁer a person with dementia has died, the diagnosis of Alzheimer's disease can be made deﬁnitively from
an autopsy, but it is extremely difﬁcult to diagnose Alzheimer's disease for an elderly person with dementia. Suppose that researchers are assessing a new test to assist clinicians in their assessment of elderly patients
with dementia. When the test was done for a sample of patients for whom the autopsy diagnosis was available,
the test gave the following results: test + test —
Alzheimer's disease 800 _ 100 900
other dementia ‘ 150 1350 1500 1. ( 4 points) Use these data to ﬁnd the following probabilities probability the test is positive given the person really has Alzheimer's disease 900
900 = .‘3‘? probability the test is negative given the person really has another dementia I350 : ﬁlo
'50“ Suppose the prevalence of Alzheimer's among this population of elderly persons with dementia is thought to
be .40 Using this prevalence and the information given above,
set up a tree diagram on the next page to help you answer the following questions: (16 points) What is the probability of Alzheimer's disease given the test is positive? N 9* T P A C: E
What is the probability of Alzheimer's disease given the test is negative? .Ho 0 “0
AV" °THER
'%'*/\“ .VVO
"TESTi 159;  TEST 1' TEST PL1E$1 e) —. P (am. Ann TEST +) ;. P‘Mnm AM“ ‘EST 0) — Puma.) Pug“ u an.) 4 p LINtan) p “I“ * I ovum) : .40 x .99 + .b0 u do
: ’35" * .0430
3 .HH. 80 PS TgsT ) ._ ‘ .th :. .‘SﬁH PLRL‘L ANb TEST +) D‘Ak1\ TEST +\ M“ PL‘TESTo)
7' .Ho 1531 — A 35L.
‘ ‘_“”——' = .95;
'H" .Hna
'PCAL‘LI 1551— .) : PU‘L‘L Ann TE51 .)
PLTES‘r3
7' M _. .OHIl
’ 1 .015 .ssq .SQH Part 2. 20 points A club has 10 members 1. (5 points) How many ways can they appoint a committee of 6 members to plan an event? ‘ 3
('0) = —'°—" : '“WS‘WU _ toxxszA'I
‘9 1155’. India: (a! ‘ “43:2
1 “’11! = 1:0 For the next 3 questions, suppose that the club consists of 5 men and 5 women. 2. (6 points) How many ways can they appoint a committee of 6 members so that there are 3 men and 3
women? MWLT‘PLQRT\ON RuLE '. NuMB‘EK oF ways ‘ Nuneea 0F wnws
*9 Pick MEN 7° pick wo MEN
5 t s!
(g) " ( 3) 5' t
3 113', 1'. 3!
Si“ * sxu
0 :
’1. 7,. l ’( M) NO 3. ( 4 points) If a committee of 6 members is appointed at random,
what is the probability that it consists of an equal number of men and women? '00 _ '0
’LIO 1 : AFN. ‘T IS HWPEKGEoMETam Bur ensma To be 3" RgﬁsonKs )
am: Tue kuSwEnsTo QMESTDONS I Am: '1. 4. (5 points ) What is the probability that a randomly appointed committee of 6 has no women? 0 THERE (we WU; 5 Man To LHooSE men MosT BE AT LEAST I \NOMRLJZ Part 3. 25 points
Is there an association between a mother's smoking during pregnancy and premature birth? The following data were obtained from a crosssectional study of 1, 300 mothers: premature birth fullterm birth
smoked throughout
pregnancy 45 355  400
I
did not smoke at all 
during pregnancy 35 865  900
80 1220 I 1300 1. (4 points) Complete the following table of expected cell counts under the assumption that there is no
association between smoking during pregnancy and premature birth. Give the details of all the calculations you need to make and explain your reasoning: premature birth fullterm birth
0k
smoked throughout ro’ F H41)
1 .
pregnancy ‘1 L 375.4 400 E “K 121:
did not smoke at all “L‘ "°° ”15
during pregnancy 55“ 844.6 900 [31 sugmnume
1f ‘tou
80 1220 1300 ‘
6x9~A\N
P(PREMA1\;.P.E) '3 $9 = aOLIS . Rmsoumtyl
1300 ‘
EM mom SMoueRS : .0515! Hoo  “M,
E19 Loom NoN—SMM<ERS ’ °"'5" (‘00 7' 55'“ 09‘ (3° ‘ 1”" o a 30 I q 00
(300
2. (4 points) Do these data satisfy the conditions required to use the )8 distribution to ﬁnd the P value? Explain why or Why not — do not actually calculate any test statistic or P value!
ﬁes ! ML 619 can. Loums 7; 5 ’THE Con‘nou For: \(1 Fort a 111 Tami 3. (4 points) What is the relative risk of premature birth? N PREMATUKENSMokER) L‘s/“W .II’AS  PtPKEMATuRE ‘ Nun SMDKERB 35/5100 .038? RR: 4. (4 points) The investigators carried out a )6 test and obtained )6 = 25.98 What are thedegrees offreedom? K NULMBER °F hws' 0 1‘ ("“MBER 0‘: (”mums ' I) L 1 — n ‘ * t 1  i)
I 3‘ I ' 
What is the P value? (It is a good idea to report the P value as an inequality.) 1ABIE F, ta” mow P’\ Y} W ‘1_ta\NbED) = °°°5 P 1,0005 5. (5 points) Do the investigators have evidence that there is an association between a mother's smoking
during pregnancy and premature birth? Use the P value to justify your answer. Write a summary sentence
that explains your conclusion in words. n: “haze Rearw is No nssoatATIoM \N "THE POPLLLA‘HON,
l“E335 fun» 3 \N 10,000 SH HPLES C’NE "Y,“ > 1J1
So, Assumwe No ﬁssouA'noN) ouK SAMPLE ham WeUth BE VEM RRRE. we coNCLubE THAT name [5 AN Assoowrrwu BETWEEN SMowMG
THRoucHouu Pkécktauc‘t PW! l’nEMATung mant. 6. (4 points) Do these data prove that a mother's smoking during pregnancy causes premature birth?
Explain your reasoning. (This question requires thinking, not calculation.) 1m; ts OSSERVA'HOHRL brunt) So 1HE answer; is NO
womeu wuq SMOKE THkonHwT PREGNANOY “9“ “”1““ FROM WNW“ WHO be ﬁt“ SMOKE A1 ALL burUNG eneeNAuq. (uneven >131) ALCOHOL °°NSKHPTIOH)..) PERHaPs LAB E‘LPEPAMEHTS um“ SHELL MHMMALS, \N WINCH
RBMMM RsstlsNMENT TO SMOkE exvosuac IS erchL, toutb ”.0me SOME CqusAL EVlhEMLE . Part 4. 15 points You are asked to assist an investigator who is working with a simple random sample of 400 men
age 40 to 49. You have calculated the body mass index (BMI) for these men and created a box plot.
The investigators have asked you to explain the plot and make some suggestions for summarizing the data. minimum 1 6.9 25th percentile 22.9 median 25.1
75th percentile 27.9 maximum 57.6 Body Mass index
41) 5 0 6 0 3‘0 20 Using the summary and the boxplot, answer the following questions. 1. (12 points) Explain clearly how stata determined which points were to be plotted as outliers in the
boxplot. Give speciﬁc numerical details, not just a general outline. IQR‘ 1”  11.“ WesQ.) 5 1.51ch = “1.5
“975K FENCE .3 Q3 4' LSIQR 7' 1".q+‘1‘5 ‘. 35'“ BM! VALuES 7 35A! WLL BE PLO‘T‘TEB As no LAN.) ukuts (Tums A115 m LERST a!) N°1E rHé Mom ts SM, LowER Fence. : (1‘  LEIQK = qua$5 : lSH km \sALuEs ‘ ISH \mu. BE none» as momre» cmLLEs we Mm IS ”9.5!, So 1HERERRE N0 ouruens AT THE Low Eu» N: THE mnmuunon. Tue Lauren wmgkgg EHDS
AT Ieﬂ 2. (3 points) Based on what you have learned in class, do you think the assumption that the distribution of
bmi for this population is approximately a normal curve? Why or why not? ‘WE MEMAN ‘5 “Emir Hethwa‘; Bewweeu 0‘ an» Q3) SiucE
MQ‘: '15.!  11.7 = 1.7. M» 023 M'— 2cm  15il= 1.8 Tu: UPPER WMSKER. \S A Bn’ mum.1 THAN THE L°WER WMSKER nus lS No‘r so 151RABLE!
Ho waven, we Saw 1HAT THE. UPPER FENCE Fon ~nu; NORMA! cums um; A 1. scoRE 3F 1.10, so Tu‘m pgomucm A, Hm” aunt]
is ABM»? .0035 WE Hem; T00 Mam; Mammals FOR 1HE BIST‘MBvTIoN To BE Rouenw NOﬂMRL cum; ! [8m °F 5'14, lS estmama \F THE ”“0“ WE“ ‘9 FE“ Ta“) nus ooaQeSPoubs To A “Just” OF HIS vouuns
PM A to Pctr mn Pek 3°" W“ WW,” 2‘00 PouMDS) 13m: qoq QOME‘UMis \NVESTKsATORS L‘MiT 3669:. OF ANALV‘S‘S «o excuse New EMKEM‘: \mwns — “we“ Maria). ‘53“‘] Part 5. 20 points Suppose that you are in charge of warranties for a manufacturer of batteries for an implanted medical device. From longterm quality studies, you know that the distribution of lifetimes for your batteries is very close to
the normal curve, with a mean of 60 months (5 years) and a standard deviation of 8 months. 1. (6 points) What proportion of your batteries last longer than 75 months? 5h; ﬁs—eO, = lg,
/ ‘4‘: 6' Q g ‘ L815  so 15 0k To use :31 0R I539 on INTERPOLATE
Aaom \ .qlo‘le : .0309
I815 2. (6 points) What is the 90th percentile of the distribution of battery lifetimes? .‘iooo ‘ \ 1.19  8 1 was” a X' (a0 FOR 1= [.18 (polLassa 3 x mun 10 rue LEF1 (.o+ norm = X 10")“ MW”: IS  2 q 1 7
3. (8 points) Suppose your research department wants to improve the quality and consistency of the manufacturing process. They have been able to modify the process so that the mean is now 67.5 months. (The consistency will be measured by the standard deviation, the spread of the battery lives around the new
mean.) If they want 99% of the batteries to last at least 75 months, what standard deviation do they have to achieve? nus ts No1 Pm A'rteMeT To Tuck ‘wu  vr was \NSTfluc‘foR eaaoaf 1; THE Mean is 51.5) matr 5F iHE Garrett‘s: mm. LAST LESS Tm‘”
v1.5 Mourns THE caKREcT 9: Fall .qcioo [s RBaUT 1.33 0: 3.1 MoNTHS womb (“we 9‘1")» 0F TREH L685 mm: '15 MONTHS, Bur TMA1'S Not A userLu. Qucs‘nou ...
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 Summer '10
 Lahiff

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