08Midterm_II_Solutions

08Midterm_II_Solutions - University of Waterloo Stat 373...

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Unformatted text preview: University of Waterloo Stat 373 Midterm II Fall 2008 Date: November 20, 2008 Duration: 90 minutes Family Name: First Name: ID. #: Si gnalure: Instructor: P. B alka Instructions: /5(, v// 71 0 This exam has 7 pages includin0 " er page. Tlfe marks for each question are indicated (total of 34). Show your work. Your grade will be influenced by how clearly you express your ideas, and l10W well you organize your solutions. Stat 373 Midterm II — Fall 2008 l) A shipping company offers a promise of on-time delivery to its customers. Orders are normally delivered by truck. which is the least expensive form of shipping for the company. However. air freight is used, at considerable expense to the company, for some orders that would not arrive on time if delivered by truck. Use of air freight is called a premium shipment. Monthly shipping data for total number of monthly shipments]number of premium shipments, and percentage 01‘ premium shipments from Jan. 2001 to Jul. 2003 (31 months) is shown below. Time series plots and correlograms for the total shipments and for the percentage of premium shipments is also given. month tota1.shipments premium.shipments percentage 1 Jan—01 5858 619 10.6 2 Feb—01 4084 451 11.0 3 Mar—01 5196 480 9.2 29 May—O3 3768 445 11.8 30 Jun—O3 4050 535 13.2 31 Jul—03 5546 654 11.8 Series total.shipments o 0 §‘— ° m {\ O ‘9 o C — o o o 1' g- o o O 0 E5 0 ‘g g _ o 0 °\ o < a 0' / / O 26—4 _ o O / O 0000 o O o o o o 0 o g — 0 V l I T- I l— ‘I Q 0 5 1O 15 2O 25 30 Index Lag Series percentage percentage Stat 373 Midterm ll — Fall 2008 a) An MA( 1) model is f it to the total shipment time series, yielding the following output: Coefficients: mal intercept —O.l628 4847.1449 s.e. 0.2083 169.0328 i) [3] Is this output consistent with what you observe on tl Explain. yr s W; M "/’/\ {/1 5.4M m (Lily/7w- .z/e {ti/J63 /,l V" 5' ’1 , [gram 9 a r" t6 W” “I” “r “We ' J l -’ A 7/1" F 1” ‘7 V:';( in /e /fl ‘b/fi AA éévffiy'fl- (“A /,\/<‘ fgj/‘l‘flf. l/r/l A /t (A 2/ I‘ [/47 1e total shipment. correlogram? ; I v j ” ' M, r ~ " / ' e x, L/ ‘ r " V r vi“ I~ f/CYC/(lp) a,“ (xiv/fly /’(/_V\ )ije‘)%j / <zt/ e- fi/a 1/. u< I I l 1/: /fly/Wmn:/fih{) t/XaXV/j m/Qjfl %/4»(@M/é”7 /LM~ /”V/(”/;l, “3 r, r I .‘ J. I I If ’// «ii/K", M L. A? I f/ )1 ’/ ‘x’t’grzm. z? m .6»: Ca (xx/Ar {/4 I f; A» ii) [2] Based only on the correlogram, Which method would you use to predict the total shipments for August, 2003? f/A ( L Q/// /éf6’//wt '//,/,';. i 57/ 3 -"-.~ 4 mm //| ’fi/J //7"/ w / Vi t/w-f 4/74. (7 J / “T {7147?sz 7 n 5‘; /\ j M/ «E b) [2] Based on the graphs, why might a AR(1) model be appropriate for predicting the percentage of premium shipments for August, 2003? 1,"; 2 " 7X 1) a ( a//g,« /J 7Z1 <"/41;< J 6 (Ly/IA 4".»“(3’2' "(its / "I fill/W m ’4 i? / m,:"v//? // c) [2] Based on the graphs, why might a AR(1) model not be appropriate for predicting the percentage of premium shipments for August. 2003? / J i f/ 'i' " / /; a ’ ,, 7/5 //9 (-411 41/ £2 1/” 0/? m . , z///4_, A A 1,,“ £f[7,') /) Aa f (4%- l '/m: /fl6/ 5 ¢,.,\V .l‘ ,Cn(/ng/n) fééffiwf /H /z{ ,fl/fléfri Stat 373 Midterm ll — Fall 2008 d) An ARtl) model is fit to the percentage time series, yielding the following output: Coefficients: ari intercept 0.8217 8.5244 s.e. 0.1047 1.3657 [3] Predict the percentage of premium shipments for August, 2003. j}; : 4 fl/ *0 5 gfgyt/ + 0,t;2/7(//?4i537}fl/) //, 52/5 ’fi’r_—flrfir e) [3] A EWMA smoothing method (0t 2 K’) yields a predicted value for the percentage of premium shipments for June, 2003 of 8.96. Use this smooth to predict the percentage of premium shipments for August, 2003. A A //5/ : ‘95400 300] I: «2&3? ; lat/Bra) My” 7: Cigar ’\ '1 - t '\ 4 :3 731 1 7”’*V“’(v"7”” "j Ky)! 7”(/"’<’Ijlz rad/m «Skim? < Wye f) [3] The sum of the squares of the differences between the observed and predicted values from the EWMA smooth in c) is 137. Calculate a 95% prediction interval for the percentage of premium shipments for August 2003. #55 Dy, 3/ 9M /1 : mag/V Varied/v5“ :C“ 6 00 /j 49/, W0?) /-\\ Stat 373 Midterm 11 ~ Fall 2008 g) The linear regression model 2) Yi : ,80 + Amour/1,. + flzmsqi + R. R. ~ G(0.0‘) is fit to the percentage series, where month is the index (1,2,.. variate squared. The (partial) output is given below ind. .,3],) and msq is the month Call: _ 1m(formu1a = percentage ~ month2 + mqu) Coefficients: Estimate Std. Error t value Pr(>lt|) (Intercept) 10.702469 0.774118 13.825 4.92e—14 *** month2 —0.729302 0.111523 -6.539 4.34e—O7 *** msq2 ********* 0.003381 7.654 2.45e—08 *** i) [2] What is the value of ,5’2 ? r : 75>? 9 fax/fl» q fat/n7 q ~.> 7mm AV 3 $2 I: . 035/8737 Afiwt/ ) / /-——-~‘ ii) [3] Predict: the percentage of premium shipments for August, 2003. ’l 4 A \ /i mg) m (W _: /fl,7flalg"t} » 729302 (.31) y»; . (721/571? (M/ [3] A shift supervisor is concerned about the late arrivals of employees. It appears that if late arrival rates are low one day, they tend to be high the next day, and high rates tend to be followed by low rates the following day. She records the proportion of late arrivals each day over a full year. A eorrelogram of the proportion of late an'ivals is consistent with her observations. As well. the sample autocorrelation functions are significant, up to and including lag 1(23. Sketcl correlogram might look like. A 1 what such a l l l i i l or I I .i t l . l l Stat 373 Midterm II — Fall 2008 3) [4] An ARU) model is fit to a time series (y1,yz,....y, ), yielding the following output: Coefficients: arl intercept 0.6881 38.401 s.e. O 1553 2.554 A simple linear regression model of the form K::/%-+/in+lfi Ri~nCK0,0) where x,- = Y” is also fit to the same data. Note that in both models, you are regressing the response variate on the previous value of the response variate, but are using different procedures to obtain parameter estimates. Use the ARtl.) output to obtain estimates of ,5” and ,8, from the linear regression model. fly a 3t.</o/(/«,6Ft/) :2: // 9773 4, fl / Stat 373 Midterm II — Fall 2008 4) [4] Recall that both the AR(1) model and the MA(1) model can be defined by a sequence of independent random variables, AI‘A2,...,AI ...,where A! ~ G(0,0). Let YLYZ....,Yr describe a AR(1) time series, with model parameters (,u,¢). and let X1.X2,...X, ...describe a MA(1) time series, with model parameters (,ufl) . Assume both series go backwards forever. In terms of q) and (9 ‘ under what conditions will Var( X I ) > Var( Y, )? . / ' , f“ "I 2/ ¢ 7,: yf ;. M f A"; 1" ¢ Ate , y] f \ W m v/Mr/ill—rm» J ' i , l _ I f / , 3 3 M74 W rm t WM— 22 f t; 4g 4,» gm : (who ...
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08Midterm_II_Solutions - University of Waterloo Stat 373...

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