Stat.285.exam2.solutions.post.2page.pdf

Stat.285.exam2.solutions.post.2page.pdf - (35 l A rimdom...

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Unformatted text preview: (35) l. A rimdom sample of 1600 weights is obtained. It is know that the population ofweights is normally distributed with mean 160 and variance 2500. Answer the tollowing questions: Rap —>_l\l(l[t>0, 501) a) What proportion ofthe population has aweight between 145 and 170'} A: B +6 A“; 1 7?} No So Nat" A! c “A”. / [”9 A) Nail) 31115 °’) 3 / l .074 I‘ll / / c ‘ 3' / :/ Z i 2 us l/SGDWD 1/, //l 6 )35w 2 (X—HJ/tr 0'1 “45.1w '_ Purim - "50 ‘ '3 Til ANSWER: ’1777’ b) How many or the 1600 weights obtained would you expect to be between 120 and 142? A: 3—6 “(WU ,2821 , / 0:406 1140!, ,1‘175 ‘36 .lqu xlwo = 234, m/5i)‘bc=‘8 <3é Q1460 {E s ”.34 L'EO/ 5° ANSWER: 234 c) “That is the probability that the average oftlle sample of 1600 weights is less than 15 X“ A) N (“1073 NCO ll) NCOt Xin “905 [email protected])J Iglyflzz A145“ ,N(lho,E>/Ll) ) Nb“) . ”A ball Afi' 5—,253l %\ fi 1 ,1 l-H’ léllto as 0 1594190 ;i X“ .78 5/4 5 - ,‘3 : Jill? ANS“ ERzi [10] (b) Using the decision rules in part a, what is the probability that one would accept Ho if the population mean was really 3.6? A: Roi Acoepi plol ll: 3)vi Pr It Zobs 7 —l.3Ll\ u: 34,} ‘i’ri >133 >’\,3L\llu~: at} \5 2.91 earl x > 31 «DH is ”MW” Li em 7 l m llkigt’a} U 7- L ‘x‘sfinofio: “(393499) A‘ 5*3 ”(34; L15 )/ \ l \/ W3 1 >_ o / 5 JV / 3 . 3% // //// 4 Hall 349 ‘11? 0 1117 \tqesdo l’L'133 ' r8480 Ts” Amer ,BQSSO d) What is the probability that the first 100 individuals weighed have a mean score above 173.7 >7“Ml\l(u,¢’_z): ’M N NOW/$9 52 /N(:0: 5‘) N(:l) / “(o/l) g .S’A nuns ”a 0 41 A:,5’<‘l’?/8 a 0 9’1 ° 2“ “.0082 VIZ-1L0 l; 5 = 2 Ll ANSWER: 10082 e) Below what number would you expect the average weight of the random sample of 1600 individuals to fall 15% of the time? M 59 _ Xiaoe Ntlho 9.0») flm’g Ell) ) /N(°i‘) .35 {i ice/.04 _ ellso «Lo s X""" X: l6!) 4.04 % Z a X/Er/ Ll g;— ( ) {Ll : [QC-[.3 : 153.7 ANSWER: ’56-7 (24) 3, A random sample of size 16 is taken from a population which is normally distributed yielding a sample mean of 110 and a sample standard deviation of 18.7\1 l L? Y (l— at loo%=%% =>o<= 01 ll;= ”0 Sslg 3) Give a 98% cont ldcncc interval for the population mean + l xl: 15 X n- we vn -°l l 1/ 0 lift" ,0! 2 L01 1% “0 -k 21.02 m \\o t ltio‘l (£19,,th ) MAXIM) ANSWER: I” (WW/WW) o = to b) Assuming that the population standard deviation is also 18, what sample size should be used so that a 96% confidence interval based on this sample would be given by the sample mean plus or minus 2.2? clb% = (lee) loo% '—'7 M10“ E522 Z 0’ _ 72r— ’ E taflifi =22 (36) 2, Four observations were selected from aflwith known population variance equal to 9. This sample yielded a sample moan equal to 3.5. RR 5 3 15 31 = q 5 3L [16] (it) Using a 9% level ufsigrliflcance, test the null hypothesis that the unknown population mean is greater than or equal to 3.7 versus the alternative that the unknown population mean is less than 3.7 by answering the following: (i) Set up the null and alternative hypotheses. clearly dcfining any unknown parameters. Note the “a value is always in the null hypothesis up. n aw ,LL: uwltnow“ rw~W 14‘; pl < 3‘7 (ii) Find atest statistic: (l) sensitive to the unknown parameter with known disuibution under the ‘5" value of the unknown parameter. First state and clearly label the distribution fact that you are using and then state and clearly label the test statistic and in distribution under Ho. XM Estimatc ofunlmoWn parameter in i : ‘=X Distribution rm: 1 MNNU’l‘) [4 >l tr/ir ”it any ‘ 3NT? " LS Tcst Statistic: iohs— N NCO 1) under Ho tiii) Look at the alternative hypothesis and decide what values oflhc estimate of the parameter in (i) would lead you to Reject Ho (general direction) Road; AB Ll Kit“ 3.7 Based on that answer, state what values orthe test statistics would lead you to Reject Ho (general direction). Then draw and label a number line refluxing this answer. Regal l—‘a \Ji gas <4 0 ‘2‘??ch M Q (iv)Using these results along with the level of significance find the cutoff point(s) tor the acceptance/rejection regions ntqn .- Alla») 5 in .0‘4 a ' O Zeb: n amt: l. 35 (vlzfilearly state your decision rules (Are they consistent with the number line In step 37—) Yes Accevvl he n" inns > 13H Ar A, skew we. (yi)Calculate your test statistic indicate whether you will accept Ho or reject 11o. Interpret your results. as -3.'l ’ — .2 ,0 l’bB Zo‘as = fi— L5 A canny lr‘u . \ _ mm l“ “mind we conclusive Using a RDA, Land of Swami unwell i- it? / V 2 3‘7 The emblem continues on the nwfl pagea Zevilv ”L3” n 1. 15 s s 10 _ . _ _ _ *(15) 4. A random sample ofsize ls is obtained from a normal population yielding a sample standard deviation of20. rest the null hypothesis that the unknown population variance is greater than or equal to 152 versus the alternative hypothesis that the unknown population variance is loss than 162 using a 5% level of significance. (i) Set up the null and alternative hypotheses) clearly defining any unknown purumemrs. Note the 1:" value is always in the null hypothesis. Ho 0’2 Z ”’7' H ‘1 61 < 1L2 (ii) Find a test statistic: (l) sensitive to the unknown parameter with known distribution under the “=" value of the unknown parameter. First state and clearly label the distribution fact that you are using and then Sufle and clearly label the test statistic and its distribution under Ho. 6‘1: unknown populoeltéy‘ variant: A - z . . . h—l 7» L Estimate of unknown parameter in (i): 0} r 5 QLSLILbfiggllEg_LI (/L: MX“_( {3‘ Z \‘-l S7“ 76' Test Statistic: Krebs : “)1 N 35 under Ho (iii) Look at the alternative hypothesis and decide what values of the estimate of the parameter in (i) vtnuld lead you to Reject Ho (general direction) Roper l—\n if 514 le Based on that answer. state what values of the test statistics would lead you to chcct Ho (general direction). Then draw and label a number line refle ting this answer. Ref} u“ l Xihfiflq Jilin,— H 'Xisb; (iv)Using these results along with thc level ofsignificanoe, find the cutoff poinl(s) for the acceptance/rejection regions /x\.+ // "’“H/ ,\e as: “15’ ///// ’Xielis 6 chl= In?“ L35) (v) (Eject y lxstaie your decision rules (Are they consistent uilh the number line in step a. Aeeepl up l5} ’Xens 7(9357l R in allwm Cm. (vi)Calcu ute your test statistic, indicate whether you will accept H0 or reject Ho. Interprct your results. X2 s \iajf; 3457 Neville C55 \b2 05mg 0 5% Laval ole Santana, l» we echelon» that 5 ZWZ inilsA lnuipo'llmflb Joel time, so“) (35) 1. A random sample of 1600 weights is obtained. It is know that the population ofwcights is normally distributed with mean 170 and vafiance 2500. Answer the following questions: d) What is the probability that the first 100 individuals weighed have a mean score above 182? a) What proportion ofthe population has a Weight between 160 and 185‘? A = “C 7‘; Xe N N ( hiya) - \\ utrto 502) N I) ’t ”(to r) . H t) A "' no 21 : 57' A\% I A\// (9 7/ Bdi‘l‘l T (2,0743 %% x\ooNN( JlDO ) A 5’“:ng / = / ‘l' ; ,t = I . 150 my rs‘S /// i t {4070151) N65“ N90) : ‘0082 12 o .3 u .3 on, // S’Agllfifl Mn \w,-£=—.z A A l ‘ 0‘ So 5° 1 l // / 155410 : L5: .3 no lag , 0 2H 0 2M So 50 ANSWER: - 1972 Z = 1-H 0' b) How many of the 1600 weights obtained would you expect to be between 135 and 152? 'L) or 1J A: 9"; \82—\1o_ ‘2’?” N(‘7° 5° °) J46 I) rlfit) E: l ‘ E" ' A t . B .1530 5 DO \ A = 4 ,s-ij" , c, .tllcl- £553 ANSWER: ' 82 z ‘ i t i/ 3 A l 7“ . [as tflno 4“) 7 36 e) Below what number would you expect the average weight of the random sample of 1600 individuals to 3? <7“ ~ ° ' 0' fall 28% ofthe time? res-no _‘ , ='.1° , . “52' - so xt~N<mnt 7° \8 '3‘ .1174x16007187.84 5 1 1 gm 1 2.: “ - 6 ‘/ 56 F)? "’ N070) ma? 6/“) ) 5" 187 84 “’°° ~ . ~ 1 “Wk— _ N < note/4) ) New Nan) W) e) What is the probability that the average of the sample of 1600 weights is less than 171‘? l f ‘ - 22 ’28 1,8 7.8 / .5 .28—. xanm’fi") ' '_ , \ ' 2 —‘ M( 5‘31 5 1 5 l ’ \ o Xingu“ ‘10)]:00‘Q) ) X V70 2 0 ’Z O ‘2 =58 2- ‘ -.58 , (HMS/tr) Nan) - -t'zo A / H mm») . _ _ —.58 4L”, g A / w 2— L517; at. Z 1&3 \11 ’/ = ’10 . 5/4 tr ‘10 o 5, u .3 X \ '58 ( ) ANSWER: 169-27 -\'lo \ : no , .73 =169.27 e ‘1‘ = — = $5 test 514 slut ANSWER: - nelb (24) 3. A random sample of size 16 is taken From a population which is normally distributed yielding a sample mean of la} and a sample standard deviation of 19. [10] 03) Using the decision rules in part a, what is the probability that one would accept 110 if the population Xle .= l'LO s = l ‘t mean was really 46? a) Give a 99% confidence interval for the population mean “7.0l A: M we at New — t 5 *e X :l'. “'W’z ’- ,.o\, 06; vi ‘1'»; ‘ 1/ 1‘ ' , Pr Kass? Allsltxs‘tbl o is t 2.41m 4 zqm ‘q ‘ , M i “*3 was \V’W '20 ’ ' W a! W - Ll 433 no i' ‘3‘“ M” v w1—\~“%L%)\W ‘ s X ' no a \‘4 4%} M6 3 5‘1 790% \l‘ ’ ‘ (106,134) ANSWER: (106,134) 7 .. 1 6' = \‘i X A) N (H .0; ) 1)) Assuming that the population standard deviation is also 19, what sample size should be used so that a “ I h 94% confidence interval based on this sample would be given by the sample mean plus or minus 2.4‘.7 (\ 49 \oo% = qt‘z 35“,» 21/ a“ in E E: 21-} 01-: . 06> 2“ N(°I\) Z 0" - FF» 5‘93 ”5'; Va " E :H’l (\.%)(ta) , 2 Ll 2%,:2‘03 71$$ W W =- W s \‘ifég 2H \'\ = 22‘}; 3 \ .5. 27,2, ANSWER: “ 27’ N (26) 2. Four observations were selected from a normal population with known population variance equal to 9. - n . . .. 1_, This sample yielded 153?: metui equal to 4.5. Xqflhg c ,9 [161 (a) Using a 7% level ofsignificanee, test the null hypothesis that the unknown population mean is greater than or equal to 4.7 versus the altemtttjve that the unknown population mean is less than 4.7 by answering the following: (i) Set up the null and alternative hypotheses. clearly defining any unknovm parameters. Note the “:" value is always in the null hypothesis. P" unknown Myra.» mm Hot N 2 DJ H.'- H < “7 (ii) Find a test statistic: (l) sensitive to the unknown parameter with known distribution under the ‘5” value of the unknown parameter. First state and clearly label the distribution fact that you are using and than state and clearly label the test statistic and its distribution under Ho. A — ’— Estimate ofunknown parameter in a); p = Xu Qisuibution Fact: D"— l M N NQI) ‘Wa “ — .7 " ~ 4.7 Test Statistic: lulu: X—i VX—Bz ~ WON under Ho 3Nfi /1 (iii ) Look at the alternative hypothesis and decide what Values of the estimate ufthe parameter in (i) would lead you to Reject Ho (general direction) 'Reaect us it? Yd Ll»? Based on that answer, state what values of the test statistics would lead you to Reject Ho (general direction). Then draw and label a number line reflecting this answer. 125ml Ho LY Zeb: <4 0 fig, 0 Zak; (iv)l.' sin g fliese results along with the level of significance, find the cutoff point(s) for the acceptance/rejection regions N (q 1 Neal) 52.07243 M L 3:3 o =t.ue Etrwg 10¢ o - Err-t v (v) Clearly state your decision rules (Are they consistent with the number line in step 3? 9g Accept l-lr. 1? a“; > —1H$ Vega} 4% other o-v 5E tvi)Cnlculatc your test statistic, indicate whether you w1ll accept Ho or reject llu. Interpret your results. gbs = its—lift g “ma? Acceptor.) 3/ uses a 1:5 trust at 33mg“? "*“M “3 m Caner-Au it“; 4?! Qéhesis-iesi‘ The emblem continues on the max! gag m) (15) 4. A random sample of size 17 is obmined from a normal population yielding a sample standard deviation of2]. Test the null hypothesis that the unknovm population variance is greater than or equal to 162 versus the alternative hypothesis that the unknown population variance is less than 162 using a 1% level of significance. at: 0.01 5:21 (i) Set up the null and alternative hypotheses, clearly defining any unknown parameters. Now the ":"’ value is always in the null hypothesis. 6:: unknow“ ”Whig,“ Via-manta Hg) 2 tr 3 \(el H t: a? s \61 (ii) Find a test statistic: (l) sensitive to the unknown parameter with known distribution under the '="’ value of the unknown parameter. First state and clearly label the distribution fact that you are using and then state and clearly label the test statistic and its distribution under l-ln. a. A n-\ 31 7. Estimate of unknown parameter in (i): (5" = 5 Distribution Fact: D? 3‘ ( ) N K rvl 51 1 (rt—l) 51, \bs" 7. Test Statistic: (Lobs : t 1 f1 ~ 25 it under Ho (iii) Look at the alternative hypothesis and decide what values of the estimate of the parameter in (i) would lead you to Reject Ho (general direction) Radiant Ho 3-; 51.“ \67— Based on that answer, state what values of the test statistics would lead you to Reject Ho (general direction). Then draw and label a number line reflecting this answer. our}; Hut 700““ it, “the“ -—-H-N——+—’— z o \6 ’Xobs (iv)Using these results along with the level of significance, mm the cutoff point(s) for the acceptance/rd ecliun regions 2. 1. X it. 0‘ -1“, “flute R°t z 2 ’X suds .. at «a» o item (v) Clearly state your decision rules (Are they consistent with the number line in step 3? its ) Aucp‘l' Ha 1‘: 'Xzohs ‘I S .6")— Rfi'zfir hr. G'l'lmyrwise (Vocalcu'file your test statistic, indicate whether you will accept H0 or reject Ho. Interpret your results. 1 X: , it, (21) = 48.67: Acwt in, z ‘ ° 5 “Oz Usm o. \% Larel ci} Sham means: new Jr'si \— tuned M390 07' 2 \taZ mg CbML\\) $91...
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