Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance
level in testing the null hypothesis that the different samples come from populations with the same mean.
-Identify the p-value. $$\begin{array} { l r c l c c }
\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \
\text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \
\text { Error } & 16 & 13.925 & 0.870 & & \
\text { Total } & 19 & 27.425 & & &
\end{array}$$
A) 5.17
B) 0.870
C) 4.500
D) 0.011

Question

Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance
level in testing the null hypothesis that the different samples come from populations with the same mean.
-Identify the p-value. $$\begin{array} { l r c l c c } 
\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \
\text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \
\text { Error } & 16 & 13.925 & 0.870 & & \
\text { Total } & 19 & 27.425 & & &
\end{array}$$
A) 5.17
B) 0.870
C) 4.500
D) 0.011

Quizplus · Accepted Answer

The p-value is given in the table as 0.011 for the factor (also known as treatment) source. This value is less than the significance level of 0.05, so we reject the null hypothesis of equal population means and conclude that at least one population mean is different from the others.

Given Below Are the Analysis of Variance Results from a Minitab