Test of Regression AnalysisTime: 9:30 am – 12:30 pm, Dec 18 (Toronto, Canada) Please check the time zone first3 hours test, may has 5-7 questionsPlease take a look at the sample I have provided first.I need someone who has learned this knowledge and has the confidence to get good gradesAlso need someone is sure not to cancel this question after I confirmMore notes will be provide laterI will send you message before I choose you, please check it.
[12] 1. In a set of observations (x1, y1), . . . , (xn, yn) the regression modelholds for (yi|xi)i.i.d. ⇠ N(g(xi), s2). In other words we can writeyi = g(xi) + ei, i = 1, 2, . . . , n,where e1i.i.d. ⇠ N(0, s2). Letg(x) =⇢b0 + b1x if x < c b0 + b2x if x ” c. In here c is a known constant. Derive the M.L.E. for b0, b1 and b2. Explain how to find an unbiased estimate for s2. Calculate the distribution for [ˆb0 b1 b2]0 the M.L.E. estimator. Hint: notice that the formulas based on matrices and vectors will be sufficient. 3 [10] 2. For the data set x 0 1 2 1 3 2 3 3 2 1 2 y 1 3 9 3 28 10 27 28 10 3 9 the following relationship holds approximately. y ⇡ aebx. Suggest a proper linear regression model to estimate a and b. Calculate ˆa and ˆb. Assuming the regular assumptions in a simple linear regression model, derive a 95% confidence interval for b. Then calculate the p-value for testin the hypothesis H0 : b = 3 against H1 : b > 3.4
The following R output considers the model with 4 predictors x1, x2, x3, x4.In other wordsyi = b0 + b1xi1 + b2xi2 + b3xi3 + b4xi4 + ei, i = 1, 2, . . . , 20.where eii.i.d. ⇠ N(0, s2).
C=solve(t(X)%*%X)Cx1 x2 x3 x40.235054177 -0.070487766 -0.011801313 0.009556365 0.005769428×1 -0.070487766 0.046549567 -0.001456669 -0.010056232 -0.011507676×2 -0.011801313 -0.001456669 0.009062628 -0.003428382 -0.001280193×3 0.009556365 -0.010056232 -0.003428382 0.013065887 -0.001222107×4 0.005769428 -0.011507676 -0.001280193 -0.001222107 0.014260505anova(m1)Analysis of Variance TableResponse: yDf Sum Sq Mean Sq F value Pr(>F)x1 1 3713.2 3713.2 598.2294 1.678e-13 ***x2 1 2175.2 2175.2 350.4406 8.208e-12 ***x3 1 1383.8 1383.8 222.9419 2.071e-10 ***x4 1 0.0 0.0 0.0016 0.9687Residuals 15 93.1 6.25summary(m1)Call:lm(formula = y ~ x1 + x2 + x3 + x4)Residuals:Min 1Q Median 3Q Max-3.0156 -1.9295 -0.1384 1.2438 5.9502Coefficients:Estimate Std. Error t value Pr(>|t|)(Intercept) -0.90379 1.20788 -0.748 0.466×1 2.96103 0.53753 5.509 6.01e-05 ***x2 3.00965 0.23717 12.690 2.01e-09 ***x3 4.23403 0.28478 14.868 2.20e-10 ***
x4 0.01188 0.29751 0.040 0.969
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 18 Use the R out put to construct a 95% confidence interval for b2 − b4in this. Use this confidence interval to testH0 : b2 = b4 against H1 : b2 6= b46at a = 5% level.78 Test the following hypothesisH0 : b1 − b2 = 0, b0 = 0against all its alternatives in model m1 (see part (i)).88 Noticing the R out-put, we decide to examine the following modelin R. Here are the results.X=cbind(x1,x2,x3)C=solve(t(X)%*%X)Cx1 x2 x3x1 0.018640682 -0.005681588 -0.008199246×2 -0.005681588 0.008400630 -0.003050782×3 -0.008199246 -0.003050782 0.012527076m2=lm(y~x1+x2+x3-1)summary(m2)Call:lm(formula = y ~ x1 + x2 + x3 – 1)Residuals:Min 1Q Median 3Q Max-3.0042 -1.9644 -0.2292 0.8901 5.9626Coefficients:Estimate Std. Error t value Pr(>|t|)x1 2.7136 0.3256 8.335 2.08e-07 ***x2 2.9667 0.2186 13.574 1.49e-10 ***9
x3 4.2743 0.2669 16.015 1.09e-11 ***
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1Residual standard error: 2.385 on 17 degrees of freedomMultiple R-squared: 0.9952,Adjusted R-squared: 0.9943F-statistic: 1163 on 3 and 17 DF, p-value: < 2.2e-16 anova(m2) Analysis of Variance Table Response: y Df Sum Sq Mean Sq F value Pr(>F)x1 1 16289.8 16289.8 2864.89 < 2.2e-16 ***x2 1 2097.4 2097.4 368.86 5.797e-13 ***x3 1 1458.4 1458.4 256.49 1.090e-11 ***
Residuals 17 96.7 5.7
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1AIC(m2)[1] 96.26738AIC(m1)[1] 99.51747108 Write the model in m2 and explain why this model m2 should bepreferred to model m1. Explain your conclusion based on numbers yousee in R out-put. Based on this model (model m2) find a 95% predictioninterval for y given that x1 = 0, x2 = 1, x3 = 1.11[5] 3. Explain if the following statement is right or wrong.”We always choose a model with smaller sum of squared of residuals(SSRes) and larger coefficient of determination (R2).12[12] 4. In a general mutlpile linear regressiony = Xb + ewhere X(n⇥(k +1) and b(k +1)⇥1, with e ⇠ Nk(0, s2 I) we deciden notto use L.S.E. or M.L.E. due to overfitting. To avoid overfitting instead wedecide to estimate b by minimizingg(b) = (y − Xb)0 (y − Xb0) + b0b.Derive the estimate for b by minimizing g(b). Denote your estimate by ˜b.Is this an unbiased estimator? Why? Find the distribution for ˜b under thisnew minimization rule.13
Consider the regular multiple regressiony = Xb + ewhere E(e) = 0 , Cov(e) = s2 I. Let A(n⇥n) be an orthogonal matrix (i.e.,AA0 = A0A = I). We can writey⇤ = Ay, X⇤ = AX, e⇤ = Ae.Prove7 E(e⇤) = 0 and Cov(e⇤) = s2 I.147 Show that ˆb⇤ = ˆb and Âni=1 e2i = Âni=1 e⇤2i15
Label each statement as True or False. Justify your answers very briefly.5 If y ⇠ Nn(μ, s2 I) and A(n ⇥ n) is an orthogonal matrix then Ay ⇠Nn(Aμ, s2 I).5 With the same assumption in part (i) with μ = 0 and s = 1 theny0By ⇠ c2(r)if and only if B is idempotent and tr(B) = 2r.5 If y ⇠ Nn(0, s2 I) theny01(101)−11y/s2 ⇠ c2(1),where 10 = [1 1 1 · · · 1](1 ⇥ n).16
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