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help me calculate the answer and give me the step A sample of 25 observations ha...
Aug 1, 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.
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