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**Unformatted text preview: **Stat 35000 Sample Final Exam Summer2016 Your name: K43)! - . Turn off your cell phone before the exam begins!
- Show your work on all questions. Unsupported work may not receive full credit.
0 Report decimal answers to three decimal places. - You are responsible for upholding IUPUI’s standard for academic integrity. This
includes protecting your work from the eyes of other students. . You are allowed the following aids: a one-page (front and back) 8.5 x 11 handwritten
formula sheet, a scientificlgraphing calculator, and pencils. . When you are done, turn in both your exam and your formula sheet. Problem Points Possible Points Earned
2 40 ......
3 40 3‘
...... 4 37 l
5 32
B ”45
7-- .18
Total-W 256" Problem 1. (6+8+7+8+9=38points) The strength of a randomly chosen thread has
mean 0.5 lbs and standard deviation 0.2 lbs. Assume that the strength of a rope is the
sum of the strengths of the individual threads. There is a rope consisting of 100 threads. 1. Can you ﬁnd the probability that the strength ofa thread exceeds 0.7 lbs, i.e.,
P(X>0.7)? If so, pEease calculate the probability; if not please explain Why. N3,- <’?r'ti'l(€3 5.2.qu Git: ﬂit-3i: {reﬁt/it,» iittf Cg». Vii/“é git/“i7 UV at" Leg ft 2 We are interested in the strength of the rope consisting of 100 threads, denoted
by T. Specify name of the approximate distribution and parameter values for T. {D{) 1- .. \-"“r t? r'x- L
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3. What theorem did you use to obtain the above approximate distribution? Under
what condition can you use this theorem? ((2 ti "it’d-i- [,5 in ii The tied-413th amkt ﬂﬁﬂﬁ”?30 4. Find the probability that the strength ofthe rope consisting of 100 threads is
greater than 54 lbs, i.e., P(T> 54). T rt“ {1-- it; .- " "if.“ --
K t E '"' fug f L n... :3 j \
. i .w =- g ‘ ’ \ "It“: , (f 2...,- EWW-m; ...... 7 wlwtwin—”WWWWWWW-w i 5. Find the 5th percentile of T distribution. That is the value that divides the T
distribution into 5% and upper 95%. “W“?
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j _ . . ' -.-.. .22 m, Mitzi Problem 2. (6+9+9+6+10=4O points) Let X = number obtained after rolling a fair six-
sided die. 1. What is the probability of getting a six after rolling the die, Le. PCX: 6).
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(3C5. >5 5 C / C 2. Let Y be the number of six that appears when rolling the die for 60 times. What is the
exact distribution of Y? Please specify name of the distribution and its parameter vaiues. 4 ,
f3 Ci. rig- gg,C{‘gj‘-g skiff C CC; CC C C. K) ijii-tft: j; Calf CsC “COCK: 5365‘” EiiibpﬁcCC {Cl 116:5} / ) Cf ----- “’Kﬁ’c I{fCC SCC) 21
..... 222-- _W.. .... mi- 3. Find the probability that no six that appears from the 60 rolls using the exact E
distribution. < — - ﬂuﬂWW.-.-----_-_‘_ ““5 ”m Clip? Ct 13;: " {3"3“ IIIII ”YET: jg ‘ :
mam t-it wit C ****** 5 ‘ i 2
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.- {ﬁéuﬂ ------ CU - ----------------- ---- ~~~~~~ 4. What Is the approximate distribution of Y? Please specify name and parameter values J M Cry/117%“ 53%,} M“ Ma (Cw/a 5. Find the probability that more than 15 six appears from the 60 rolls using the ap-
proximate distribution of Y. VC'TCC>'$ (W PCT§i§> _ . CM ,7 (C55 \
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3:. {we FC a; ﬁ {UK/‘3 Problem 3. (10+3+3+2+12+3+7=40 points) A biologist is studying the cellutose content
of a variety of alfalfa hay Suppose the cellulose content comes from a Normal
population with unknown mean but known standard deviation 0: 8mg In a sample of 25 cuttings the mean cellulose content of 145 mg.A AMA” WA “1W” """"" a) Find a 95% confidence interval for the mean cellulose content. It ‘1‘: 7%; ﬁn @533"! 02.2.. M33059: iffﬁéﬁi‘tf-I’KFE” Rig U53 IW SQIJVHLPEIEI (:35 III “(15:33 {3:435 ”325?": (23:, ‘3?“ ----- >5 I <4 Ila”? 03W?
5 J {a :{ 3C3” # ti (‘5 f'w “ I M w w WW Hm - "“3
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I! 35732, 3” gee??? v l 75 I} I444 3I33 I III4 ,NI4;4WMW;;WmmeWWM5M» b) A previous study ciaimed that the mean cellulose content was 140 mg Perform a
hypothesis test to determine ifthe mean cellulose content IS greater than 140 mg. 1. State the null and alternative hypotheses.
HI IIIIo II nIcIM>Wa>
2. What does the parameter p in the hypotheses represent?
{5% 3 [email protected][email protected]’3n (I; {M .EO‘ -5 I‘jaI’ITlgéfI/‘j‘ 3 Which method would you use? 4 M “\gfm’e III-I2 {enema 37:22:44 kwméliﬂw .w [’4 {"L M Li) LIL it. { 1&3" tulip “If H
I, 3; 5 III:
II”'~nMW-?‘} M
7 ffémwigﬁ mmmmmmmm m E: I g ’3 (3 5. State your conclusion in the context of the problem, i.e. whether mean cellulose
content is greater than 140 mg. (assume a = 0 05) $555653 {5...}! pat-{55; :TQ emf <1. Xguuﬂ hint He
ft? {eve/t; $3“ U CU shit? twee wirttgwmﬁg é; x’rrttaww Effie» ‘ I
531 WILLQ fﬂtﬁif’ct‘t (£1 Jay/535‘“ f/{ﬂtt‘e} ii“ ’53.. [LEE/rt.” ‘I‘g‘fﬁ u ‘2‘”‘hﬂ :35 5 it 2;; 4/5 U ., 6. Someone might argue that the above testing procedure (calculating a p— v—aiue etc)
is unnecessary. Their reasoning goes as follows: the sample produced an average
cellulose content of 145 mg, which is more than 140mg, so conclude mean
cetluiose content is greater than 140mg. Do you agree with this fine of reasoning?
Explain. N90» X ..... (Qt: '5]; {555 0‘6““ [if ”gs/{£9 WEI}- ('3? the” {X if} {_’_:1f55’{ju~gtf3’3{jf__ ( 0%. {1,5 tier}; {71:5 £552 «tats-mt sift t” ["t? t’t’fatﬁtt’L It 51 5*- Cttt 55-in W5 out av (55. 5.55 weave [rim/45553 We. 555553-45 (i’ctteté’é‘tt .45... ‘IW somatic”; meanest“ i“"’?w 52055127Warmer/t HQ whet 0.15»an wigs/t» £95; at V54 [55% [5255 “UL/5:55,“; ‘QCE‘M’f Problem 4. (6+4+5+10+6+6= —37 points) An investigator theorizes that peopte who
participate in a regutar program of exercise will have levels of systolic blood pressure that are sgﬁ fIca YIXI deereanrom that of people who dommndeartipapate
in a regularmprogra exeét‘seé‘ WW5 Ideatlﬁ“ InvestIgator randomly”
assigns 21 subjects to” “ash exercise program for 10 weeks and 21 subjects to a
non exercise comparison group. After ten weeks the sample mean systolic blood
pressure and sampie standard deviation for the two groups are given below. r group I Sample Size Sample Mean ‘ Sample Standard deviation
‘ exercise 21 137 t 10
‘ Non-exercise 21 127' t 9 1. State the null and alternative hypotheses. a 1 (it; ”(Lt/3:“ O t} C. H, ; it; H2 '75 O 2. What do the parametets a and 512 in the hypotheses represent. Fl 5” gt” {‘4' , ‘7 _
m c 5 {a >UL’51‘Q 73v 555ij 755% w“ a, \
{ti/(t «4 m‘” ‘1 E { Ii. 5‘ _ 3. Which method would you use? W“‘”\.
(a) 1-sarnp1e T (b) 1—sample Z /(c) 2 sampte T ‘5 (d) 2-samp1e Z .2 . f 3". t- . . . .. r“ "
SEXY 9 wit}? {it/x; V?- iﬁf hit-0‘42 it“; “3““ J: ‘32 4. Assuming the population standard deviations are unknown but equal calculate the 955% _ _,.. ....~
conﬁdence interval for the mean difference. -‘ (”55151 (:1; {151/ 1511 My ié {<1 1:. 555 ,, 25.1525 - 1 mi ,: W. 1 .4... w — M-.. M... m “M
w1f~§w1:i2ﬂi11nil5?5.€5 $4131 5 >1 1; 23445;
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141-1111 311:. 51 - 35331111141111 1111.}
II t3 {2: E.
5. Use the calculated conﬁdence interval above to test hypotheses given in pang. Does \33
exercise have an effect on the systolic blood pressure?
13“.." 1111-); 15511. {1‘23 1?”? 110111 11111“ 1 1 1151111119,. Cw 1511536§35 {#15131 127171. 111.31”: “15111,, Mjﬁ3ffv1111175 131115191“? :2: w— (“)3 CF12 E (x/ ‘
13131;- 31131115 5151,. 1!”) .955 3331115- 7517-1511135 {3 K11,“ (>11... 511111.151: (.3111. {riff-{3:135 3:). 1’1 55 1“: 0511121! 5 32115551355151.- 6 How would you g1aph1caliy check the assumption that the two populations are normal? /
( 55155011351111.1141 01121125 {5111“ 11111.15 3.2112111? 5‘ 513513“ 3155055 Problem 5. (8+3+3+2+2+11+3=32 points) Twelve cars were equipped with radiai tires and driven
over a test course. Then the same 12 cars (with the same drivers) wereequipped with reguiar
beited tires and driven over the same course. After each run, the cars' gas economy (in kmfi)
was measured. is there evidence that radial tires produce better fuel economy? (Assume
normality of data.) Data can be found in the table below. Gaseco.(km/I) 1 2 3 4 5 6 7 3 9 10 in 12 Mean|s.0. gelted 4.1 4.9 6.2 6.9 6.8 14.4 5.7 ‘58 6.9 4.7 16.0 4.9 550810394
Madmgelted 0.1 —0.2 0.4 0.1 0110.1 0.0102 0.5 0.2 0.1 03 1114210193 1
Radial 4.2 4.7 6617.0 6.7 14.5 5.7 6.0 7.4 4916.1 5.2 5.750l1.053l a) Find a 90% conﬁdence interval for the mean gas economy difference between cars equipped
with radial tires and cars with belted tires. 55‘55‘75 5 555 55555” 55555675555515 M {161111 5.51.11’1-131’9151‘73’. F Q3553}: 3 f, 53;: 35251 “/5956, 1 C; -
"11' a.
1...- , I (3.11:5 :5 A f." (3 5518 $51 5 E} 5?
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11151.1 1 1 55 i Z .
.... m 41.1 W .5 ”:1 ,,,,,,,,, 3 I ﬂ M 4 a... ~. 3 _. .. l
........ 1.21142: .1... i 1’31; .3113” 1 33.33. 33 1.33 -5 % -1. 1.13 11 Mg 15
“ ’ “‘1‘ --------- ~12 /
.:6.1111 "’4: 51111.5?» 56111-5555 H1? 3
.. L12» {1 5 1 3 9.151%-l * b) Based on the above confidence interval, do we find any evidence that radial tires, on
average, produce better fuel economy? 5% are (<5: 555a aux.) 5.53; newt-6545?"? 5 63-5553? [as 55515555. 555555.55..-
{133,55 {35g [ﬁg/U Lﬂisﬂﬁimi ”55.55;? 55755515655 "5"??? ‘3 {5WD 65595;; Attittjr’ “5055/5 53: diet"? «iﬁ’t’i- Z5 3 t; 5/1. (3% artifaﬁﬁ “ 0) Next, we perform the hypothesis test to determine whether radial tires, on average produce
better fuel economy. 1. State the null and alternative hypotheses. it. fitrwtitﬁrc v a. n: it Mum 2. What doesthe arameterintheh otheses re resent? . x K
D W p W 5551 Varﬁﬁ‘wﬁ. “fry/€15; . 5953 55955 35g“!- :: 5’35? €0.55 9 act; (it {metres} 59w {TQMS‘ ﬁg 5:: R . .......... 3. Which method would you use?
( (a) 1 sampleT (b) 1 sample Z (c) 2-sample T (d) 2-sample Z ”hm-A M 4. Calculate the test statistic and find the Rejection Region, using 0: = 0.01. 7-1.. wwwwwww t- a a t a 5. State your conclusion and interpret the result in the context for the researchers. .
,_. u . 77 r. . . ’ 5)" (“a .- :5 .1“. .. “5“} g s" %
(Fiﬁ... .,,N.. A{§F‘a{ Wu"? LE]: 53‘;- L? .51; (2:: 1’? g {a} [5 «Er {“ng is gr} 5% Tat/5:1: f 153% 5’; fit... {:1 V W ,/s. . . . f w pm M i u " 3
15555 tile 5 5;, t? e 5.5“ n...) 55% 05 e L. "t 5;}
[315-353; (”A at Vijtgf dig; {7 t3% 5151559 194552 (New)? {3 .595 55515 (ti-:2
9555;522:2355" 1555:1355“ \‘tkcégatf 55555“; 5* ,- 5?) t“: (Leciéﬁj 5’ - g}? affix/{(lﬂ 54715—3/ 554/ iii/Lt? {i {2” {n.- €155 {.45}? k} Problem 6. (5+10+5+5+10+10=45 points) The owner of a towing and garage service
wants to know how the lowest temperature on a winter day affects the number of
emergency road service calls he receives. He records the lowest temperature (in 0F)
and the number of emergency road service calls he received on 30 randomly selected
winter days. Summary statistics of data are: E2l4, SJ. 216.6, ﬁ=18.6, s), 28.3, ”:30 1. In view of the study the owner wants to answer, what is the independent variable, and what
is the response variable? X‘ 2:: 1512222 3% 1’2: 2<’2--<2{2.2222:2t%2-232 X” *1“?- 212+ 6? 22’“: “$622213 2/222 2 21;. 2% “~22” 225‘s 21>: 27‘ <72 .2
-- - “ 2 <1 i 2. Suppose that the correlation between the number of emergency road service calls and the
lowest temperature r = —0.89. Find the least squares regression equation. - “7 L 53’”- 722
ft 4-2 2334 W‘ 2.. .I’ mow-«NWM ‘""" ~ " ‘ "ﬂ“
{5‘2 :: '1" 2 “T "”22 W ' 2 <2 “21’3- WK
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3. What is the predicted value ofthe number of calls for a day when the lowest temperature is
16°F. ‘ ti £2 n27) :— 2222. 2’22 2- (22:2 5242:???) (7/22 ‘2
1:: {if a 7 g 4. Is the model overestimating or underestimating for that day when the lowest temperature is
16°F and the actual number of calls is 13? "N1. {3 yeﬁ2< ek/iwlFYlCtbf?tt<§ 2 gv2ta.rsf $1 7'5 f? 5 x [.F"
\. 5. Answer the following questions by filling in the blanks based on the interpretations of 0: and a . 0 When the lowest temperature increases by 1°F, on average the number of emergency alt-’22:". 2.222722. is 232. L212 £5 road service calls the owner receives will - When lowest temperature is 0°F, on average the number of emergency road service calls the owner receives is ,2 A?" g; S Find the 95% confidence interval for the slope ﬂ ofthe regression iine. Does the 95% CI.
support the claim that )3 < O? - {:1 g“ ..
(Ti-Vii if?“ "g 2 ,2" 5:“ iii 21:2th (>2: C 0 f 2, 2?va "73“” I "" i M M 5:36“
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x: k , (0 O"? m 211.3 2"" W ““'"""7"':"'§"
{7 U” L” "2““ (2 H > i <2 ) : “5’2 r“: 23 ....... 25:2 3“
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W mm W 2 2 ii ,1 2 1‘” 5%: Mpgum 222 ) fawn {032?}
:pc-ﬁrw> ﬂow?) < < 0000;" Problem 7. (6+12=18 points) A random sample of 100 data points was collected to
test if the population mean, u, is different from 240 at a =0.05|evel. That is, Ho:#=240 vs. H12y¢240 The sample resulted in a mean of 235. Assume the sample is from a Normal
population with known standard deviation 0:40. 1. Whatdoesarepresent?Howdoyouinterpretit? _
at: PCT WEI meow r: [3: MIT-{- H l HO mime )
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whale it t; actuallﬁ eeuai “Eb 314C). 2. Suppose that the true population mean is actually 236. For 1the test above, find the probability [3 of committing a Type ll error. ”a
b , . x
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