2013 STAT3613 Past Paper 1

THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT2313/3613 MARKETING ENGINEERING December 18, 2013 Time: 2:30 p.m.- 4:30 p.m. Only approved calculators as announced by the Examinations Secretmy can be used in this examination. It is candidates' responsibility to ensure that their calculator operates satisfactorily, and candidates must record the name and type of the calculator used on the front page of the examination script. Answer ALL FIVE questions. Marks are shown in square brackets.

S&AS STAT2313/3613 Marketing Engineering 1. A study was conducted to demonstrate the ways cruise lines and their ships can be differentiated. The data set is SHIP Eight characteristics of 224 ships were obtained as follows. X1 Tonnage X2 Age of Ship_ X3 Total numbers of crews X4 Total number of Berths X5 Total number of cabins X6 Overall ship evaluation (1 =not satisfied, ... , 1 O=satisfied) X7 Accommodation (1 =not satisfied, ... , 1 O=satisfied) X8 Entertainment ( 1 =not satisfied, ... , 1 O=satisfied) a) A SAS program and part of the SAS output are given for a statistical analysis. proc distance data=ship method=euclid out=dist; var ratio(xl-xB/std=std); run; proc cluster data=dist method=ward nonorm nosquare; run; Cluster History Between Cluster T Sum of i NCL ----Clusters Joined---- Freq SpRSq RSq Squares e 6 CL16 CL18 46 0,0146 .536 5 '7399 5 CL? CL 13 74 0.0199 .516 7.8042 4 CL9 CL10 60 0.0217 .495 6.5261 3 CL8 CL4 100 0.0609 .434 23.903 2 CL5 CL6 120 0' 1077 ,326 42.305 1 CL3 CL2 220 0,3262 .000 126.13 1. What statistical method IS employed? Describe how the method IS specified by the program. n. Draw a part of a dendrogram IN SCALE based on the output. iii. How many groups of ships do you suggest? State your reason. b) Another SAS program and part of the output are given for a statistical analysis. proc standard data-ship m-0 std=l out=shipl; run; proc fastc1us data=shipl maxc=4; run; Maximum Distance RMS Std from Seed Radius Nearest Cluster Frequency Deviation to Observation Exceeded Cluster 77 0.5046 2.4928 3 2 0 4 3 50 0.4735 2.3084 4 92 0.6274 3.7049 P.2 of 11

S&AS STAT2313/3613 Marketing Engineering Summary Statistics for the Row Points Quality Mass Inertia Barcelona ? 0.2020 0.1245 Budapest ? 0.1313 0.2120 London 0.9158 0.1960 0.1280 Munich 0.9998 0.1535 0.1941 Paris 0. 9738 0.1758 0.1586 Zurich 0.9964 0.1414 0.1828 Partial Contributions to Inertia for the Dim1 Dim2 Barcelona ? 0.2144 Budapest ? 0.5273 London 0. 1279 0.0913 Munich 0.2208 0.1217 Paris 0.2037 0.0035 Zurich 0.2290 0.0419 Squared Cosines for the Dim1 Row Points Barcelona ? Budapest ? London Munich Paris Zurich 0.7537 0.8574 0.9688 0.9443 Dim2 0.3913 0.5652 0.1621 0.1424 0.0051 0.0521 Column Coordinates Arousing Gloomy Pleasant Sleepy Summary Arousing Gloomy Pleasant: Sleepy Dim1 -0.2575 0.0943 0.4420 -0.1870 Statistics for Quality 0.9588 0.9093 0.9946 0.9994 Dim2 0.2545 0.1562 -0.0608 ·0.1166 the Column Mass 0. 1152 0.2283 0.2081 0.4485 Points Inertia 0.1799 0.0955 0.4758 0.2489 Row Points Partial Contributions to Inertia for the Column Points Arousing Gloomy Pleasant Sleepy Squared Arousing Gloomy Pleasant Sleepy Dim1 0.1157 0.0308 0.6160 0.2375 Cosines for the Dim1 0.4851 0.2430 0.9761 0.7195 Dim2 0.3749 0.2799 0.0387 0.3066 Column Points Dim2 0.4737 0.6663 0.0185 0.2800 1. What statistical method is employed? n. Find the values marked by"?". 111. Evaluate the fitness of the model. State the statistics you used. IV. Describe the row and column dimensions. State the statistics you used. v. Produce a perceptual map and describe the map. P.4 ofll

S&AS STAT2313/3613 Marketing Engineering b) A data set DIST contains the Euclidean distance among the cities calculated based on the four descriptions. A SAS program and part of the SAS output are given for a statistical analysis. proc mds data=dist dirn=2 forrnula=l pfit pconfig level=ratio; var Barcelona Budapest London Munich Paris Zurich; id city; run; Configuration Barcelona Budapest London Munich Paris Zurich _MATRIX_ Dim1 Dim2 18.60 -18. 91 12.14 -9.97 11 . 35 -13.20 Number of Nonmissing Data 15 -5.51 6.80 7.73 -6.75 1 .62 -3.89 Weight 1.00 Badness-of- Fit Criterion 0.04 Distance Correlation 1.00 uncorrected Distance Correlation 1.00 1. What statistical method is employed? Describe how the method is specified by the program. n. Comment on the fitness of the model. State the statistics you used. 111. Produce a map of the cities and describe the map. [Total: 30 marks] 3. Consumers' preferences on attributes of broiler meat were investigated. Three attributes (Form, Price and Taste) were used in a conjoint study. The average ratings of the preferences of the profiles are given as follows (I = least preferred to 5 =most preferred) Form Price Taste Preference rating Parts $6 Tasty 4 Parts $6 Not Tasty 5 Parts $10 Tasty 3 Parts $10 Not Tasty 2 Dressed whole $6 Tasty 5 Dressed whole $6 Not Tasty 3 Dressed whole $10 Tasty 2 Dressed whole $10 Not Tasty 1 Live bird $6 Tasty 3 Live bird $6 Not Tasty 3 Live bird $10 Tasty 1 Live bird $10 Not Tasty 1 a) Based on an additive AN OVA model, find the part-worth of each level of each factor. b) Find the importance of each factor. Which is the most important factor? c) Which is the most preferred profile under the model? P.5 of 11

S&AS STAT2313/3613 Marketing Engineering e) Based on the model estimate the market shares of the profiles given as follows ' Form Price Pro xi Parts $6 Far Dressed whole $6 Near Live bird $6 Near [Total: 14 marks] 5. This study intended to help producers of organic food products to understand the importance of various factors on overall satisfaction towards organic food products. Respondents rated eight items on a 7-point Likert scale (!=strongly disagree, ... , 7=strongly agree) in a questimmaire. The genders of the respondents were also recorded. The data are stored in a data set ORGANIC and the variabl · £ 11 es are g1ven as o ows. XI Safe for consumption X2 Free from contamination X3 Nutrient rich X4 Expense X5 Ready to pay premium X6 Product display influences choice towards organic food products X7 Purchase more during promotional offer period X8 Sales promotion offers are significant GENDER Male, Female a) A SAS program and part of the SAS output are given for a statistical analysis. proc factor data=organic n=3 rotate=varimax ; var xl-x8; run; Eigenvalue Difference Proportion 1 4. 67247585 3.07893743 0.5841 2 1 . 59353841 0.15034832 0.1992 3 1. 44319009 1.35643478 0.1804 4 0.08675531 0.01656049 0.0108 5 0.07019482 0.01114608 0.0088 6 0.05904873 0.01535450 0.0074 7 0.04369424 0.01259168 0.0055 8 0.03110255 0.0039 Factor Pattern Factor1 Factor2 Factor3 x1 0.80629 0.53926 ·0.11210 x2 0.80220 0.54616 ·0.16433 x3 0.82306 0.49725 -0.18227 x4 -0.58157 0.50989 0.60928 x5 -0.56477 0.48816 0.64284 x6 0.84940 ·0.20766 0.42482 x7 0.83458 ·0.32082 0.41679 x8 0.79134 -0.33592 0.48139 Variance Factor1 4.6724758 Explained by Each Factor Factor2 Factor3 1.5935384 1.4431901 P.7ofll Cumulative 0.5841 0.7833 0.9637 0.9745 0.9833 0.9907 0.9961 1 . 0000

S&AS STAT2313/3613 Marketing Engineering Final Communality Estimates: Total- 7.709204 x1 x2 x3 x4 0.95347762 0.96882434 0.95790946 0.96943877 x5 x6 x7 x8 0.97051540 0.94507067 0. 97317081 0.97079728 Rotated Factor Pattern Factor1 Factor2 Factor3 x1 0.94131 0.24235 -0.09313 x2 0.95479 0.20327 -0.12604 x3 0.93646 0.22466 -0.17457 x4 -0.13531 -0.18887 0.95680 x5 -0.14738 -0.14766 0.96280 x6 0.31014 0.91176 -0.13260 x7 0.21964 0.94156 -0.19592 x8 0.16678 0.96090 -0.14021 1. What statistical method is employed? Describe how the method is specified by the program. 11. Suggest the number of factors based on (A) the latent root criterion and (B) scree plot. State clearly the reasons. 111. Comment on the fitness of the three-factor model. 1v. Describe the rotated factors. b) Another SAS program and the output are given as follows. proc factor data=organic n=2 rotate=varimax; var xl-x8; run; x1 x2 x3 x4 x5 x6 x7 x8 Orthogonal Transformation Matrix 2 1 2 0.71088 -0.70331 0.70331 0.71088 Rotated Factor Pattern Factor1 Factor2 y y 0.18615 0.95245 0.23538. 0.93235 -0.77204 -0.04655 -0.74482 ·0.05018 0.74987 0.44977 0.81892 0.35891 0.79881 0.31775 1. Calculate the values marked by "Y". 11. Calculate the variance explained by each rotated factor. m. Plot the variables based on the rotated factor loadings. Describe the plot. 1v. The inverse of the sample correlation matrix is as follows. x1 x2 x3 X4 xs x6 x7 x8 x1 10.21 -6.40 -3.32 -0.01 -0.54 0.50 -1 .24 0.14 x2 -6.40 14.23 -7.64 1 .27 -1.29 -0.74 0.33 0.72 x3 -3.32 -7.64 11 .90 -1 .43 2.24 -1 . 14 0.43 0.30 x4 -0.01 1 .27 -1 . 43 9.19 -8.43 ·0.32 2.41 -1 . 41 x5 -0.54 -1.29 2.24 ·8.43 9.02 -0.06 -1 . 08 0.81 x6 0.50 -0.74 -1 . 14 -0.32 -0.06 9.05 ·4.59 ·3.48 x7 -1 . 24 0.33 0.43 2.41 ·1. 08 -4.59 1 9. 1 1 -13.5 x8 0.14 0.72 0.30 - 1 . 41 0.81 -3.48 -13.5 16.57 P.8 of 11

S&AS STAT2313/3613 Marketing Engineering Formulae Market response model " SSE = L (y,- );,) 2 i=l n SST = l:(y,- y)' i=l R' =l- SSE SST Cluster analysis Single linkage d(uv)w = min {duw, dvu,} Complete linkage d(uv)w = max{durv,dr'W} Average linkage 2:: l:d" d _ le(UV)keW NuNwdurv + NvNwdvw (uv)w - N(,w)Nw N(,w)Nw Centroid linkage NuNvd~v (Nu +Nu)' Ward's method p ESS = LLL(xiiK -x·JK f K ieK j=l BSS = ~(x,-x f L.,; } .I i=l Factor analysis Conununality h' =£ 2 +···+!!' 1 il im Specific variance lJit =a it -h;2 Total variances explained p '\' n2 n2 0 2 L..'u =<,J+···+<PJ i=l Principal component factor analysis L=~V 1 vi +A 2 V 2 V~ +···+APVPV~ L = [Avl Fmvm] '¥ = diag(l:- LL') Factor scores Weighted least squares t 1 = (F'P- 1 tf 1 Fl¥- 1 (xi- x) Regression t = Fs- 1 (xi- x) Multidimensional Scaling P.lOofll

S&AS STAT2313/3613 Marketing Engineering MetricMDS B = (b,k) = UAUr b = -~S2 +~ ~Sfk +~ ~Sfk -~ ~ ~Sfk tk 2 tk 2 L n 2 L n 2 L L n 2 i k i k Coordinates X= UA 112 MDPREF model S = UAVr, SSr = UA 2 Ur, srs = VA 2 Vr Y = V A, ideal vector X= U, coordinates of the objects In SAS Y = (n-1)- 112 VA X= (n-1 ) 112 U Correspondence analysis Mass D,,,= J;.ln DCJ 1 =f. 1 /n Chi-square value fu- eii ,re: - J;.f.; eiJ- n M = n- 1 1 2 X = UQVr, MMr = UQ 2 Ur, MrM = VQ 2 Vr Coordinates Y = D~"'UQ X= D~ 112 VQ Inertia Q2 Partial contribution to inertia 2 2 Uti' vij Quality Sum of the squared cosines Squared cosine y[j xfJ "'k z' "'k z .L.j=1YiJ L.j=1xiJ Conjoint analysis Part-worth= level average rate -"overall average rate Bradford-Terry-Luce (BTL) model share(x j) = u(x, y;;, u(x,) Logit model share(x 1 )= Exp [uk )YJ; Exp[U(x, )] **************END OF PAPER************** P"ll of 11