Answer
(a) $β_0=3.6$ and $β_1=-1.8$.
(b) $s_e=0.5164$
(c) $s_{b_1}=0.1633$
(d) $t_0\lt -t_{\frac{α}{2}}$: 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 10d in section 4.2, $b_0=3.6$ and $b_1=-1.8$.
$b_0$ is the estimate for $β_0$ and $b_1$ is the estimate for $β_1$.
(b) $ŷ =−1.8x+3.6$
$s_e=\sqrt {\frac{Σ(y_i-ŷ_i)^2}{n-2}}=\sqrt {\frac{[7-(−1.8(-2)+3.6)]^2+[6-(−1.8(-1)+3.6)]^2+[3-(−1.8\times0+3.6)]^2+[2-(−1.8\times1+3.6)]^2+[0-(−1.8\times2+3.6)]^2}{5-2}}=0.5164$
(c) According to problem 10d 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)
$s_{b_1}=\frac{0.5164}{\sqrt {5-1}\times1.581}=0.1633$
(d) $H_0: β_1=0$ versus $H_1: β\ne0$
$t_0=\frac{b_1}{s_{b_1}}=\frac{-1.8}{0.1633}=-11.023$
$n=5$, 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)
Also, $-t_{\frac{α}{2}}=-3.182$
Since $t_0\lt -t_{\frac{α}{2}}$, we reject the null hypothesis.
