Answer
$F_0=5.81$ with a P-value $\ltα$. Reject the null hypothesis.
There is enough evidence to conclude that the mean amount dispensed from the machines is not the same.
Work Step by Step
$H_0:µ_1=µ_2=µ_3=µ_4$ versus
$H_1:$ at least one population mean is different from the others
In MINITAB, enter the Machine I values in C1, the Machine II values in C2, the Machine III values in C3 and in C4 enter the Machine IV values.
Select Stat -> ANOVA -> One-Way
Select Response data are in a separate column for each factor level.
In Responses enter C1 C2 C3 C4
Click Ok.
$F_0=5.81$ with a P-value $=0.004\ltα=0.05$
