STAT 3A03 Applied Regression With SAS Fall 2017

Transcription

1 STAT 3A03 Applied Regression With SAS Fall 2017 Assignment 2 Solution Set Q. 1 I will add subscripts relating to the question part to the parameters and their estimates as well as the errors and residuals. a) The model is y i = β 0a + β 1a x 1i + ε a;i. From class notes the parameter estimates are ˆβ 1a = S 1y ˆβ0a = y S 1y x 1. Thus the residuals are e a;i = y i ŷ i = y i ˆβ 0a + ˆβ 1a x 1i ) ( = y i y S 1y x 1 + S ) 1y x 1i = (y i y) + S 1y (x 1i x 1 ) [4 marks] b) Now we have the model x 2i = β 0b + β 1b x 1i + ε b;i. The slope of the least squares line is then ˆβ 1b = S 12 Since we know that we have that S 12 = 0 and so r 1,2 = S 12 S11 S 22 = 0 ˆβ 1b = 0 Since ˆβ 1b = 0, we have that ˆβ 0b = x 2 and ˆx 2i = x 2, (i = 1,..., n). Hence the residuals from this model are e b;i = x 2i x 2. 1

2 c) Now we have the model e a;i = β 0c + β 1c e b;i + ε c;i. First we note that e a;i = (y i y) + S 1y e b;i = (x 2i x 2 ) = 0 (x 1i x 1 ) = 0 Hence ˆβ 0c = 0 and ˆβ 1c = ea;i e b;i e 2 b;i Next we have e 2 b;i = e a;i e b;i = = (x 2i x 2 ) 2 = S 22 [ (y i y) + S ] 1y (x 1i x 1 ) (x 2i x 2 ) (x 2i x 2 )(y i y) S 1y (x 1i x 1 )(x 2i x 2 ) = S 2y S 1yS 12 = S 2y r 1,2 S 1y = S 2y Hence the slope of the least squares line is ˆβ 1c = S 2y S 22 which we note is exactly the slope that would be found from fitting the simple regression model with response Y and covariate X 2. [5 marks] d) From (b) we have that the slope of the line is and so the intercept is Hence the residuals become ˆβ 0d ˆβ 1d = r 1,2 = x 2 r 1,2 x 1. e d;i = (x 2i x 2 ) r 1,2 (x 1i x 1 ) To prove that all of the e d;i = 0 we will show that e 2 d;i = 0. Since every e2 d;i 0 then the only way that e 2 d;i can be 0 is if every e2 d;i = 0 and hence e d;i = 0, i = 1,..., n. 2

3 e 2 d;i = = [ (x 2i x 2 ) r 1,2 (x 1i x 1 ) (x 2i x 2 ) 2 2r 1,2 ] 2 (x 1i x 1 )(x 2i x 2 ) + r1,2 2 S 22 (x 1i x 1 ) 2 = S 22 2r 1,2 S 12 + r 2 1,2S 22 S 12 = S 22 2r1,2S r S11 S 1,2S = S 22 2S 22 + S 22 (since r 2 1,2 = 1) = 0 Hence e d;i = 0, i = 1,..., n. [6 marks] Now we recall that the slope of the regression from part (c) requires division by the sum of squares of the residuals from part (b) but we just showed that, when r 1,2 = 0, those residuals satisfy e 2 d;i = 0 and so the slope of the regression line is indeterminate since it requires d division by 0. Q. 2 Textbook 3.5 Here is the SAS code that I used. You can also find this code in the SAS file posted on the website. Note that, for simplicity of code, I am using a single PROC REG to fit all 5 models but it is perfectly acceptable to fit each separately also. The output will be identical other than the naming of the models. Libname S3A3 "D:\STAT 3A03\Fall 2017\Data"; PROC IMPORT out=s3a3.exam datafile="d:\stat 3A03\Fall 2017\Data\Examination.txt" DBMS=DLM REPLACE; Getnames=yes; Datarow=2; run; PROC REG Data=S3A3.exam plots=none; Model F=P1 P2; Model F=P1; Model F=P2; Model P1=P2; Model P2=P1; run; 3

4 SAS Output Page 32 of 36 The SAS output for each of the five models is: Model: MODEL1 Dependent Variable: F Model <.0001 Error Corrected Total Root MSE R-Square Dependent Adj R-Sq Coeff Var SAS Output s Intercept P P Page 33 of 36 Model: MODEL2 Dependent Variable: F Model <.0001 Error Corrected Total file:///c:/users/angelo/appdata/local/temp/sas%20temporary%20files/_td9652_l... Root MSE R-Square Dependent Adj R-Sq Coeff Var s Intercept P <

5 SAS Output Page 34 of 36 Model: MODEL3 Dependent Variable: F Model <.0001 Error Corrected Total Root MSE R-Square Dependent Adj R-Sq Coeff Var s Intercept P <.0001 SAS Output Page 35 of 36 Model: MODEL4 Dependent Variable: P1 Model <.0001 Error Corrected Total file:///c:/users/angelo/appdata/local/temp/sas%20temporary%20files/_td9652_l... Root MSE R-Square Dependent Adj R-Sq Coeff Var s Intercept P <

6 SAS Output Page 36 of 36 Model: MODEL5 Dependent Variable: P2 Model <.0001 Error Corrected Total Root MSE R-Square Dependent Adj R-Sq Coeff Var s Intercept P <.0001 Based on these outputs we have the following estimates using the notation of the question ˆβ 0 = ˆβ 1 = ˆβ 2 = ˆβ 0 = ˆα 0 = ˆβ 1 = ˆα 2 = ˆβ 0 = ˆα 0 = ˆβ 2 = ˆα 1 = a) For this part we have file:///c:/users/angelo/appdata/local/temp/sas%20temporary%20files/_td9652_l ˆβ 1 + ˆβ 2 ˆα 1 = = = ˆβ 1 b) For this part we have [5 marks] ˆβ 2 + ˆβ 1 ˆα 2 = = = ˆβ 2 [5 marks] 6

7 Q. 3 a) b) The fitted values are ˆβ = (X t X) 1 X t Y = = The error sum of squares is then Ŷ = X ˆβ = SSE = (5 2.5) 2 + (2 3.95) 2 + (6 5.4) 2 + (5 6.85) 2 + (9 8.30) 2 +(0 1.9) 2 + (4 3.35) 2 + (5 4.80) 2 + (8 6.25) 2 + (7 7.7) 2 = Hence an unbiased estimate of σ 2 is ˆσ 2 = c) The standard errors for ˆβ 1 and ˆβ 2 are SSE n p 1 = = s.e.( ˆβ 1 ) = = s.e.( ˆβ 1 ) = = The appropriate critical value is t 7;0.025 = 2.36 Hence the confidence interval for β 1 is ˆβ 1 ± 2.36s.e.( ˆβ 1 ) = 0.60 ± = ( 3.243, ) and that for β 2 is ˆβ 2 ± 2.36s.e.( ˆβ 2 ) = 1.45 ± = ( 0.516, ) 7

8 d) The test statistic we must use to test this hypothesis is Hence the observed test statistic value is t = ˆβ 2 1 s.e.( ˆβ 2 ). t obs = = If β 2 = 1 then this would come from a t 7 distribution. The 5% critical value for a two-sided test is then t 7;0.025 = 2.36 and since t obs < t 7;0.025 we fail to reject H 0 and conclude that there is insufficient evidence against the null hypothesis for us to conclude it is false. Hence it is perfectly possible that the true value of β 2 is equal to 1, given the observed data. e) This is estimation of the mean response when x 0 = (1, 1, 3) t. Hence the point estimate is The standard error of ˆµ is s.e.(ˆµ) = ˆσ x t 0 (Xt X) 1 x 0 ˆµ = ˆβ 0 + ˆβ ˆβ 2 = 4.8 [1 mark] = ( ) + 1 ( ) + 3 ( ) = = Hence a 95% confidence interval is ˆµ ± 2.36s.e.(ˆµ) = 4.8 ± = ( 2.931, ) [1 mark] f) In this case we are predicting a single value with x 0 = (1, 0, 5) t. The point predictor is ŷ 0 = ˆβ ˆβ 2 = 8.3 The standard error of prediction is s.e.(ŷ 0 ) = ˆσ 1 + x t 0 (Xt X) 1 x 0 [1 mark] = ( ) + 0 ( ) + 5 ( ) = = Hence a 95% prediction interval is ŷ 0 ± 2.36s.e.(ŷ 0 ) = 8.3 ± = ( 3.355, ) [1 mark] 8

9 Q. 4 a) a = p = 6 b = n p 1 = = 17 c = n 1 = 23 d = a = e = = f = e/b = /17 = g = /f = / = b) The null and alternative hypotheses are [7 marks] c) H 0 : β tax = β age = β bed = β bath = β space = β lot = 0 H 1 : At least one of these parameters is non-zero The F statistic found above has F obs = If H 0 is true then this should be an observation from an F 6,17 distribution. From the table we know F 0.05;6,17 < F 0.05,6,15 = Since F obs > 2.79 we reject H 0 and conclude that at least one of the 6 variables is related to sale price. ˆσ = MSE = = d) The required correlation is the square root of the coefficient of determination, R 2, and must be positive. r = R 2 = = e) The null and alternative hypotheses being tested here are H 0 : β age = β bath = β space = β lot = 0 H 1 : At least one of these parameters is non-zero The test statistic is F = (SSE red SSE full )/(df red df full ) MSE full = ( )/ = If H 0 is true then F F 4,17. From the table we see that the critical value is F 0.05;4,17 > F 0.05;4,15 = Since < 3.06 we cannot reject the null hypothesis and so we conclude that taxes and number of bedrooms alone are sufficient in the model for sale price of a house. 9