1. The following model was fitted to a sample of 30 students using data from the end of their second year in university. The aim was to explain students' weight gains. Y = 7.35 + .653 X1 - 1.345 X2 + .613X3 where Y = Weight gain in pounds, during the sophomore year X1 = average number of meals eaten per week X2 = average number of hours of exercise per week X3 = average number of caloric beverages consumed per week a. Interpret the coefficients of each explanatory variable. b. Suppose that the standard errors associated with the coefficients above were respectively, sb1=0.189, sb2=0.565, and sb3=0.243. Test the hypothesis that all else being equal, hours of exercise do not influence weight gain. =.05 Find the 90% and 99% confidence intervals for beta1. 2. The file Germany.xls contains twenty two annual observations from Germany on percentage change in wages and salaries (y), percentage change in productivity from year to year (x1), and percentage change in the rate of inflation from year to year (x2). Use SPSS to Estimate the least squares line. Construct the 90% confidence interval for beta1 and beta2. Use the Sum of Squares Table (ANOVA Section) to verify the R 2 and R2adj obtained for the regression. Interpret the b1 Create a new variable, lagged productivity (x3). Estimate a new model containing three explanatory variables. Does x3 contribute anything to the regression model? Cut and paste relevant portions of the output. 3. A computer repair service is examining the time taken on service calls to repair computers. Data obtained for 30 service calls are found in comprep5.xls. Information obtained includes: o x1 = number of machines to be repaired (NUMBER) o x2 = years of experience of service person (EXPER) o y = time taken (in minutes) to provide service (TIME). Based on these or a subset of these variables or suitable transformations of them develop a model to explain the time taken to provide service. Provide justification for including your choice of variables. (Test for linearity.) Based on your final model how long would it take to provide service if the number of machines to be repaired is 12 and the service person has 10 years of experience? Note that your final model may not contain all available variables. Cut and paste relevant portions of the output. NUMBER 1 1 3 4 6 6 7 8 9 11 11 12 13 14 15 16 17 20 19 20 22 22 23 24 25 26 27 28 29 30 EXPER 9 11 11 8 9 9 10 9 8 10 10 7 9 9 10 11 10 9 9 9 9 9 10 10 11 8 10 9 9 10 TIME 66 74 88 99 134 120 178 139 187 227 225 270 265 301 343 383 383 515 474 495 628 636 660 731 752 800 863 918 976 1027 Wages 9 6 8.9 9 7.1 3.2 6.5 9.1 14.6 11.9 9.4 12 12.5 8.5 5.9 6.8 5.6 4.8 6.7 5.5 4 3.3 Productivity Inflation 3.5 2.8 6.3 4.5 3.1 1.5 7.6 6.7 4.2 2.7 3.5 5 2.3 1.5 6 2.9 2.8 2.6 0.9 0.6 0.7 3.1 4.5 3 3.1 3.8 3.8 1.1 2.3 3.6 7.5 8 6.3 6.1 6.9 7.1 3.1 3.7 3.9 3.9 4.8 4.3 4.8 3.2 Regression Analysis We\thave\talready\tseen\tthat\tassociated with each\tregression\tparameter is a\tstandard error\t(sb). Coefficientsa Model 1 (Constant) EDUCAT EXPER TIME Unstandardized Coefficients B Std. Error 3179.744 383.485 139.618 27.716 1.481 .697 20.633 6.155 Standardized Coefficients Beta .449 .190 .298 t 8.292 5.037 2.124 3.352 Sig. .000 .000 .036 .001 a. Dependent Variable: SALARY Salary\t= 3179.7\t+ 139.6\tEduc\t+ 1.4\tExper + 20.6\tTime (383.5) (27.7) (0.7) (6.2) Knowing their values permits\tus\tto \tConstruct\tConfidence\tIntervals\tfor\tParameters Test Various Hypotheses & Conduct Diagnostic Tests. Formulate\tStrategy Confidence\tIntervals: Let Y\t=\t5.32\t+\t0.31X1\t+\t1.05x2 and n=26 (1.25) (0.09) (0.34) If n\tis\tsmall,\tthen we use\tthe\tt-distribution to construct confidence\tintervals. The\tnumber of degrees of freedom\t(DOF)\tis given by DOF= n-p-1, where\tp\t=\tnumber\tof\texplanatory\tvariables. The\t(1-\ta)%\tconfidence\tinterval\tfor\tb\tis\tthen, b \tta/2,\tn-p-1\tsb Example:\tConstruct\tthe\t95 %confidence\tinterval\tfor\tb1 then 0.31 t.025,23\t(0.09) 0.31 (2.069)\t(0.09) 0.31 0.19 [0.12,0.50] If\tn\twere\tlarge,\tthen\tthe\t5%\tconfidence\tinterval\tfor\tb2\twould\tbe 1.05\t z.025 (0.34) 1.05\t\t1.96\t(0.34) 1.05\t\t0.67 [0.38,\t1.72] Tests of Significance: Suppose that after estimating the regression equation, we see that b 1 is very small. If it were zero it would imply that X 1 does not have much of an impact on Y. Should b 1 be kept in the model? What is small? How can we test the hypothesis that b 1is zero? Calculate the test statistic, t = b1\t/\tsb1 If calculated I t I> 2, then b 1 is statistically different from zero. acceptance region. True\tRegression: Sales\t=\ta\t+b1Advertising\t+b2Competition\t+\te Estimated\tRegression: Sales\t=\ta\t+\tb\t1\tAdvertising\t+ b 2Competition\t+\te Test\tthe\thypothesis\tthat\tb1\tis\tstatistically\tinsignificant. H0\t: b1=0 H1\t:\tb10 Compare\t(calculate) = #$ %&$ '($ = #$ %* '($ = #+ '($ with\ttabulated\tvalues\tof\tt\ta/2,\tn-p-1. If\tthe\tabsolute\tvalue\tof\tthe\tcalculated\tt\tis\t\tt\ta/2,\tn-p-1,\tthen\tb1 statistically\tsignificant\tat\tthe\t(1-a)\tlevel.\tDo\tnot\taccept\tH0. Why\t|t|\t\t2? Let y\t=\t5.32\t+\t0.31\tx1\t+\t1.05\tx2 n=26 Test\tthe\thypothesis\tthat\tb1\t=\t0.25;\tlet\ta=0.05. H0\t: b1=0.25 H1\t:\tb10.25 Calculate = #$ %&$ '($ = *.-+%*../ *.*0 = *.*1 *.*0 =\t0.67 Compare\twith\tthe\ttabulated\tvalues\tof t\t0.025,\t23 =\t2.069 Our\tcalculated\tt\tis\tin\tthe\tacceptance\tregion. Let y\t=\t5.32\t+\t0.31\tx1\t+\t1.05\tx2 n=26 Could\tb1\t<0.25?\tLet\ta=0.05. H0\t: b1=0.25 H1\t:\tb1<0.25 Calculate = #$ %&$ '($ = *.-+%*../ *.*0 = *.*1 *.*0 =\t0.67 Compare\twith\tthe\ttabulated\tvalues\tof t\t0.05,\t23 =\t-1.714 Our\tcalculated\tt\tis\tin\tthe\tacceptance\tregion