Question

Q3. 47 ptsSuppose the weight (y) of babies depends on their height (xi) and age (x2). The following table presents the data for a sample of 5 babies and some results of analysis of the data are also given below. 2 | 4 | 5 | 7|8| Given that (XX) xi | 1.5 | 2 | 2.5 | 3.5 | 3 | [16.4286-14.2857 4.428571 Corr matrix 1.00 0.985 0.985 1.000 0.8285 X2 1 35 7 614.2857 13.25714.285710.3142 3) 0.7142 4.42857-4.2857 1.42857] 0.8285 0.3142 Using above information: 2 )/(9 pts] Find the least-squares estimates of the regression coefficients and thefittedregression) model. Interpret the estimated values of the parameters 3 b) [5 pts] Find the adjusted R2 and interpret the reşsult. 0.05 and draw the (3 S15 pts] Test the hypothesis Ho :B-/32 0, Hi : β, * 0,/-1, 2, at α conclusion d) 17 pts] Find the standardized residual d,, -1, 2, 3, 4, 5 and identify whether there is any outlier or leverage points in the given data set. (C>3 [8 pts] Calculate the Cook' Di and identify whether there is any influential point in the data set 16 pts] What do you mean by multicolinerity problem in regression analysis? What are the primary sources of multicollinearity? Give the name of three techniques for detecting mulicollinearity 13 pts] Find the variance inflation factors VIF, J-1, 2 and determine the presence of multicollinearity in the given data set. a h) [4 pts] The eigenvalues of the XX matrix are given by λ2-1 .701 and λ.-0.033. Determine the condition number and condition 157.016, indices and comment on the presence of multicollinearity.

Answer 1

(a) Consider the regression, ECY) oB1 B2 x2, 1,2,..,5 Then 1 1.5 1 1 2.0 3 X-1 2.5 5 1 3.5 7 1 3.0 6 (2,4,5,7,8), Ху-(26, 72, 136), β (X,X)-1X'y = (.8571, .2286, .8571), Fitted regression equation is y-857 +.2286x1 +.8571x2 The estimated coefficient for Height is .2286, which means, on an average, baby weight is likely to increase by 2286 unit for one unit increase in baby height. The estimated coefficient for Age is.8517, which means, on an average, baby weight is likely to increase by .8517 unit for one unit increase in baby age. The estimated intercept coefficient is .857, which means, on an average, for a baby before birth (i.e. age 0) with height 0, Weight is expected to be.857 units. However, the last interpretation is absurd.

(n-1)(1-32) (b) Rad/ = 1 . n-k-1' n-number of observations 5 K-no. of predictors-2 R2-SS(Regression)/SS(Total)-20.1143/22.8-8822 Rad.7644 Adjusting for the number of predictors, 76.44% variation in the data is explained by the fitted regression line. (c) The test statistic is F-(SS(Regression)/2)/(SSE/2) F22 We calculate SS(Regression)-20.1143, SSE-2.6857. Therefore, wefail toreie, pvalue-P(F>7.489), 0.1178 Therefore, we fail to reject the null. That is predictors are not significant in prediction at all. Hence F-7.489, Fos,22-19, value-P(F>7.489)-0.1178

(d) Standardised residual-Residual/Estimated SD(Residual) We compute the residuals as 0.05714286, 0.11428571, -0.71428571,-0.65714286, 1.31428571 Since Var(Residual)-1 - h), Estimated SD(Residual)-\MSE (1 -hu) and we get the standardized residuals as 0.1190983 1190983,-1.1531640,-1.3696301, 1.3696301 Since all the standardized residuals are less than 3, there are no outlier and influential points (e) First we calculate the predicted values 2.057143, 3.885714, 5.714286,7.657143, 6.685714 We calculate the Cook's distances as 0.022852640, 0.002167061, 1.108156028, 3.022261623, 0.286593775 We find that the distance for the 3rd and 4th observations are more than unity and hence are influential

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