91影视

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

Here,$\hat{\mathbf{y}}\mathbf{=}\mathbf{80}\mathbf{+}\mathbf{16}\mathbf{.}\mathbf{8}{\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathbf{40}\mathbf{.}\mathbf{4}{\mathit{x}}_{\mathbf{2}}\mathbf{+}\mathit{蔚}$

${\mathit{尾}}_{\mathbf{1}}$and${\mathit{尾}}_{\mathbf{2}}$ denotes difference between the mean levels for different dummy variables.

This means ${\mathit{尾}}_{\mathbf{1}}\mathbf{=}{\mathit{渭}}_{\mathbf{2}}\mathbf{-}{\mathit{渭}}_{\mathbf{1}}$while${\mathit{尾}}_{\mathbf{2}}\mathbf{=}{\mathit{渭}}_{\mathbf{3}}\mathbf{-}{\mathit{渭}}_{\mathbf{1}}$

$\mathbf{}{\mathit{H}}_{\mathbf{0}}\mathbf{}\mathbf{:}{\mathit{尾}}_{\mathbf{1}}\mathbf{=}{\mathit{尾}}_{\mathbf{2}}\mathbf{=}\mathbf{0}$

${\mathit{H}}_{\mathbf{a}}\mathbf{}\mathbf{:}$At least one of parameters${\mathit{尾}}_{\mathbf{1}}$ and${\mathit{尾}}_{\mathbf{2}}$ differs from 0

Here, the null hypothesis becomes that the means for the three groups are equal meaning ${\mathit{渭}}_{\mathbf{1}}\mathbf{=}{\mathit{渭}}_{\mathbf{2}}\mathbf{=}{\mathit{渭}}_{\mathbf{3}}$while the alternate hypothesis implies that at least two of the three means role="math" localid="1649851966245" ${\mathbf{\text{(渭}}}_{\mathbf{1}}\mathbf{,}{\mathit{渭}}_{\mathbf{2}}\mathit{a}\mathit{n}\mathit{d}{\mathit{渭}}_{\mathbf{3}}\mathbf{)}$differ

$\mathbf{}{\mathit{H}}_{\mathbf{0}}\mathbf{}\mathbf{:}{\mathit{尾}}_{\mathbf{1}}\mathbf{=}{\mathit{尾}}_{\mathbf{2}}\mathbf{=}\mathbf{0}$

$\mathbf{}{\mathit{H}}_{\mathit{a}}\mathbf{}\mathbf{:}$At least one of parameters${\mathit{尾}}_{\mathbf{1}}$or${\mathit{尾}}_{\mathbf{2}}$is non zero

Here, F test statistic$\mathbf{=}\frac{\mathbf{S}\mathbf{S}\mathbf{E}}{\mathbf{n}\mathbf{-}\mathbf{k}\mathbf{+}\mathbf{1}}\mathbf{=}\mathbf{24}\mathbf{.}\mathbf{72}$and the p-value is 0

$\mathbf{}{\mathit{H}}_{\mathbf{0}}$is rejected if$\mathit{P}\mathbf{-}\mathit{v}\mathit{a}\mathit{l}\mathit{u}\mathit{e}\mathbf{}\mathbf{<}\mathit{a}$For$\mathit{伪}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$since p- value is less than 0.05

Sufficient evidence to reject$\mathbf{}{\mathit{H}}_{\mathbf{0}}\mathbf{}$at 95% confidence interval.

Therefore, ${\mathit{尾}}_{\mathbf{1}}\mathbf{鈮�}{\mathit{尾}}_{\mathbf{2}}\mathbf{鈮�}\mathbf{0}$Hence two of the three means differ in the model