Answer
(a) $β_0=1.2$ and $β_1=2.2$.
(b) $s_e=0.8944$
(c) $s_{b_1}=0.2829$
(d) $t_0\gt t_{\frac{α}{2}}$: null hypothesis is rejected.
There is enough evidence to conclude that a linear relation exists between x and y.
Work Step by Step
(a) According to problem 9d in section 4.2, $b_0=1.2$ and $b_1=2.2$.
$b_0$ is the estimate for $β_0$ and $b_1$ is the estimate for $β_1$.
(b) $ŷ =2.2x+1.2$
$s_e=\sqrt {\frac{Σ(y_i-ŷ_i)^2}{n-2}}=\sqrt {\frac{[-4-(2.2\times(-2)+1.2)]^2+[0-(2.2\times(-1)+1.2)]^2+[1-(2.2\times0+1.2)]^2+[4-(2.2\times1+1.2)]^2+[5-(2.2\times2+1.2)]^2}{5-2}}=0.8944$
(c) According to problem 9d in section 4.2, $s_x=1.581$
$s_{b_1}=\frac{s_e}{\sqrt {∑(x_i-x ̅)^2}}=\frac{s_e}{\sqrt {n-1}s_x}$ (see note on page 684)
$s_{b_1}=\frac{0.8944}{\sqrt {5-1}\times1.581}=0.2829$
(d) $H_0: β_1=0$ versus $H_1: β\ne0$
$t_0=\frac{b_1}{s_{b_1}}=\frac{2.2}{0.2829}=7.777$
$n=5$, so:
$d.f.=n-2=3$
$t_{\frac{α}{2}}=t_{0.025}=3.182$
(According to Table VI, for d.f. = 5 and area in right tail = 0.025)
Since $t_0\gt t_{\frac{α}{2}}$, we reject the null hypothesis.
