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solutions2011 - Stat 665 (Spring 2011) Kaizar Midterm Name...

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Unformatted text preview: Stat 665 (Spring 2011) Kaizar Midterm Name Sow/770M! There are three problems in. this test. To receive credit or partial credit, please be sure to ShOW all your work. The exam is closed book / closed notes. You may bring a single 8.5”x11” page of formulae to this exam. A calculator is allowed a personal digital assistant or cellphone is not. The exam will last for 68 minutes. Good luck! Question 1 (21 points total) Based on the Analysis Part I, answer the following. (a) (6 points) \Vhat do you think the most reasonable sampling distribution for the number of orders using loyalty cards that Bob completed? Calculate the maximum likelihood estirnate(s) of the pararneter(s) for this san'ipling distribution. Lil” “‘1: fl 0PM OSlS carols Egéa Cmplblfae flu’V 1>O;SSM 944"» WWW (b) (2 points) Estimate the proportion of orders completed during the one—hour period that were corrrpleted by Bob. A 0: Tr: 217:; = 0.9:}? (c) 5 points) A statistics student who works at the store correctly calculated the following 95% Wald confidence interval for the proportion of orders completed by Bob: (0.5055862, 0.8515567). Explain using language that the store owner could understand Why this interval is not as good as a score—type or exact-type interval for showing that the two applicants have dillerent check—out 51:)eeds. Wald beltrvwcfltI Govern? f9 usual smalllr “ll/um “Hal. Mimi/Mi 7M- 5) lenM wstmvt A “'90 (112% W” 4w 4: ml' at my all at camera 1‘4ka will val’an an w W or: made. WI m'lk WM ;A~LWJS, umtej law “hum k 0187,; 41* "AWS‘ (Lax/ole?“ vl’lu. ‘l‘lme ‘Tlm‘s MUM-I vl’lm'l: m afar-JAM ltd-trng is +00 mam»). :1. Question incontinued (d) (8 points) Estimate the odds ratio 1:>etween applicant. and loyalty card use. Find a 95% confidence interval for this odds ratio, and interpret it in the context of this problem. 43- M“ = 94—73—— MAB/37' ‘ nil—“u ‘ l3"? ’ gm 4L“ R N‘LPO-(Z, a». mac? qu gm 33'- [Doll/.0517 / an m, MIL-M 4m 41,; mm am am lags, mm 0.01! ad 1.19:1. 724414 ,3, /am 752, 'wA-M 4% #u was afi arm/2'77 a $747!? Cd/A 55067. {S 0-01 7"» [05' 19mg: {mwr 74f 8&5 flaw” 4W xvi/tea. Question 2 (37 points total) Answer the following questions based on Model A of Part II of the the grocery store analysis: (a) (6 points) Write down the generalized linear model that the owner fit, including the random component, systematic component and link function. we = ‘3 Wig/1:1 CMSW - Y3” Poisum ' r av‘im ’ koflgufir—i o(+ @X; I )(L‘= it? 09% reglslts at 05W L (b) (3 points) Report the estiniiated prediction equation. : 2.8", *0.§7‘XL‘ (C) (6 points) Calculate a 95% confidence interval for the openregs coefficient, and interpret it in the context of the problem. A )5: «43.5? §%(IEB= 0-073 7370 Cf: fit (.96 66‘ng (440537] enf/Bé?) I 4M 79% “£102? 11% % expevlwfl dam-ease In. Io? a numbor all 001W ‘fi‘rbr ewe/{S @zc-(m affilslv enamel is how 0.6? and g‘qq' 4 Question 2, continued ((1) (10 points) Is the number of staffed" registers associated with the number of customers waiting? Report the mantle, null hypothesis, value of the test statistic, the distribution of your test statistic, and conclusions of your test. WAA'PQG itifiro flyfi%0 Tcst sin’ifi'S‘ix'c e ~I331— = Z 1': U0 4‘5 “Hue, Z~ N‘[OII\ “Flynn/naive? “F'Wlflue. A 2,4049 LET = tt’ ‘0 “fl’éio T— w “i u w «r; T” $ \ \ ‘2 L: n ~ E “T (\u Coming»: flexed i-ib. m Mamba- A: mrkna CM‘S‘IfiN‘thS SioymchAn‘ib dilemmas th ewfl mcrram ,'4 of,“ rtflo'flcrx. (6 points) Estimate the number of customers you would expect to be waiting on a. . randomly s<'31(‘,(:te<i day if 4 of the registers are stafied. Show your work. ,(20 (A); 2.3% — 0.5100 J/M = 0.37? SE fif— eo $13“; '2}! Question 2, continued (f) (6 points.) Do you think this model fits well? Briefly explain (1—7 sentences). If you can not the fit of the. model with the provided output, explain What you would do if you had access to the. data. and a computer. 72 euwmk m 41+ cfiphlwue, ( CMsiob-r' (D overcasmm @ Residuafls @ Lew/mag . @% VH4. reside lek, l sa— 4qu m. a? m “Slim/Q: Art My ’fLmn. 4900+ 2’ but name. m “mm 3 aer "hm ML no «MusMQ. oLWWJA‘ms 60 94W:- 714 Q’QQLOWL docs pul— skmo ans Wa‘b'm W @ Ll‘hwlfll “Le Leng flak (“Chablis 2(5-W/ (Zack‘s dirlunce, 412%, W! emu-ls) 060 “0+ slaw ma glad!» closervmfims ol' Coma/n, ‘53»- pWMA m. “IL Cara/fend» ra‘l‘n‘o 4S0 putt «of Show dtj ole-fmf “Far Pym/IA l. 0 WC Low/i W +0 6J9, flnL W Var/ma rs. Samp/é Sampla mans W as egb'm‘lrd 74m ogwnfi‘m: W by 415 bl: 0W prank/S. ' 6 Question 3 (12 points total) Answer the follcwing questions based on all analyses in Part II of the the grocery store analysis: (a) (8 points) Conduct a Likelihood Ratio Test comparing models A and D. Be sure to state your null hypothesis, value of the test statistic, the distribution of your test statistic; and conclusions of your test in the context of this problem. Show your work. E95. barium.“— A= Ea. Damn“ “D —. maze LKT = /3/.39— @539 1 23.0; W Myra : 0 HA: a w w it Wag M 5% w. 2. [p W, LET ~ 7L2. . . '2 , gm 23.0; >> 9 = mm mm. a @71; ML 0’ f rrjecl- ’11" “we? h7po7l7’4835 My? cur/uer ‘7’Z\a‘i‘ WM A 1‘: nor bold“. 41W mm, D. (b) points) Consider all 11 models. VVhichmodel would you choose to use? Briefly explain. 9402 “1‘24, wwMS an: nol'fllrmsw, / Cam, 40%“ ax LfiT' '17» 6Mde “flaw. luv/tape, / can VS! a ,4ch = 4/592 4165 a 53%;! 4146 = 5979476 firch = sat/m smiles} 141C is best, So I cLoosz malice D- flflg Grocery Store Data and Analysis A small grocery store owner is attempting to improve his customer service. He conducts two analyses. Part I: Check—Out Speeds The owner plans to hire one new person to check out groceries. He has narrowed his choices down to two applicants ~ Bob and Mikey. One of his consider— ations is the speed of the applicant in conmleting orders. On a very busy day (where there is always someone in line), he asks both Bob and Mikey to check out customers in the two express lanes (15 items or less) for one hour. He counts the number of orders each applicant completes. He also notes Whether the applicant convinced the customer to use a store loyalty card when checking out. The results of this experiment are reported in the table below: Loyalty Card Yes iio l Total Bob 6 13 19 Mikey 7 2 I 9 ’ “otal 13 15 I 28 Here is some output from analyses conducted in R: Pearson’s Chi-squared test X~squared = 5.2406, df = 1, p—value = 0.02207 Log likelihood ratio (G-test) test of independence without correction Log likelihood ratio statistic (G) = 5.4397, X—squared df = 1, p-value = 0.01968 Fisher’s Exact Test for Count Data p—value = 0.04182 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 0.01121498 1.05120975 Part II: Customer Waiting Times A small grocery store owner is studying the flow of customers through the store’s check-out lines. The store has 5 cash registers, but does not always have enough staff to use them all, To study the flow, the store owner randomly selects 100 times and days to measure the number of customers in line at that time (variable ‘inline’). He also records the number of staffed cash registers (variable ‘openregs’), whether or not he measured on a weekday (l\/I(:)nday—Friday) or weekend (Saturday—Sunday) (variable‘weekday‘), and whether or not. the store is offering sale prices that day (variable ‘sale’). (Note that since the day and time are randomly selected, it is reasonable to assume that the counts are independent from. one another.) The data is plotted here: Number of Customers 12345 No Yes Number of Staffed Registers Weekday Sale The owner tries three models to describe the number of customers waiting at his store. These are detailed on the next few pages. ii Model A: Call: glm(formu1a = inline ‘ openregs, family = poisson, data = grocery) Deviance Residuals: Min 10 Median SQ Max —2.51890 —1.18828 0.01005 0.71525 2.50611 Coefficients: Estimate Std. Error 2 value Pr(>lzl) (Intercept) 2.84136 0.09741 29.17 <2e-16 *** openregs —O.57029 0.04266 -13.37 <2e-16 *** Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 (Dispersion parameter for poisson family taken to be 1) Null deviance: 347.64 on 99 degrees of freedom Residual deviance: 131.39 on 98 degrees of freedom AIC: 413.72 Number of Fisher Scoring iterations: 5 Residuals vs Fitted U. Normal Q—Q (‘0 'a 2 9 N g r .‘9 cu o w ,_ .S g ' w '0 N to 5' l | 4—1 03 0.0 1.0 2.0 Predicted values Theoretical Quantiles ; Scale—Location U. Residuals vs Leverage '0‘; ._ 9 i 8 O, c N :: .,.. o .9 E > (3 O % 0‘.) . _ N :9 O. 1:: | Q 0 c7) 0.0 1.0 2.0 Predicted values iii 0.02 Leverage Cook's distance g|m1covratio 0.10 0.00 Cook’s distance 0 20 4O 60 80 Obs. number covariance ratio 0 20 4O 60 80 Index g|m1dfbeta[, 1] glm1diff 0.02 —0.04 0.4 —0.4 0.0 dfbeta (Intercept) O 20 40 60 80 Index dffits 0 20 4O 60 80 Index iv g|m1dfbeta[, 2] 0.010 —0.010 dfbeta openregs O o o 00 a? $8 W 000 °Q> fiwfiw 0Q9°° ®0°c9° <5 0 20 4O 60 80 Index Model B: Call: glm(formula = inline “ weekday, family = poisson, data = grocery) Deviance Residuals: Min IQ Median 30 Max -3.8448 -1.5230 —0.2090 0.8239 3.2901 Coefficients: Estimate Std. Error 2 value Pr(>|z|) (Intercept) 2.00030 0.07670 26.081 < 26—16 *** weekday -0.78343 0.09863 ~7.943 1.988-15 *** Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 (Dispersion parameter for poisson family taken to be 1) Null deviance: 847.64 on 99 degrees of freedom Residual deviance: 289.17 on 98 degrees of freedom AIC: 571.51 Number of Fisher Scoring iterations: 5 Residuais vs Fitted '0' v -- <1- (D m 8 7o 93 a N g a: N :3 O 79 Q g o m .— m 0 5 c: o 73 T 590 13' v a-n I 09 1.2 1.4 1.6 1.8 2.0 Predicted values Theoretical Quantiies Scale—Location 3 . '— 79 i’.» 5 v § 5 w cu 9 'S m (D a) 13 0. cu 2' rs ' Q as i .2 1.4 1.6 i .8 2.0 0.00 0.02 0.04 Predicted values Leverage \f Cook’s distance glm2covratio V. o 0.2 0.0 1.02 0.96 Cook’s distance 0 20 4O 60 80 Obs. number covariance ratio O 20 40 60 80 Index glm2dfbeta[, 1] glm2diff dfbeta (Intercept) RT 4 N o g Q 0. .o O ._ o 1: m g a O _ . . o C: °’ ' 0 20 4O 6O 80 Index O 20 4O 60 80 index Vi dfbeta weekday 0 20 40 60 80 Index Model C: Call: glm(formula - inline “ sale, family = poisson, data = grocery) Deviance Residuals: Min 10 Median SQ Max ~3.2591 ~1.5293 -0.2162 0.8160 3.2672 Coefficients: Estimate Std. Error 2 value Pr(>|zl) (Intercept) 1.22090 0.05998 20.357 <2e-16 *** sale 0.91261 0.10088 9.047 <2e-16 *** Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 (Dispersion parameter for poisson family taken to be 1) Null deviance: 347.64 on 99 degrees of freedom Residual deviance: 274.64 on 98 degrees of freedom AIC: 556.98 Number of Fisher Scoring iterations: 5 Residuals vs Fitted '0 V as 0 g ‘1‘ .3 N g a, N = S g o o g o 0 o a n: 0 g .3 <r '0 ‘3’ ' 5 1.2 1.6 2.0 Predicted values - Theoretical Quantiles E Scale—Location U. Residuals vs Leverage g . A .3 <r \ <1) 0 C N a 2 'S m d) d) '0 CL N 2‘ Q a ' (.0. 0 a 1.2 1.6 2.0 0.00 0.02 0.04 Predicted values Leverage vii Cook’s distance gim3covratio 0.2 0.0 1.02 1.08 0.96 Cook’s distance 45 0 20 4O 60 80 Obs. number covariance ratio 0 20 4O 60 80 index gldefbetaL 1] g|m3diff 0.01 ~0.01 0.4 —0.4 0.0 dfbeta (Intercept) 0 20 40 60 80 Index dffits 0 20 4O 60 80 Index viii glm3dfbeta[, 2] —0.06 0.00 0.06 dfbeta sale O 20 40 60 80 Index Model DZ Call; glm(formula = inline " openregs + weekday + sale, family = poisson, data = grocery) Deviance Residuals: Min 10 Median SQ Max —2.38377 —1.15428 0.07962 0.67986 2.15945 Coefficients: Estimate Std. Error 2 value Pr(>lzl) (Intercept) 3.20936 0.20668 15.528 < 2e—16 *** openregs -0.54128 0.04642 -11.661 < 29—16 *** weekday -0.58363 0.16556 -8.525 0.000423 *** sale -0.12973 0.17940 —0.723 0.469594 Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 (Dispersion parameter for poisson family taken to be 1) Null deviance: 347.64 on 99 degrees of freedom Residual deviance: 108.36 on 96 degrees of freedom AIC: 394.69 Number of Fisher Scoring iterations: 5 Residuals vs Fitted 6 Normal Q—Q N g N 2 23 (U 5 o g o U) '5 ‘1’ m 0: N 'o 33‘ OD —2 0 1 2 Predicted values Theoretical Quantiles 5- Scale—Locatlon U. Residuals vs Leverage L 9—) N ‘D o c F' 8 .S a o (D (D 13 CL 2‘ c2 :9 s! Q o a) 0.0 1.0 2.0 0.00 0.10 0.20 Predicted values Leverage ix Cook’s distance glm4dfbeta[, 3] glm4diff 0.10 0.20 0.00 —0.05 0.05 —0.6 0.0 0.6 Cook’s distance 35 51 0 20 40 60 80 Obs. number dfbeta weekday 0 20 40 60 80 Index O 20 40 60 80 Index gIm4dfbeta[, 1] gIm4dfbeta[, 4] —0.10 0.00 0.10 —0.05 0.05 dfbeta (Intercept) 0 20 4O 60 80 dfbeta sale 0 20 40 60 80 Index Index dfbeta openregs a? w” m 33 8. “.5 O E o a '5 C? o 20 4o 60 80 Index covariance ratio .9 o n: O *5 o o 00 00300 8 0 O©%OCDe E [email protected]%o Q) a 00 00 o O 0 20 4O 60 80 Index ...
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This note was uploaded on 07/26/2011 for the course STA 665 taught by Professor Staff during the Spring '10 term at Ohio State.

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solutions2011 - Stat 665 (Spring 2011) Kaizar Midterm Name...

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