Homework #4

1. Chapter 7 Problems 7

load data

library(car)
data618 <- read.table("./data/CH06PR18.txt", col.names = c("Y", "X1", "X2", "X3", "X4"))

Problem 7.7(a)

\(SSR(X_4)\): only variable X4 in the model
- Method 1

anova_x4 = Anova(lm(Y~ X4, data=data618))
SSR_X4 = anova_x4[1][1,1]
SSR_X4

## [1] 67.7751

  - Method 2

SSR <- anova(lm(Y~X4, data=data618),lm(Y~1, data=data618))
SSR_X4 = SSR$RSS[2]-SSR$RSS[1]
SSR_X4

## [1] 67.7751

Hence, \(SSR(X_4) = 67.775098\).

\(SSR(X_1|X_4)= SSR(X_1,X_4)-SSR(X_4).\) \(X_1\) is added to the previous model
- Method 1

anova_x1_x4 = Anova(lm(Y~X4+X1, data=data618))
SSR_X1_X4 = anova_x1_x4[1][2,1,drop=FALSE]
SSR_X1_X4

##    Sum Sq
## X1 42.275

  - Method 2

RSS <- anova(lm(Y~X1+X4, data=data618),lm(Y~X4, data=data618))
SSR_X1_X4 = RSS$RSS[2]-RSS$RSS[1]
SSR_X1_X4

## [1] 42.27457

Hence, \(SSR(X_1|X_4)= SSR(X_1,X_4)-SSR(X_4) = 168.782402 -126.5078337= 42.2745683\).

\(SSR(X_2|X_1, X_4) =SSR(X_1, X_2, X_4)-SSR(X_1,X_4)\)
- Method 1

anova_x2_x1x4 = Anova(lm(Y~X4+X1+X2, data=data618))
SSR_X2_X1X4 = anova_x2_x1x4[1][3,1,drop=FALSE]
SSR_X2_X1X4

##    Sum Sq
## X2 27.858

  - Method 2

RSS = anova(lm(Y~X4+X1+X2, data=data618),lm(Y~X1+X4, data=data618))
SSR_X2_X1X4 = RSS$RSS[2]-RSS$RSS[1]
SSR_X2_X1X4

## [1] 27.85749

Hence, \(SSR(X_2|X_1, X_4) =SSR(X_1, X_2, X_4)-SSR(X_1,X_4) = 126.5078337- 98.6503402= 27.8574935\).

\(SSR(X_3|X_1, X_2, X_4) = SSR(X_1,X_2,X_3,X_4)-SSR(X_1,X_2,X_4)\)
- Method 1

anova_x3_x1x4x2 = Anova(lm(Y~X4+X1+X2+X3, data=data618))
SSR_X3_X1X2X4 = anova_x3_x1x4x2[1][4,1,drop=FALSE]
SSR_X3_X1X2X4

##     Sum Sq
## X3 0.41975

    - Method 2

RSS = anova(lm(Y~X1+X2+X3+X4, data=data618),lm(Y~X1+X2+X4, data=data618))
SSR_X3_X1X2X4 = RSS$RSS[2]-RSS$RSS[1]
SSR_X3_X1X2X4

## [1] 0.4197463

Hence, \(SSR(X_3|X_1, X_2, X_4) = SSR(X_1,X_2,X_3,X_4)-SSR(X_1,X_2,X_4) = 98.6503402- 98.2305939= 0.4197463\)

Problem 7.7(b)

Hypothesis
- \(H_0\): \(\beta_3\) = 0, \(H_{\alpha}\): not all \(\beta_3\neq 0\)
- test statistic (\(\alpha =0.01, n=81, p=5\)): \[ F^* = \frac{SSR(X_q,\cdots,X_{p-1}|X_1,\cdots, X_{q-1})}{p-q}\div \frac{SSE(X_1,\cdots,X_{p-1})}{n-p}\]
  - If \(F^* \leq F(1-\alpha; p-q, n-p) = F(0.99, 1, 76)\), conclude \(H_0\),
  - If \(F^* > F(1-\alpha; p-q, n-p) = F(0.99, 1, 76)\) conclude \(H_a\).
\(F^*\) and \(F\)
Since \[ F^* = \frac{0.4197463}{5-4}\div \frac{98.2305939}{81-5}= 0.3247534.\] And
\[ F = F(1-\alpha, p-q, n-p) = F(0.99,1,76) = 6.9805778 \]

\(F^* < F\), hence we conclude \(H_0\).

p-value

pvalue = pf(q=0.3247534, df1=1, df2=76, lower.tail=FALSE)
pvalue

## [1] 0.5704457

Hence the p-value from the F test socre 0.3247534 is 0.5704457 .

2. Chapter 7 Problems 8

Hypothesis
- \(H_0\): \(\beta_2=\beta_3\) = 0, \(H_{\alpha}\): not both \(\beta_i\) = 0 (i=2,3)
- test statistic (\(\alpha =0.01, n=81, p=5\)): \[ F^* = \frac{SSR(X_q,\cdots,X_{p-1}|X_1,\cdots, X_{q-1})}{p-q}\div \frac{SSE(X_1,\cdots,X_{p-1})}{n-p}\]
  - If \(F^* \leq F(1-\alpha; p-q, n-p) = F(0.99, 2, 76)\), conclude \(H_0\),
  - If \(F^* > F(1-\alpha; p-q, n-p) = F(0.99, 2, 76)\) conclude \(H_a\).
\(F^*\) and \(F\)

\(SSR(X_3|X_1, X_2, X_4)\) =

SSR_X3_X1X2X4

## [1] 0.4197463

\(SSR(X_2|X_1, X_4)\) =

SSR_X2_X1X4

## [1] 27.85749

\(SSR(X_2,X_3|X_1, X_4)= SSR(X2|X_1,X_4) +SSR(X_3|X_1,X_2,X_4)=\)

SSR_X2X3_X1X4 = SSR_X2_X1X4+SSR_X3_X1X2X4
SSR_X2X3_X1X4

## [1] 28.27724

\(SSE(X_1,X_2,X_3,X_4)\) =

anova_x3_x1x4x2 = Anova(lm(Y~X4+X1+X2+X3, data=data618))
SSE = anova_x3_x1x4x2[1][5,1,drop=FALSE]
SSE

##           Sum Sq
## Residuals 98.231

Since \(SSR(X_2,X_3|X_1,X_4) = 28.2772397\) and \(SSE(X_1,X_2,X_3,X_4)= 98.2305939\),

\[F^*= \frac{28.2772397}{5-3}\div \frac{98.2305939}{76}=10.9389047\] while

Fvalue = qf(0.99, 2,76)
Fvalue

## [1] 4.89584

Since \(F^*> 4.8958399\), hence we conclude \(H_\alpha\).

Conclusion:
- \(F^*\) = 10.9389 > F(.99, 2, 76), conclude \(H_\alpha\). Not both \(\beta_2\) and \(\beta_3\) = 0
- Not both \(X_2\) and \(X_3\) can be dropped from the regression model given that \(X_2\) and \(X_4\) are retained

3. Chapter 7 Problems 10

Hypothesis

Alternatives:
- \(H_0\): \(\beta_1\) = -0.1 and \(\beta_2\) = 0.4
- \(H_\alpha\): not both equatities are true
Decision rules (\(df_R=81-3=78,df_F=81-5=76\)):
- if \(F^* \leq F(1-\alpha,df_R-df_F, df_F)=F(.99, 2, 76)\), conclude \(H_0\)
- if \(F^* > F(1-\alpha,df_R-df_F, df_F)= F(.99, 2, 76)\), conclude \(H_\alpha\)

Full model

\[ Y = \beta_0 + \beta_1\cdot X_1 + \beta_2\cdot X_2 +\beta_\cdot X_3+\beta_4\cdot X_4\]

Reduced model

The reduced model \(Y' = Y + 0.1\cdot X_1 - 0.4\cdot X_2\)
The data set for reduced model is

Y = data618$Y + data618$X1*.1 - data618$X2*.4
data618_reduced = cbind(Y, data618[,c(4,5)])
data618_reduced

##         Y   X3     X4
## 1  11.592 0.14 123000
## 2  10.124 0.27 104079
## 3  10.900 0.00  39998
## 4  11.120 0.05  57112
## 5  11.512 0.07  60000
## 6   8.220 0.24 101385
## 7  11.000 0.19  31300
## 8  13.952 0.60 248172
## 9  15.120 0.00 215000
## 10 12.588 0.03 251015
## 11 12.352 0.08 291264
## 12 12.080 0.03 207549
## 13 12.348 0.00  82000
## 14 12.604 0.04 359665
## 15 12.621 0.03 265500
## 16 12.296 0.00 299000
## 17 12.904 0.14 189258
## 18 11.693 0.12 366013
## 19 11.832 0.00 349930
## 20 11.020 0.03  85335
## 21 12.040 0.13 235932
## 22 11.968 0.00 130000
## 23 11.092 0.05  40500
## 24 11.512 0.00  40500
## 25  9.752 0.00  45959
## 26 10.600 0.33 120000
## 27 11.544 0.05  81243
## 28 10.898 0.06 153947
## 29 11.016 0.22  97321
## 30 12.308 0.09 276099
## 31 10.770 0.00  90000
## 32 12.285 0.00 184000
## 33 11.105 0.03 184718
## 34 11.228 0.04  96000
## 35 11.360 0.04 106350
## 36 12.330 0.10 135512
## 37 14.150 0.21 180000
## 38 10.988 0.03 315000
## 39 11.184 0.04  42500
## 40  9.720 0.00  30005
## 41 11.898 0.00  60000
## 42 13.672 0.00  73521
## 43 10.500 0.00  50000
## 44 11.710 0.00  50724
## 45 12.200 0.16  31750
## 46 12.152 0.00 168000
## 47 12.776 0.00  70000
## 48 13.300 0.00  27000
## 49 12.166 0.03 129614
## 50 12.050 0.00 129614
## 51 11.850 0.00 130000
## 52 12.398 0.00 209000
## 53 14.704 0.57 220000
## 54 11.652 0.27  60000
## 55 10.648 0.00 110000
## 56 12.192 0.05 101206
## 57 12.180 0.00 288847
## 58 13.276 0.14 105000
## 59 12.060 0.05 276425
## 60 11.434 0.06  33000
## 61 14.604 0.73 210000
## 62 16.318 0.22 240000
## 63 14.706 0.00 281552
## 64 15.206 0.00 421000
## 65 15.470 0.04 484290
## 66 11.368 0.00 234493
## 67 11.168 0.03 230675
## 68 14.412 0.08 296966
## 69 11.922 0.00  32000
## 70 11.300 0.03  38533
## 71 12.956 0.02 109912
## 72 12.748 0.23 236000
## 73 11.458 0.05 243338
## 74 11.672 0.04 122183
## 75 12.262 0.00 128268
## 76 11.164 0.00  72000
## 77 10.532 0.00  43404
## 78 11.860 0.08  59443
## 79 11.712 0.14 254700
## 80 11.842 0.03 434746
## 81 10.828 0.03 201930

Calculate SSE(reduced model): SSE(\(X_1,X_4\)) given known beta2 and beta3

anova_reduced_model = Anova(lm(Y~X3+X4, data=data618_reduced))
anova_reduced_model

## Anova Table (Type II tests)
## 
## Response: Y
##            Sum Sq Df F value    Pr(>F)    
## X3          6.600  1  4.6738   0.03369 *  
## X4         31.872  1 22.5713 9.058e-06 ***
## Residuals 110.141 78                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

SSE_reduced_model = anova_reduced_model[1][3,1,drop=FALSE]
SSE_reduced_model

##           Sum Sq
## Residuals 110.14

SSE(full model): \(SSE(X_1,X_2,X_3,X_4)\) =

SSE

##           Sum Sq
## Residuals 98.231

calculate \(F^*\)

F_star = ((SSE_reduced_model-SSE)/2)/(SSE/76)
F_star[,,drop=TRUE]

## [1] 4.60764

calculate F(.99, 2, 76)

F_2_76 = qf(.99, 2, 76)
F_2_76

## [1] 4.89584

Conclusion

F_star = 4.6076402 < F(.99, 2, 76) = 4.8958399, conclude \(H_0\): \(\beta_1\) = -0.1 and \(\beta_2\) = 0.4.