91Ó°ÊÓ

From the excel printout, the values of the coefficients can be used to write the least square prediction equation for the model

Here,

$y=80+16.8x_1+40.4x_2+\mathrm{Î\mu }$

${\mathrm{β}}_1$and${\mathrm{β}}_2$denotes difference between the mean levels for different dummy variables.

This means ${\mathrm{β}}_1={\mathrm{μ}}_2-{\mathrm{μ}}_1$while${\mathrm{β}}_3={\mathrm{μ}}_3-{\mathrm{μ}}_1$

$H_0:{\mathrm{β}}_1={\mathrm{β}}_2=0$

$H_a:$At least one of parameters${\mathrm{β}}_1$ and ${\mathrm{β}}_2$differs from 0

Here, the null hypothesis becomes that the means for the three groups are equal meaning${\mathrm{μ}}_1={\mathrm{μ}}_2={\mathrm{μ}}_3$ while the alternate hypothesis implies that at least two of the three means$\left({\mathrm{μ}}_1,{\mathrm{μ}}_2,{\mathrm{μ}}_3\right)$ differ.

$H_0:{\mathrm{β}}_1={\mathrm{β}}_2=0$

$H_a$At least one of parameters${\mathrm{β}}_1$or role="math" localid="1651888100746" ${\mathrm{β}}_2$is non zero

Here, F test statistic$=\frac{SSE}{\left(n-k\right)}+1$and the p-value is 0

$H_0$is rejected if p- value$<\mathrm{Î\pm }$,For since $\mathrm{Î\pm }=0.05$then $p$is less than 0.05

Sufficient evidence to reject$H_0$ at 95% confidence interval.

Therefore,Hence ${\mathrm{β}}_1≠{\mathrm{β}}_2$two of the three means differ in the model