Homework 9 Solution (Due Nov 23th, Monday)Problems:1) 8.4a) The estimated regression function isˆY = 82.93575 − 1.18396X�1+0.01484X2 �1where X�1= X1−59.9833333. The plot below shows the data and fitted values. Yes, the quadratic function appears tobe a good fit here, and R2= 0.7632, which is fairly high.b) See the SAS results bel ow. Here, we are testing H0: β1= β11=0vs. Ha: at least one is notzero. Here, we can use the general linear regression results from the ANOVA table from SAS. F* =MSR/MSE = 5915.31/64.41 = 91.84. The degrees of freedom = (2,57). The p-value from SAS <0.0001. Since the p -value < 0.05, we reject the null and conclude that at least on e of β1, β11is notzero.e)See the SAS output below. Here we are testing H0: β11=0vs. Ha: β11�= 0 . For this, we canuse a t-test, for which the t* value is of the form: t∗ = b11/s(b11) , with degrees of freedom = (n-p) =57. The SAS output shows the t* value = 0.01484/0.00836 = 1.78. The p-value = 0.0811. Since thep-value is not less than 0.05, we cannot reject the null, and we conclude that we can drop the quadraticterm. This is telling us that a linear model is adequate, and we do not need the quadratic term.Analysis of VarianceSum of MeanSource DF Squares Square F Value Pr > FModel 2 11831 5915.31037 91.84 <.0001Error 57 3671.31259 64.40899Corrected Total 59 15502Root MSE 8.02552 R-Square 0.7632Dependent Mean 84.96667 Adj R-Sq 0.7549Coeff Var 9.44549Parameter EstimatesParameter StandardVariable DF Estimate Error t Value Pr > |t| Type I SSIntercept 1 82.93575 1.54315 53.74 <.0001 433160c_age 1 -1.18396 0.08863 -13.36 <.0001 11627c_age2 1 0.01484 0.00836 1.78 0.0811 203.13491242) 8.9a) See the plot below. The resulting lines are: X2=3: E(Y ) = 37+ 7.5X1and X2=6: E(Y ) = 49+ 12X1. The interactioneffect of X1and X2on Y is apparent since the lines ar e not parallel, which indicates that the effect ofX1for a given level of X2depends on the level of X2. This is a reinforcement effect, since β1and β2are both positive and β3is also positive in the response function: E(Y ) = 25 + 3X1+4X2+1.5X1X2.3) 8.15a) For this problem, Y is the total number of minutes spent by the service per son on the copier. Themodel (8.33) for this problem is: Yi= β0+ β1Xi1+ β2Xi2+ �i. The response function for companiesusing the large mode l is E(Y )=β0+ β1Xi1. The response function for companies using the smallmodel is E(Y )=(β0+ β2)+β1Xi1. Here β1indicates the increase in service time for each increase inX1(number of copiers ser v ic ed ) , for both types of companies. β0is the y-intercept, and β2indicateshow much higher (lower) the response is for the companies using the small model versus the largemodel, for any given level of X1(number of copiers serviced).b) The SAS output is below. The estimated regression function isˆY = −0.92247 + 15.0461X1+0.75872X2.Analysis of VarianceSum of MeanSource DF Squares Square F Value Pr > FModel 2 76966 38483 473.94 <.0001Error 42 3410.32825 81.19829Corrected Total 44 80377Root MSE 9.01101 R-Square 0.9576Dependent Mean 76.26667 Adj R-Sq 0.9556Coeff Var 11.81513Parameter EstimatesParameter StandardVariable DF Estimate Error t Value Pr > |t| 95% Confidence LimitsIntercept 1 -0.92247 3.09969 -0.30 0.7675 -7.17789 5.33294number 1 15.04614 0.49000 30.71 <.0001 14.05728 16.03500type 1 0.75872 2.77986 0.27 0.7862 -4.85125 6.36870c) T he 95% CI is b2± t(0.975, 42)s(b2). From the SAS output, this is 0.7587 ± 2.018(2.77986) =(-4.8514,6.3688). We can be 95% confi d ent that with the number of copiers to be serviced held fixed,the average difference between small and large copier service times is in this interval.d) The analyst is interested in the effect of copier size on average service time with the numb er ofcopiers to be serviced held fixed. Controlling for the number of copiers being serviced achieves thesecond part of that goal.254) 8.21a) Whe n a hard hat is worn, the response function is E(Y )=(β0+ β2)+β1X1. When a bump cap isworn, the response function is E(Y )=(β0+ β3)+β1X1. When no cap is worn, the response functionis E(Y )=β0+ β1X1.b) (1) H0: β3=0vs. Ha: β3< 0, (2) H0: β2− β3=0vs. Ha: β2− β3�= 0.5) This problem is based on Ch. 6, problem 15 data. See SAS output below.a) T he estimated regression functi on isˆY = 190.52 + 0.79X1− 3.15X2− 14.41X3+0.02X1X2−1.20X1X3+0.93X2X3. For the overall F test, H0: β1= β2= β3= β4= β5= β6=0vsHa: not all βsin H0equal 0. From the SAS output, F*=14.86, with df=(6,39) (from ANOVA table). The p-value <0.0001, which indicates that we would reject the null and conclude that at least one of the parametersin the null is not equal to z er o.b) For this, we use the extra sums of squares from the SAS output, which are labeled Typ e I SS in SASoutput. The hypotheses are H0: β4= β5= β6=0vsHa: not all βsinH0equal 0. The partial F teststatistic requires that we have SSR(X1X2,X1X3,X2X3| X1,X2,X3) = 8.92+133.42+38.08=180.42.The degrees of freedom=3. The F* value =180.423÷4068.4239=0.58, with degrees of freedom = (3,39).The critical F(0.95,df=(3,39)) value = 2.85, which is calculated using SAS code. Since F* < F-critical,we do not reject the null hypothesis, and conclude that no interaction terms are needed in the model.Analysis of VarianceSum of MeanSource DF Squares Square F Value Pr > FModel 6 9300.88929 1550.14821 14.86 <.0001Error 39 4068.41506 104.31833Corrected Total 45 13369Root MSE 10.21363 R-Square 0.6957Dependent Mean 61.56522 Adj R-Sq 0.6489Coeff Var 16.58994Parameter EstimatesParameter StandardVariable DF Estimate Error t Value Pr > |t| Type I SSIntercept 1 190.51810 117.37011 1.62 0.1126 174353age 1 0.79293 3.15488 0.25 0.8029 8275.38885illness 1 -3.14572 3.26554 -0.96 0.3413 480.91529anxiety 1 -14.40686 70.96754 -0.20 0.8402 364.15952ageill 1 0.01565 0.06396 0.24 0.8080 8.92314ageanxiety 1 -1.19694 0.93509 -1.28 0.2081 133.41875illanxiety 1 0.93330 1.54466 0.60 0.5492