IE4521_midterm2

IE4521_midterm2 - Problem I (14 points) IE 4521 Fall...

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Unformatted text preview: Problem I (14 points) IE 4521 Fall Semester 2009 — 2"“ Mid-Term Name In a random sample of 100 batteries, the average lifetime was 150 hours and the standard- deviation was 25 hours. " (4 points) (4 points) (4 points) (2 points) (:9 "‘ {QEJMSI a) Find a 95% confidence interval for the mean life time of the batteries. b) An engineer claims that the mean life time is between 147 and 150 hours. With What level of confidence can this statement be made? fie c) Approximately how many batteries must be sampled so that a 95% confidence interval will specify the mean within 1 2 hours? cl) State the test statistic you are using while solving parts a. b, and 0 above and also what assumptions, if any, you are making. W i : m I “3’03; 3:13- ) 01:001— ._ 1c S“ fetvkjmf 3 if??? “a e (“H—.95; E) Y—Joi s/m A W M‘T‘JM C/fc) 2 : PHI? < w «z r Ff?) 1' i?< "MSW 56w <2 em) :F('T>f/;mfl42 4% m raft-dam saw is BMW: C, . @314ch I 'fl-e Mfume?! Samplr 3329 (“cm be (gram cad 150 b6 about _ QXMK'L /”‘“ l- *5) =62: .0 fi‘UKLfi-l L713}th- nufan/fj. $429 iffr/rj’é/w are all fight) CD EMU“), ttr'wljelmev CS‘liiQ (TF3 unknl’Wi/t «item, M Mfr-2 43-ing 9m dents Meiiigirzeprgerwf we pry/1r rm BOCULGQ as n Inge, t-dfsfifbubc‘n =>n0hw{ Purl-(la) 2 —- {Fi’l’tQX‘l/Cll gamer-51 name? oil‘sarfbufié paw-(c) T-— lfimefl’fi' Problem 2 (13 points) Fit! in the blanks in 1 & 2. (1 point) . 1. The value measures the plausibility of H0. (Based on the given data). (1 point) 2. The smaller the P-value the stronger is the evidence against H0" . 3. A certain type of automobile engine emits a mean of 100 mg of oxides of nitrogen (NOX) per second at 100 horsepower. A modification to the engine design has been proposed to reduce NOx emissions. The new design will be put into production if it can be demonstrated that _theimean __ernission rate is less than 100 mg/s. L Hal“: .1 diff! L" hit: J4” 55;! L LU! ié “1' A sampie of 50 modified engines are built and tested. The sample mean NOx emission is 92 mg/s and a standard deviation is 21 mgls. The manufacturers are concerned that the modified engines might not reduce emissions at all. ____.-—-- in $315475“ NJ“: ; ,a (loo; HAflafi’ (2 points) a) Formulate the problem as a testing of hypotheses (Define Ho an'd HA) ” 3 .19 tie in?» regime [‘5 "’0, nab S9" Ha“ M3!“ . Hi; : fl d“? We” ,wéci‘ mm; Ho (2 points) b) Give the test statistic you would use. (g Hawaiian , (Last. 990% evident of H. (2 points) 4. Draw the sampling distribution of E, when H0 is true. -_ {93° "' Ha {aging ) Hflfiqq Y (a sample size of 50 is large enough to assume normality of X ) {WC ,mm { weaker W 9'? oint' F ‘ on HUI»? evrtfence q 8 3 P 1C y Pavel/He g g a”? d E A . - . ' ‘7‘“ - m IQ‘MQIW) ;_,{,(p [30 - 811'? fix]; (9 rim-fl (Ht/Q Oman/er (1 point) 5. Show the observed vaiue of 92 on the graph .in 4. (2 points) 6. Shade the region which represents the P value in 2. above. (2 points) 7. If in your decision making a = 0.01, what inference would you draw in the above case. I Problem 3 (8 points) (2 points) (4 points) (2 points) a) X is a normal distribution with mean u and variance oz. Both parameters are unknown. A small random sample of size n, with values X1, X2, , Xn is taken from the above population. it and s are computed from the data. Then the statistic X135 followsa afar/)5szin- s n _ Write the sampling distribution. b) Let X be the octane ratings for a particular type of gasoline in (a) above. The results (in %) for a sample of size 5 are 87.0, 86.0, 86.5, 88.0, 85.3 Find a 99% confidence interval for the mean octane rating for this type of gasoline. 0) Given below are 4 curves (density functions) representing tm, t4, t1, and Z. Labeieachcurve. 4'5) {1 87i86+%~§t 9'5Jr3’5~5 ' g— < Sara 32'; 5231(me :l 0mg! 1 MED/l «(Moi Wit?!) 33”. 5:3) Problem 4 (10 points) Fifi in the blanks with the words provided out of the list given below. I- ~'\ \ "I. ‘ 'H—E’uiir". I «(if ‘ «I ‘i (1 Point)- 1. A $1; 5126 Ha random vanale that depends only on the observed values of a random sample. ' (1 Point) 2. A Ffimlfiiflconsists of the totality of the observations with which we are conce ned. . (1 Point} 3. The probability distribution of a statistic is called a fiancfi'bt‘ru? distribution. (1 Point) 4. The standard deviation of the sampling distribution of a statistic is called them gaggng of the statistic. (2 points) 5. - Statistical inference may be divided into two major areas: Qm‘mgfifm - and :t m‘ 43 MW 93;}: (2 points) 6. in estimation if e is the unknown parameter of the population. and 6 is its estimator. then unbiasedness of 8 is a very desirable property oi a good point estimator. Write the missing left hand side in the following equation which makes 9 unbiased. ,- 1R = 3 (1 point) '7. If we consider all possible unbiased estimators of so ‘parameter s, the estimator with the smallest variance is called the most (2%?(5MT estimator of e. (1 point) ‘ 8. Both mean (denoted by 11' ) and median (denoted by 3?) are unbiased estimators of the population mean p, which of the two estimators is more efficient1 and why? lea it List of words D: ’ 1+5 ’V‘JlHHEerCQ, emit/lid 'bQ % WWW/l” (mean, E(6' )lefficient. population, statisticsI standard error, sampling.'estimation, testing of hypothesis) Problem 5 (12 points) The mean sitting height of adult males may be assumed to be normaliy distributed, with mean 36" and standard deviation 1.3”. For a sample size of n = 100 men, determine .the. foliowing probabilities. (i). P (S s 1.00”) (ii) P (S .>. 1.4") (iii) P (1.25” s S s 1.35") (HINT? *There is a weii known distribution connected with 32 * Also use Central Limit Theorem {2310- 9% 6" a}; glw )w 36 L F17 a" "" [ {J g E, i) 00”) wal we M56 Cefif‘i’flé? a 9— . I + [7(5 5 Lab) , "5'7 (2)?)7}OXFM flT-e 3(9):“ 67%6 L ( l x (1mm!) ‘9 I V” i~3L A 1/ degrees GAE—$306!“? mim‘bms :— 36; <“ :KXfffijaiJ) 0%“ 0"" {56 Rppm (r‘mafeiy by 2 rim a MV’ M “WW” ‘ KO) 2% : 4>(-T_“3*53’?r 9’ @ W51???) _ )7 <1 ) :- ¢§~2_~37}J7 ._ VH1 __ _ l:‘ r. I" I ‘J _F .1' {AUDZTH ! Ft 5L «L/noir .i-;';..’: S‘ 2" \E-“i la \{i‘i/ ifii’lei-U: ’- [‘([U P( SQCLEJLQ‘J //K _' I. i TEN/vb?” Six-9&9- “ l . ,' f . . :- P(')(Wz4 "‘ p()(oql7t2§:.z.flig Problem 6 (8 points) A good estimator is unbiased and has minimum variable. Suppose X1, X2, X3 denote a random sample from an exponential distribution with density function ie—J‘m, x > 0 6‘ for) = elsewhere 0 Consider the following four estimators of 6' ' él=XI 6A1 : X1+X2 " 2 53 = X1+X2 3 634:} (4 points) a) Which of the above estimators are unbiased estimators of 9. (4 points) b) Among the unbiased estimators of 9., which has the smallest variance. 00. Bahama 9 WWW: a} :1) bum/aimed ix ; 2 3 to) 3 MR Moved _. — K1 - K may: mg") + HE“ “Qt; 59‘ :9 13361594 $ “ 3 Jr}— 7‘9’ 5- firDi/mbi‘ased- ...
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IE4521_midterm2 - Problem I (14 points) IE 4521 Fall...

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