Answer
(a) $β_0=-3.3$ and $β_1=0.780$.
(b) $s_e=1.7416$
(c) $s_{b_1}=0.1101$
(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 12d in section 4.2, $b_0=-3.3$ and $b_1=0.780$.
$b_0$ is the estimate for $β_0$ and $b_1$ is the estimate for $β_1$.
(b) $ŷ =0.780x−3.3$
$s_e=\sqrt {\frac{Σ(y_i-ŷ_i)^2}{n-2}}=\sqrt {\frac{[2-(0.780\times5−3.3)]^2+[4-(0.780\times10−3.3)]^2+[7-(0.780\times15−3.3)]^2+[11-(0.780\times20−3.3)]^2+[18-(0.780\times25−3.3)]^2}{5-2}}=1.7416$
(c) According to problem 12d in section 4.2, $s_x=7.906$
$s_{b_1}=\frac{s_e}{\sqrt {∑(x_i-x ̅)^2}}=\frac{s_e}{\sqrt {n-1}s_x}$ (see note on page 684)
$s_{b_1}=\frac{1.7416}{\sqrt {5-1}\times7.906}=0.1101$
(d) $H_0: β_1=0$ versus $H_1: β\ne0$
$t_0=\frac{b_1}{s_{b_1}}=\frac{0.780}{0.1101}=7.084$
$n=5$, 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.
