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help me calculate the answer and give me the step A sample of 25 observations ha...
Aug 3, 2024
help me calculate the answer and give me the step
Solution by Steps
step 1
Identify the given values: β^1=0.14\hat{\beta}_{1} = 0.14, MSE=0.0361MSE = 0.0361, and (X′X)22−1=0.2500(\mathbf{X}'\mathbf{X})^{-1}_{22} = 0.2500
step 2
Calculate the standard error (SE) for β1\beta_{1}: SE(β1)=MSE⋅(X′X)22−1=0.0361⋅0.2500=0.009025=0.095SE(\beta_{1}) = \sqrt{MSE \cdot (\mathbf{X}'\mathbf{X})^{-1}_{22}} = \sqrt{0.0361 \cdot 0.2500} = \sqrt{0.009025} = 0.095
step 3
Determine the critical value for a 95%95\% confidence interval with n−k−1=25−3−1=21n - k - 1 = 25 - 3 - 1 = 21 degrees of freedom. Using a t-table, t0.025,21≈2.080t_{0.025, 21} \approx 2.080
step 4
Calculate the margin of error: ME=t0.025,21⋅SE(β1)=2.080⋅0.095=0.1976ME = t_{0.025, 21} \cdot SE(\beta_{1}) = 2.080 \cdot 0.095 = 0.1976
step 5
Construct the confidence interval: CI=β^1±ME=0.14±0.1976=(−0.0576,0.3376)CI = \hat{\beta}_{1} \pm ME = 0.14 \pm 0.1976 = (-0.0576, 0.3376)
Answer
The 95%95\% confidence interval for β1\beta_{1} is (−0.0576,0.3376)(-0.0576, 0.3376).
(b) Find the test statistics for testing H0:β2=2H_{0}: \beta_{2} = 2.
step 1
Identify the given values: β^2=0.45\hat{\beta}_{2} = 0.45, MSE=0.0361MSE = 0.0361, and (X′X)33−1=5.0625(\mathbf{X}'\mathbf{X})^{-1}_{33} = 5.0625
step 2
Calculate the standard error (SE) for β2\beta_{2}: SE(β2)=MSE⋅(X′X)33−1=0.0361⋅5.0625=0.18275625=0.4275SE(\beta_{2}) = \sqrt{MSE \cdot (\mathbf{X}'\mathbf{X})^{-1}_{33}} = \sqrt{0.0361 \cdot 5.0625} = \sqrt{0.18275625} = 0.4275
step 3
Calculate the test statistic: t=β^2−β2SE(β2)=0.45−20.4275=−1.550.4275=−3.625t = \frac{\hat{\beta}_{2} - \beta_{2}}{SE(\beta_{2})} = \frac{0.45 - 2}{0.4275} = \frac{-1.55}{0.4275} = -3.625
Answer
The test statistic for testing H0:β2=2H_{0}: \beta_{2} = 2 is −3.625-3.625.
(c) Construct a 95%95\% confidence interval for 3β0+5β1+2β23\beta_{0} + 5\beta_{1} + 2\beta_{2}.
step 1
Identify the given values: β^0=−4.04\hat{\beta}_{0} = -4.04, β^1=0.14\hat{\beta}_{1} = 0.14, β^2=0.45\hat{\beta}_{2} = 0.45, MSE=0.0361MSE = 0.0361, and (X′X)−1(\mathbf{X}'\mathbf{X})^{-1}
step 2
Calculate the linear combination: L=3β^0+5β^1+2β^2=3(−4.04)+5(0.14)+2(0.45)=−12.12+0.7+0.9=−10.52L = 3\hat{\beta}_{0} + 5\hat{\beta}_{1} + 2\hat{\beta}_{2} = 3(-4.04) + 5(0.14) + 2(0.45) = -12.12 + 0.7 + 0.9 = -10.52
step 3
Calculate the variance of the linear combination: Var(L)=MSE⋅(3,5,2)⋅(X′X)−1⋅(3,5,2)′Var(L) = MSE \cdot (3, 5, 2) \cdot (\mathbf{X}'\mathbf{X})^{-1} \cdot (3, 5, 2)'
step 4
Compute the variance: Var(L)=0.0361⋅(3,5,2)⋅[188.9832amp;0.8578amp;−28.02750.8578amp;0.2500amp;−0.6−28.0275amp;−0.6amp;5.0625]⋅(3,5,2)′Var(L) = 0.0361 \cdot (3, 5, 2) \cdot \left[\begin{array}{ccc}188.9832 & 0.8578 & -28.0275 \\ 0.8578 & 0.2500 & -0.6 \\ -28.0275 & -0.6 & 5.0625\end{array}\right] \cdot (3, 5, 2)'
step 5
Simplify the matrix multiplication: Var(L)=0.0361⋅(3,5,2)⋅[188.9832⋅3+0.8578⋅5−28.0275⋅20.8578⋅3+0.25⋅5−0.6⋅2−28.0275⋅3−0.6⋅5+5.0625⋅2]Var(L) = 0.0361 \cdot (3, 5, 2) \cdot \left[\begin{array}{c} 188.9832 \cdot 3 + 0.8578 \cdot 5 - 28.0275 \cdot 2 \\ 0.8578 \cdot 3 + 0.25 \cdot 5 - 0.6 \cdot 2 \\ -28.0275 \cdot 3 - 0.6 \cdot 5 + 5.0625 \cdot 2 \end{array}\right]
step 6
Continue simplifying: Var(L)=0.0361⋅(3,5,2)⋅[566.9496+4.289−56.0552.5734+1.25−1.2−84.0825−3+10.125]Var(L) = 0.0361 \cdot (3, 5, 2) \cdot \left[\begin{array}{c} 566.9496 + 4.289 - 56.055 \\ 2.5734 + 1.25 - 1.2 \\ -84.0825 - 3 + 10.125 \end{array}\right]
step 7
Finalize the variance calculation: Var(L)=0.0361⋅(3,5,2)⋅[515.18362.6234−76.9575]Var(L) = 0.0361 \cdot (3, 5, 2) \cdot \left[\begin{array}{c} 515.1836 \\ 2.6234 \\ -76.9575 \end{array}\right]
step 8
Compute the final variance: Var(L)=0.0361⋅(3⋅515.1836+5⋅2.6234+2⋅−76.9575)=0.0361⋅(1545.5508+13.117−153.915)=0.0361⋅1404.7528=50.7126Var(L) = 0.0361 \cdot (3 \cdot 515.1836 + 5 \cdot 2.6234 + 2 \cdot -76.9575) = 0.0361 \cdot (1545.5508 + 13.117 - 153.915) = 0.0361 \cdot 1404.7528 = 50.7126
step 9
Calculate the standard error: SE(L)=Var(L)=50.7126=7.12SE(L) = \sqrt{Var(L)} = \sqrt{50.7126} = 7.12
step 10
Determine the critical value for a 95%95\% confidence interval with 2121 degrees of freedom: t0.025,21≈2.080t_{0.025, 21} \approx 2.080
step 11
Calculate the margin of error: ME=t0.025,21⋅SE(L)=2.080⋅7.12=14.8096ME = t_{0.025, 21} \cdot SE(L) = 2.080 \cdot 7.12 = 14.8096
step 12
Construct the confidence interval: CI=L±ME=−10.52±14.8096=(−25.3296,4.2896)CI = L \pm ME = -10.52 \pm 14.8096 = (-25.3296, 4.2896)
Answer
The 95%95\% confidence interval for 3β0+5β1+2β23\beta_{0} + 5\beta_{1} + 2\beta_{2} is (−25.3296,4.2896)(-25.3296, 4.2896).
Key Concept
Confidence Intervals and Hypothesis Testing in Multiple Linear Regression
Explanation
The problem involves calculating confidence intervals and test statistics for regression coefficients in a multiple linear regression model. This requires understanding the standard errors of the coefficients, the t-distribution, and how to construct confidence intervals and test hypotheses.
Solution by Steps
step 1
Identify the given values: MSE=5MSE = 5, n=43n = 43, R2=0.75R^2 = 0.75, and Var⁡(yi)=19.047\operatorname{Var}(y_i) = 19.047
step 2
Calculate the total sum of squares (SST): SST=Var⁡(yi)⋅(n−1)=19.047⋅42=799.974SST = \operatorname{Var}(y_i) \cdot (n - 1) = 19.047 \cdot 42 = 799.974
step 3
Calculate the regression sum of squares (SSR): SSR=R2⋅SST=0.75⋅799.974=599.9805SSR = R^2 \cdot SST = 0.75 \cdot 799.974 = 599.9805
step 4
Calculate the residual sum of squares (SSE): SSE=SST−SSR=799.974−599.9805=199.9935SSE = SST - SSR = 799.974 - 599.9805 = 199.9935
step 5
Calculate the degrees of freedom for regression (df_{reg}): dfreg=k=2df_{reg} = k = 2 (number of predictors)
step 6
Calculate the degrees of freedom for error (df_{err}): dferr=n−k−1=43−2−1=40df_{err} = n - k - 1 = 43 - 2 - 1 = 40
step 7
Calculate the mean square for regression (MSR): MSR=SSRdfreg=599.98052=299.99025MSR = \frac{SSR}{df_{reg}} = \frac{599.9805}{2} = 299.99025
step 8
Calculate the mean square for error (MSE): MSE=SSEdferr=199.993540=5MSE = \frac{SSE}{df_{err}} = \frac{199.9935}{40} = 5
step 9
Form the ANOVA table: Sourceamp;Sum of Squaresamp;dfamp;Mean Squareamp;FRegressionamp;599.9805amp;2amp;299.99025amp;299.990255=59.99805Erroramp;199.9935amp;40amp;5amp;Totalamp;799.974amp;42amp;amp; \begin{array}{|c|c|c|c|c|} \hline \text{Source} & \text{Sum of Squares} & \text{df} & \text{Mean Square} & F \\ \hline \text{Regression} & 599.9805 & 2 & 299.99025 & \frac{299.99025}{5} = 59.99805 \\ \hline \text{Error} & 199.9935 & 40 & 5 & \\ \hline \text{Total} & 799.974 & 42 & & \\ \hline \end{array}
Answer
The ANOVA table is formed as shown above.
Question (b) Calculate the standard errors of the coefficients.
step 1
Identify the given values: MSE=5MSE = 5, and (t(t-ratios in parentheses) for β^0\hat{\beta}_0, β^1\hat{\beta}_1, and β^2\hat{\beta}_2 are 1.6, 5, and 9 respectively
step 2
Calculate the standard error for β^0\hat{\beta}_0: SE(β^0)=β^0tβ^0=51.6=3.125SE(\hat{\beta}_0) = \frac{\hat{\beta}_0}{t_{\hat{\beta}_0}} = \frac{5}{1.6} = 3.125
step 3
Calculate the standard error for β^1\hat{\beta}_1: SE(β^1)=β^1tβ^1=75=1.4SE(\hat{\beta}_1) = \frac{\hat{\beta}_1}{t_{\hat{\beta}_1}} = \frac{7}{5} = 1.4
step 4
Calculate the standard error for β^2\hat{\beta}_2: SE(β^2)=β^2tβ^2=39=0.3333SE(\hat{\beta}_2) = \frac{\hat{\beta}_2}{t_{\hat{\beta}_2}} = \frac{3}{9} = 0.3333
Answer
The standard errors of the coefficients are SE(β^0)=3.125SE(\hat{\beta}_0) = 3.125, SE(β^1)=1.4SE(\hat{\beta}_1) = 1.4, and SE(β^2)=0.3333SE(\hat{\beta}_2) = 0.3333.
Question (c) Given an out-of-sample observation xh=(1amp;200amp;300)′\mathbf{x}_h = \left(\begin{array}{lll}1 & 200 & 300\end{array}\right)^{\prime}, calculate the prediction of the corresponding YY value.
step 1
Identify the given values: y^=5+7X1+3X2\hat{y} = 5 + 7X_1 + 3X_2, and xh=(1amp;200amp;300)′\mathbf{x}_h = \left(\begin{array}{lll}1 & 200 & 300\end{array}\right)^{\prime}
step 2
Substitute the values into the regression equation: y^h=5+7(200)+3(300)\hat{y}_h = 5 + 7(200) + 3(300)
step 3
Calculate the prediction: y^h=5+1400+900=2305\hat{y}_h = 5 + 1400 + 900 = 2305
Answer
The predicted YY value for the given out-of-sample observation is 23052305.
Key Concept
ANOVA table, standard errors, and prediction in regression analysis
Explanation
The ANOVA table helps in understanding the variance explained by the model. Standard errors of coefficients are crucial for hypothesis testing, and predictions are made using the regression equation with new data points.
Solution by Steps
step 1
To test H0:β1=β2=β12=0H_{0}: \beta_{1}=\beta_{2}=\beta_{12}=0 versus H1:H_{1}: At least one of the βj\beta_{j}'s is nonzero, we use the F-test
step 2
Calculate the total sum of squares SST=20856SS_T = 20856
step 3
Calculate the error sum of squares for Model I, SSE=80.02×(35+27−4)=80.02×58=4641.16SS_E = 80.02 \times (35 + 27 - 4) = 80.02 \times 58 = 4641.16
step 4
Calculate the regression sum of squares for Model I, SSR=SST−SSE=20856−4641.16=16214.84SS_R = SS_T - SS_E = 20856 - 4641.16 = 16214.84
step 5
Calculate the F-statistic: F=SSR/3SSE/(n−p)=16214.84/34641.16/58=5404.9580.02=67.55F = \frac{SS_R / 3}{SS_E / (n - p)} = \frac{16214.84 / 3}{4641.16 / 58} = \frac{5404.95}{80.02} = 67.55
step 6
Compare the F-statistic to the critical value from the F-distribution table with df1=3df_1 = 3 and df2=58df_2 = 58
Part (b)
step 1
To test H0:β2=β12=0H_{0}: \beta_{2}=\beta_{12}=0 versus H1:H_{1}: At least one of the βj\beta_{j}'s is nonzero, we use the F-test
step 2
Calculate the error sum of squares for Model II, SSE=4676SS_E = 4676
step 3
Calculate the regression sum of squares for Model II, SSR=SST−SSE=20856−4676=16180SS_R = SS_T - SS_E = 20856 - 4676 = 16180
step 4
Calculate the F-statistic: F=SSR/2SSE/(n−p)=16180/24676/59=809079.25=102.08F = \frac{SS_R / 2}{SS_E / (n - p)} = \frac{16180 / 2}{4676 / 59} = \frac{8090}{79.25} = 102.08
step 5
Compare the F-statistic to the critical value from the F-distribution table with df1=2df_1 = 2 and df2=59df_2 = 59
Part (c)
step 1
Calculate RAdj2R_{Adj}^{2} for Model I: RAdj2=1−SSE/(n−p)SST/(n−1)=1−4641.16/5820856/61=1−80.02342.72=0.766R_{Adj}^{2} = 1 - \frac{SS_E / (n - p)}{SS_T / (n - 1)} = 1 - \frac{4641.16 / 58}{20856 / 61} = 1 - \frac{80.02}{342.72} = 0.766
step 2
Calculate RAdj2R_{Adj}^{2} for Model II: RAdj2=1−4676/5920856/61=1−79.25342.72=0.768R_{Adj}^{2} = 1 - \frac{4676 / 59}{20856 / 61} = 1 - \frac{79.25}{342.72} = 0.768
step 3
Calculate RAdj2R_{Adj}^{2} for Model III: RAdj2=1−13434/6020856/61=1−223.9342.72=0.346R_{Adj}^{2} = 1 - \frac{13434 / 60}{20856 / 61} = 1 - \frac{223.9}{342.72} = 0.346
step 4
Model II should be most preferred as it has the highest RAdj2R_{Adj}^{2}
Part (d)
step 1
The prediction equation for males (where X2=0X_2 = 0) is Y=β0+β1X1Y = \beta_0 + \beta_1 X_1
step 2
Using the given matrices, solve for β0\beta_0 and β1\beta_1: X′Xβ=X′y\mathbf{X}^{\prime} \mathbf{X} \mathbf{\beta} = \mathbf{X}^{\prime} \mathbf{y}
step 3
β=(X′X)−1X′y=[0.83amp;−0.03amp;0.025−0.03amp;0.003amp;−0.0050.025amp;−0.005amp;0.33][2252539−18]\mathbf{\beta} = (\mathbf{X}^{\prime} \mathbf{X})^{-1} \mathbf{X}^{\prime} \mathbf{y} = \left[ \begin{array}{ccc} 0.83 & -0.03 & 0.025 \\ -0.03 & 0.003 & -0.005 \\ 0.025 & -0.005 & 0.33 \end{array} \right] \left[ \begin{array}{c} 225 \\ 2539 \\ -18 \end{array} \right]
step 4
Calculate β0\beta_0 and β1\beta_1: β0=0.83×225−0.03×2539+0.025×(−18)=186.75\beta_0 = 0.83 \times 225 - 0.03 \times 2539 + 0.025 \times (-18) = 186.75, β1=−0.03×225+0.003×2539−0.005×(−18)=7.62\beta_1 = -0.03 \times 225 + 0.003 \times 2539 - 0.005 \times (-18) = 7.62
step 5
The prediction equation for males is Y=186.75+7.62X1Y = 186.75 + 7.62 X_1
step 6
Estimate the mean systolic blood pressure for males age 35: Y=186.75+7.62×35=453.45Y = 186.75 + 7.62 \times 35 = 453.45
step 7
Determine the 95%95\% confidence interval for the mean systolic blood pressure: CI=Y^±tα/2,n−p⋅SE(Y^)CI = \hat{Y} \pm t_{\alpha/2, n-p} \cdot SE(\hat{Y})
Part (e)
step 1
Determine the prediction interval for the systolic blood pressure of a male age 35: PI=Y^±tα/2,n−p⋅SEpred(Y^)PI = \hat{Y} \pm t_{\alpha/2, n-p} \cdot SE_{pred}(\hat{Y})
Answer
Part (a): F=67.55F = 67.55
Part (b): F=102.08F = 102.08
Part (c): Model II is most preferred with RAdj2=0.768R_{Adj}^{2} = 0.768
Part (d): Prediction equation for males: Y=186.75+7.62X1Y = 186.75 + 7.62 X_1, Mean systolic blood pressure for males age 35: 453.45453.45
Part (e): Prediction interval for systolic blood pressure of a male age 35: PI=453.45±tα/2,n−p⋅SEpred(Y^)PI = 453.45 \pm t_{\alpha/2, n-p} \cdot SE_{pred}(\hat{Y})
Key Concept
Regression Analysis
Explanation
Regression analysis involves fitting models to data to understand relationships between variables and make predictions. The F-test is used to compare models, and RAdj2R_{Adj}^{2} helps in model selection.
Solution by Steps
step 1
To test whether Model 1 fits significantly better than Model 2, we need to perform an F-test. The F-statistic is calculated using the formula: F=(SSE2−SSE1)/(df2−df1)SSE1/df1 F = \frac{(SSE_2 - SSE_1) / (df_2 - df_1)}{SSE_1 / df_1} where SSE1 SSE_1 and SSE2 SSE_2 are the sum of squares of error for Model 1 and Model 2, respectively, and df1 df_1 and df2 df_2 are the degrees of freedom for Model 1 and Model 2, respectively
step 2
Calculate the degrees of freedom for each model. For Model 1, there are 50 observations and 8 parameters (including the intercept), so df1=50−8=42 df_1 = 50 - 8 = 42 . For Model 2, there are 50 observations and 4 parameters (including the intercept), so df2=50−4=46 df_2 = 50 - 4 = 46
step 3
Calculate the F-statistic: F=(900.96−307.64)/(46−42)307.64/42=593.32/4307.64/42=148.337.33=20.23 F = \frac{(900.96 - 307.64) / (46 - 42)}{307.64 / 42} = \frac{593.32 / 4}{307.64 / 42} = \frac{148.33}{7.33} = 20.23
step 4
Compare the calculated F-statistic to the critical value from the F-distribution table with df1=4 df_1 = 4 and df2=42 df_2 = 42 at α=0.05 \alpha = 0.05 . The critical value is approximately 2.61. Since 20.23 > 2.61, we reject the null hypothesis and conclude that Model 1 fits significantly better than Model 2
Part (b)
step 1
To test whether Model 3 fits significantly better than Model 4, we need to perform an F-test. The F-statistic is calculated using the formula: F=(SSE4−SSE3)/(df4−df3)SSE3/df3 F = \frac{(SSE_4 - SSE_3) / (df_4 - df_3)}{SSE_3 / df_3} where SSE3 SSE_3 and SSE4 SSE_4 are the sum of squares of error for Model 3 and Model 4, respectively, and df3 df_3 and df4 df_4 are the degrees of freedom for Model 3 and Model 4, respectively
step 2
Calculate the degrees of freedom for each model. For Model 3, there are 50 observations and 4 parameters (including the intercept), so df3=50−4=46 df_3 = 50 - 4 = 46 . For Model 4, there are 50 observations and 5 parameters (including the intercept), so df4=50−5=45 df_4 = 50 - 5 = 45
step 3
Calculate the F-statistic: F=(941.25−707.42)/(45−46)707.42/45=233.83/1707.42/45=233.8315.72=14.88 F = \frac{(941.25 - 707.42) / (45 - 46)}{707.42 / 45} = \frac{233.83 / 1}{707.42 / 45} = \frac{233.83}{15.72} = 14.88
step 4
Compare the calculated F-statistic to the critical value from the F-distribution table with df1=1 df_1 = 1 and df2=45 df_2 = 45 at α=0.05 \alpha = 0.05 . The critical value is approximately 4.06. Since 14.88 > 4.06, we reject the null hypothesis and conclude that Model 3 fits significantly better than Model 4
Part (c)
step 1
To compute R2 R^2 for Model 2, we use the formula: R2=1−SSESST R^2 = 1 - \frac{SSE}{SST} where SSE SSE is the sum of squares of error for Model 2, and SST SST is the total sum of squares
step 2
Given that Var⁡(yi)=2000 \operatorname{Var}(y_i) = 2000 , we can calculate SST SST as: SST=Var⁡(yi)×(n−1)=2000×(50−1)=2000×49=98000 SST = \operatorname{Var}(y_i) \times (n - 1) = 2000 \times (50 - 1) = 2000 \times 49 = 98000
step 3
Calculate R2 R^2 : R2=1−900.9698000=1−0.00919=0.99081 R^2 = 1 - \frac{900.96}{98000} = 1 - 0.00919 = 0.99081
Answer
Model 1 fits significantly better than Model 2. Model 3 fits significantly better than Model 4. The R2 R^2 of Model 2 is 0.99081.
Key Concept
F-test and R2 R^2 calculation
Explanation
The F-test is used to compare the fits of two models by examining the ratio of their mean squared errors. R2 R^2 measures the proportion of variance in the dependent variable that is predictable from the independent variables.
Solution by Steps
step 1
We start with the given information: n=20n=20, k=3k=3, R2=0.5R^2=0.5 for the original model, and R2=0.52R^2=0.52 for the new model
step 2
The formula for adjusted R2R^2 is RAdj2=1−((1−R2)(n−1)n−k−1)R_{Adj}^2 = 1 - \left( \frac{(1-R^2)(n-1)}{n-k-1} \right)
step 3
For the original model, RAdj,original2=1−((1−0.5)(20−1)20−3−1)=1−(0.5⋅1916)=1−0.59375=0.40625R_{Adj, \text{original}}^2 = 1 - \left( \frac{(1-0.5)(20-1)}{20-3-1} \right) = 1 - \left( \frac{0.5 \cdot 19}{16} \right) = 1 - 0.59375 = 0.40625
step 4
For the new model, RAdj,new2=1−((1−0.52)(20−1)20−4−1)=1−(0.48⋅1915)=1−0.608=0.392R_{Adj, \text{new}}^2 = 1 - \left( \frac{(1-0.52)(20-1)}{20-4-1} \right) = 1 - \left( \frac{0.48 \cdot 19}{15} \right) = 1 - 0.608 = 0.392
step 5
The percentage change in RAdj2R_{Adj}^2 is calculated as RAdj,new2−RAdj,original2RAdj,original2×100=0.392−0.406250.40625×100=−3.51%\frac{R_{Adj, \text{new}}^2 - R_{Adj, \text{original}}^2}{R_{Adj, \text{original}}^2} \times 100 = \frac{0.392 - 0.40625}{0.40625} \times 100 = -3.51\%
Answer
-3.51%
Question 6
step 1
We are given SSE=220SS_E = 220 for Model A, FF-ratio for testing β2=0\beta_2 = 0 from Model A to Model B is 30.83, and FF-ratio for testing β3=0\beta_3 = 0 from Model B to Model C is 12
step 2
The FF-ratio for testing β2=0\beta_2 = 0 is given by F=(SSE,A−SSE,B)/1SSE,B/(n−k−1)F = \frac{(SS_{E, \text{A}} - SS_{E, \text{B}}) / 1}{SS_{E, \text{B}} / (n - k - 1)}
step 3
Rearranging for SSE,BSS_{E, \text{B}}, we get SSE,B=SSE,A−F⋅SSE,Bn−k−1SS_{E, \text{B}} = SS_{E, \text{A}} - F \cdot \frac{SS_{E, \text{B}}}{n - k - 1}
step 4
Solving for SSE,BSS_{E, \text{B}}, we get SSE,B=SSE,A1+Fn−k−1=2201+30.8320−3−1=2201+1.925=75.29SS_{E, \text{B}} = \frac{SS_{E, \text{A}}}{1 + \frac{F}{n - k - 1}} = \frac{220}{1 + \frac{30.83}{20 - 3 - 1}} = \frac{220}{1 + 1.925} = 75.29
step 5
Similarly, for SSE,CSS_{E, \text{C}}, we use the FF-ratio for testing β3=0\beta_3 = 0: F=(SSE,B−SSE,C)/1SSE,C/(n−k−1)F = \frac{(SS_{E, \text{B}} - SS_{E, \text{C}}) / 1}{SS_{E, \text{C}} / (n - k - 1)}
step 6
Rearranging for SSE,CSS_{E, \text{C}}, we get SSE,C=SSE,B−F⋅SSE,Cn−k−1SS_{E, \text{C}} = SS_{E, \text{B}} - F \cdot \frac{SS_{E, \text{C}}}{n - k - 1}
step 7
Solving for SSE,CSS_{E, \text{C}}, we get SSE,C=SSE,B1+Fn−k−1=75.291+1220−4−1=75.291+0.8=41.83SS_{E, \text{C}} = \frac{SS_{E, \text{B}}}{1 + \frac{F}{n - k - 1}} = \frac{75.29}{1 + \frac{12}{20 - 4 - 1}} = \frac{75.29}{1 + 0.8} = 41.83
step 8
The test statistic for testing β2=β3=0\beta_2 = \beta_3 = 0 is F=(SSE,A−SSE,C)/2SSE,C/(n−k−1)=(220−41.83)/241.83/(20−3−1)=178.17/241.83/16=89.0852.614=34.08F = \frac{(SS_{E, \text{A}} - SS_{E, \text{C}}) / 2}{SS_{E, \text{C}} / (n - k - 1)} = \frac{(220 - 41.83) / 2}{41.83 / (20 - 3 - 1)} = \frac{178.17 / 2}{41.83 / 16} = \frac{89.085}{2.614} = 34.08
Answer
34.08
Key Concept
Adjusted R-squared and F-ratio in regression analysis
Explanation
Adjusted R-squared accounts for the number of predictors in the model, and the F-ratio tests the significance of additional predictors.
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