Answer
$ŷ=279.0+0.07x_1+1.146x_2+0.390x_3-2.938x_4$
There is enough evidence to conclude that exists a linear relation between at least one of the explanatory variables and the response variable.
The slope coefficients of $x_1$ and $x_3$ are not significantly different from zero.
Work Step by Step
In MINITAB, enter the $x_1$ values in C1, the $x_2$ values in C2, the $x_3$ values in C3, the $x_4$ values in C4 and the $y$ values in C5.
Select Stats -> Regression -> Regression -> Fit Regression Model
Enter C5 in "Responses" and C1 C2 C3 C4 in "Continuous Predictors"
The least-squares regression line will be shown in "Regression Equation", where C5 is $ŷ$, C1 is $x_1$, C2 is $x_2$, C3 is $x_3$ and C4 is $x_4$
$ŷ=279.0+0.07x_1+1.146x_2+0.390x_3-2.938x_4$
$H_0: β_1=β_2=β_3=0$ versus $H_1: at~least~one~β_i\ne0$
$F_0=40.90$ with a P-value $\lt0.001\ltα$. Reject the null hypothesis.
1) $H_0: β_1=0$ versus $H_1: β_1\ne0$
$t_0=0.06$ with a P-value $=0.955\gtα$. Do not reject the null hypothesis.
2) $H_0: β_2=0$ versus $H_1: β_2\ne0$
$t_0=7.41$ with a P-value $\lt0.001\ltα$. Reject the null hypothesis.
3) $H_0: β_3=0$ versus $H_1: β_3\ne0$
$t_0=1.48$ with a P-value $=0.183\gtα$. Do not reject the null hypothesis.
4) $H_0: β_4=0$ versus $H_1: β_4\ne0$
$t_0=-3.57$ with a P-value $=0.009\ltα$. Reject the null hypothesis.
