Answer
(a) $β_0=116.6$ and $β_1=-0.72$.
(b) $s_e=3.2863$
(c) $s_{b_1}=0.1039$
(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, $b_0=116.6$ and $b_1=-0.72$.
$b_0$ is the estimate for $β_0$ and $b_1$ is the estimate for $β_1$.
(b) $ŷ =−0.72x+116.6$
$s_e=\sqrt {\frac{Σ(y_i-ŷ_i)^2}{n-2}}=\sqrt {\frac{[100-(−0.72\times20+116.6)]^2+[95-(−0.72\times30+116.6)]^2+[91-(−0.72\times40+116.6)]^2+[83-(−0.72\times50+116.6)]^2+[70-(−0.72\times60+116.6)]^2}{5-2}}=3.2863$
(c) According to problem 11d in section 4.2, $s_x=15.811$
$s_{b_1}=\frac{s_e}{\sqrt {∑(x_i-x ̅)^2}}=\frac{s_e}{\sqrt {n-1}s_x}$ (see note on page 684)
$s_{b_1}=\frac{3.2863}{\sqrt {5-1}\times15.811}=0.1039$
(d) $H_0: β_1=0$ versus $H_1: β\ne0$
$t_0=\frac{b_1}{s_{b_1}}=\frac{-0.72}{0.1039}=-6.930$
$n=5$, 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.
