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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.
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