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Unformatted text preview: QUESTIONS L4: To gauge how your firm’s advertising and other promotional
activities affect your ﬁrm’s sales, you collected data on your ﬁrrn’s sales (in thousands of
dollars) for 51 months and the firm’s promotional expenditures (in thousands of dollars. Graphs and statistics support the regression model: .. 1..
Sales m 25.1264 + 0.7623PromoExpend. ll «’3’ M” 5—0 1. To test whether the simple linear model is useful, we test
a Whether the y~intercept is different from 0 1
19. Whether the yintercept is different from 1. Qf‘l
0. Whether r2 is large; M ,... ' ..... MW... “mum... a... MW
.1. a». 1....1 __..... .....,.___ 11:: Weather the slope parameter is different from MD. . _ . MW”? e. Whether the slope parameter is differentWi. “P mm 15 :25? 9.111111) 2 Which of the following is a correct interpretation of the parameter estimates?
a. As sales increase by $1, promotional expenditures increase on average by $25.12 WW___W_M.__V b AS Promomnal expenditures increase by $1000 sales increase on average W3,
11 ”$762 30 .. MM MMMMM ..__.._..._.. 2,112,111,1111... .11 M. c. MAS promotional expenditures increase by $1000, sales decrease on average by
$762. 30 d. When promotional expenditures are $0, sales are predicted to be $25,126.40 per
month on average. e. When sales are $0, promotional expenditures are predicted ‘00 be $25.12 during
that month on average. 19.9%.2‘133‘3 (2.130%)(0129‘5‘) 3. If Sm I 0.1209, which of the following is correct? :31 Q 1 211. 3”“
a. We are 95% conﬁdent that as sates increase by $1, promotional expenditures 1
Mjgggegﬁgm 0n achrage between $1 23 to $51392.“ MM 11...“... m1...“ , ﬁr; 52'th 1 no 5:13 1
b We are 95% conﬁdent that afproﬁoiional expenditures 1ncrease by $1000 ‘
Lincrease on average between 519 40 and $1005.20. MM c. We are 95% conﬁdent that as promotional expenditures increase by $1, sales
decrease on average between $0.52 and $1.01. (1. We are 95% conﬁdent that as sales increase by $1, promotional expenditures
increase on average between $519.40 and $1005.20. 4. If r = 0.673, which of the following is correct? Q 2
a. The correlation between sales and promotional expenditures is O. 4529. 1 "" O UT 326132?
b. If we use the model to predict, we can expect 45. 29% of our actual sales values to
be within one standard error of their predicted values 1111+ osm ”9’
c. We get 45 29% more prediction error by simply using the sample average to
predict sales, rather than using the. model to predictsales (“a "l" .. 4: ”m” “A (61.15. 219% of the variability in monthly sales can be explained by monthly 0) “Ma—1,91.“ pgghhsw
promotional expenditures. 31 {if i e. Bothb andc (2:313) QUESTlONS 5 8: A real estate agent would like to predict the selling price of
single family homes In her small town She takes a random sample of 15 recently
sold homes and creates a regression model using the size of the home (in 100 ftz) to
predict the selling price (in thousands of dollars). Summary results are given: xx = 272.6, 2y «7— 13326, SSxx = 268.19, SSyy = 6230.24,
SSXy w 1040.18, r = 0.8047, sK = 4.3768, sy W 21.095,
SSE = 21943416 5. The least squares regression line IS closest to which equation?
1! 2‘, (301 5 a J? 11.49+4.55x bx ’ s3}, {3 %$%q)l\ LL37???) r" 3 3'78
b 37m14.82+3.88x
C j) 17.20e4.55x @ be :" (l ’ b, “‘2 : Wmm @WXW’5) e. j):20.12+3.65x .:' 1‘63? 6. What is the standard error of the residuals? b. 11.76 .1 “s: was... M6341 ..
C. 12.68 C ‘d. 12.99 L" _. e. 13.68 .._ Rolcl 7. What is the residual for a home that was 22 hundred square feet and sold fer $118,900? /\
a. 8.54 (@VL u? :: @333 Jr 338""{3‘251 103, ”N
b. «6.39
C . 1 :1)
2. 6:5399 (wig. : ama\~g(ereglxal‘e&
6 4519 ;. weft—— \e 3m : \El‘l 8. What are error degrees of freedom?
a. 3
b. 4 d. 14
e. 15 QUESTIONS 912: Is economic growth in the developing world related to growth in
the industrialized countries? Data is gathered over the past 38 years. A regression
predicting annual GDP growth of developing countries is run. Output given below. If needed for any question, use 01 m Regression Analysis: Developing versus Developed Predictor Coef SE Coef
Constant 3.4558 0.4496
Deveioped 0.4327 0.1407 0.01. T P
7.69 0.000
3.08 0.004 s = 1.2421111 R—Sq = 20.8% Rquladj)
Obs Fit: SE Fit 95% Interval
3.5 4.970 0.221 (4.522, 5.419) 18. 95% 6% Interval (2.407, 7.533) diodeleye—CQ 9 What 15 the correlation between developing nations GDP and developed nations GDP? a. 0.3821
b. “0.3821
d. 0.4313
c. 0.4561 (F5 @5935: o 4%? 10. Suppose the researchers want to test if there is evidence of a linear relationship
between developing nations GDP and developed nations GDP. Which are the correct _,_,,n,,ﬁ,_wwm . H r1,=1; H: 131:1 C. Hui/30 20;}111180 350 d. 11,: ,6,=1; Hazﬁoil 11 Is there strong evidence of a linear relationship between developing nations GDP and deve10ped nations GDP? a. Yes. Because r2 is large conclude there 15 a relationship wanﬂ" Qt”
b. No. _Becanse..the. pavaluej is small, "concludeehere' 1s notarelationship. ?W .Yes. Because the value 15 small conclude there 13 21%
d Yes Because s is small, conclude there IS a relations 1p. e. No. Because the puvalue is large, conclude there is not a relationship. l2. A 95% interval for
nation GDP was 3 5 is a (4 522, 5419)
are; (2. 407, 7535))
c (0.1513, 07141)
d. (—1.0673,1.9327) qua individual developing nation GDP when the developed \léﬁlx’rml m.;'\' QUESTIONS 1315: Manufacturers of frozen pizzas reformuiate their products to maintain and increase customer satisfaction and sales. To help determine what might
need to be changed, they frequently use the prestigious Consumer‘s Union rated frozen
pizza for ﬂavor and quaiity. An overail score is assigned to each of 29 brands tested. A
regression model to predict the CU score from calories, type, and fat content gives the foliowing result. If needed, use 0. = 0.05. 'HisiogéaiaoiREin j ; . Regression Analysis: Score versus Caiories, Fat, Type The regression equation is Score m — 149 + 0.743 Calories — 3.89 Fat + 15.6 Type S m 19.7902 R—Sq = 28.7% R—Sq(adj} = 20.2% Predictor Coef SE Coef T P
Constant ~148.82 77.99 —l.92 0.068
Calories 0.7430 0.3066 1.42 0.073
Fat —3.893 2.138 —1.82 0.081
Type 15.634 8.103 1.93 0.065 Analysis of Variance Source DE 83 MS Regression 3947.3 1315 3.3gﬁ
Residual Error 23' ﬁqﬁj.ﬁ' 3ilhﬁﬂé Total 28 13738.7 13. Interpret the two plots above §
0.035 ' ._j'; ScatterplotofRESIpilALvsFit_""' ' a. The histogram is approxirnateiy normai but the residual plot 18 not linear. This reveais a problem with using linear regression. b The histogram is not approximately normal but the residual plot has no pattern. This reveais a problem with using linear regression. c. The histogram is approxirnately normal but the residual piot is linear. This reveais a problem with usmg linear regression .. 1..— m ,..'d The histogram is approxrmately normal and the residual plot has; no pattern. It /) is appropriate to use iinear regression techniques 4
\iQK‘SlW l 14. Regardless of your decisiori in the question above, what test statistic would be used
to determine the overall usefulness of the model? a. 2 4
Mb. .359
' c. 4.222
d. 20.20 ' Egg? Mia waddle  i i la .’IT
15. Does there appear to beamulticollinearity problem here? mmgai \3 ”Q’gua :9“ l
‘ Vie—(ML. Rmxm‘hw we
Q I “mkmﬁglaﬁ We“. SEE/agiég'toazl‘ c. Not enough information to ansWer. 16. The least squares regression equation minimizes the sum of the a. differences between actual and predicted y values.
b. absolute deviations between actual and predicted y values.
0. absolute deviations between actual and predicted 3g vaiues. {gigs uared‘"d'iffe“féﬁde”‘:bfelfwe7e’ﬁ acfiial and Wredicted y valﬁe‘s.
e. squared differences between actual and predicted x values. 17. A scatterplot reveals a linear pattern so a linear regression is run. If the correlation
coefcient is equal to one then a. the yintercept must equal zero.
‘0. the explained variation of y approximately equals the unexpiained
variatiarrairhwmm 0 ﬂaw Ml» r al ’i/lxeva. v: saw:7
d. there 13110 exp alne var1at10n of y. 3 e. none of the above. % & S t Q {9‘9“ng (”ab mowed QUESTIONS 1822: Insurance companies base their premiums on many factors, but
basically all the factors are variables that predict life expectancy. Using data from aii 50
states, conditions were checked and regression analysis is appropriate. Given below are
regression results that predict life expectancy using murder rate (per 100,000 people),
high school graduation rate, median income level, and illiteracy rate. (Remember to
round down on (if if necessary.) Ufag at :g) £35 Regression Analysis: Life Expectancy versus Murder, HS Grad, Income, lliiteracy The regression equation is:
LifeExpect: 69.5 — 0.262Mnrder + 0.0461HSGrad + 0.000125Income + 0.276Iliitercy Predictor ‘ Coeff SE Coeff T P Constant 69.483 1.325 2.93 0.001 . ‘ ‘
Murder ~0.26194 0.04447 ”2.79 opozw B‘Khﬁwﬁsﬁ
HS Grad 0.04614 0.02185 2.11 0.040 ... 3‘0 mike«cam.
chome 0.0001249 0.0002422 0.52 0.508,» ij:
Illiteracy 0.2761 0.3105 0.89 0.379 —— “CA( S 3 0.804929 R—Sq : 67.0% R—Sq(adj) = 64.0% Analysis of Variance Source DF SS MS F P
Regression 4 59.143 14.786 22.82 0.000
Residual Error 45 29.156 0.648 Total 49 88.299 i8. if we wish to test the hypotheses H0: ﬂ} = 182 m 163 ”—" 134 = 0 3 the correct test
statistic and puvalue are: ”its? 5 2.93, = 0.001
/ b. F m 22.83, p m 0.
c. t= ~2.79, p m 0.002
d. F =16.99, p = 0.002
e. Coeff = 69.483, p m 0.001 19. Based on the above analyses, we can conciude a. Income is not usefui as a predictor variabie unless ali other variables are present.
b. Income is useful as a predictor variable even when all other variables are present. _.a....._Ar........_... "w«Wmmu m... __......,,....... .. “~le ... “rm—Income 1s notiiEéiﬁl‘as a predié‘to'r vet‘iabiemhenili other variablesma‘femm.
7%... __
d. HSGrad is not usefui as a predictor variable uniess amm are present.
e. HSGrad is not useful as a predictor variable when all other variables are present. 6 _
\ie/gtm i 20. If using Backward Regression to ﬁt a better model, which of the following is true?
a. All 4 explanatory variables contribute to the model. Make no changes.
13. Remove "33 354943 «\4 iﬁoﬁﬁd, then rerun. the regression.
c. Remove Illiteracy, then rerun the regression. d. R‘ emove Income, then rerun the re ressro e. Remove HS Grad, Income and illiteracy. That is the ﬁnal model. &§: 4&3“ 3’0 1. A96% conﬁdence interval for is ive b which of the followin ? ....
Lew) s seaxw was we iEQgmé y % g ﬁne; 35; .. age
a. (0.1019, 0.4503) ' b "it E s egg.
b. (0.024I,0.068‘1) 9 b! h :2. R3 ‘. ‘0i0‘”0i§0.092
3T: ‘3???) Q l 5% let i: (2.\2§)C©.0213§3 l e: (~0:1439,0.6961) + 0 04493:} é 22. The proportion of variability in Life Expectancy that is explained by the model is: afwaﬁo b. 0.804929
c. ‘04.0 6/2 :: CHINE :. game d. 0.648
e. 38.299 ...
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 Spring '08
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