When \(\bar{x}=0\) in simple linear regression, \(\hat{\beta}_0=\bar{Y}\) and the formula for \(\hat{\beta}_1\) simplifies to \(\hat{\beta}_1=\frac{\sum x_iY_i}{\sum x_i^2}\).
If the observed value of \((\hat{\beta}_0,\hat{\beta}_1)\) is \((0,-0.11)\), then \((a,b)=(-4.6,4.3)\)
If \((a,b)=(0,1)\), then the observed value of \((\hat{\beta}_0,\hat{\beta}_1)\) is \((0.26,0.20)\)
\(Cov(\overline{Y},\hat{\beta}_0)=7.2\), where \(\overline{Y}=\dfrac{1}{5}\sum_{i=1}^{5}Y_i\)
\(Var(\hat{\beta}_0)=2Var(\hat{\beta}_1)\)
Show Solution
The Correct Option isA, C, D
Solution and Explanation
Step 1: Use $\bar x=0$.
The $x$ values $-2,-1,0,1,2$ give $\bar x=0$ and $S_{xx}=10$. Then $\hat\beta_0=\bar y$ and $\hat\beta_1=\frac{\sum x_iy_i}{10}$.
Step 2: Check (A).
From $\hat\beta_0=0$ we get $a+b=-0.3$; from $\hat\beta_1=-0.11$ we get $b-a=8.9$. Solving, $a=-4.6,\,b=4.3$. (A) holds.
Step 3: Check (B).
With $(a,b)=(0,1)$, $\hat\beta_0=0.26$ but $\hat\beta_1=\frac{1-10}{10}=-0.9$, not $0.20$. (B) fails.
Step 4: Check (C) and (D).
Since $\hat\beta_0=\bar Y$, $\mathrm{Cov}(\bar Y,\hat\beta_0)=\mathrm{Var}(\bar Y)=\frac{36}5=7.2$, so (C) holds. Also $\mathrm{Var}(\hat\beta_1)=\frac{36}{10}=3.6$ and $\mathrm{Var}(\hat\beta_0)=\frac{36}5=7.2=2(3.6)$, so (D) holds.
