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. $$\begin{array} { l | r | c | c | c | c }
\text { Source } & \boldsymbol { D } \boldsymbol { F } & \boldsymbol { S S } & \boldsymbol { M S } & \boldsymbol { F } & \boldsymbol { p } \
\hline \text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \
\hline \text { Error } & 16 & 13.925 & 0.870 & & \
\hline \text { Total } & 19 & 27.425 & & &
\end{array}$$
A) 0.011
B) 0.870
C) 5.17
D) 4.500

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. $$\begin{array} { l | r | c | c | c | c } 
\text { Source } & \boldsymbol { D } \boldsymbol { F } & \boldsymbol { S S } & \boldsymbol { M S } & \boldsymbol { F } & \boldsymbol { p } \
\hline \text { Factor } & 3 & 13.500 & 4.500 & 5.17 & 0.011 \
\hline \text { Error } & 16 & 13.925 & 0.870 & & \
\hline \text { Total } & 19 & 27.425 & & &
\end{array}$$
A) 0.011
B) 0.870
C) 5.17
D) 4.500

Quizplus · Accepted Answer

The p-value for the Factor is given in the table as 0.011. This is the probability value associated with the F-statistic and measures the strength of evidence against the null hypothesis. Since the significance level is 0.05, and the p-value is less than 0.05, we reject the null hypothesis and conclude that the different samples come from populations with different means. Thus, the P-value is 0.011.

Given Below Are the Analysis of Variance Results from a Minitab