Chapter 12.1, Problem 11E

(a)

Section 1:

To determine

To find: The degrees of freedom for the provided statistic.

Answer to Problem 11E

Solution: The degrees of freedom for groups is 4 and for observations is 25.

Explanation of Solution

Calculation: The degrees of freedom for groups is calculated as,

$\begin{array}{c}\text{DFG}=I-1\\ =5-1\\ =4\end{array}$

And the degrees of freedom for errors is computed as,

$\begin{array}{c}\text{DFE}=N-I\\ =\left(5\times 6\right)-5\\ =25\end{array}$

Section 2:

To determine

The critical values corresponding to calculated degrees of freedom.

Answer to Problem 11E

Solution: The $F$ statistic value will lie in between $\underset{\_}{4.18<F<6.49}$.

Explanation of Solution

From the table E, provided in Appendix of the textbook the F-statistic value will lie in between 4.18 and 6.49.

(b)

To determine

To graph: A distribution plot.

Explanation of Solution

Graph: To draw a probability distribution curve follow the below mentioned steps in Minitab,

Step 1: In a Minitab worksheet go to ‘Graph’ and click on Probability Distribution Plot.

Step 2: In the dialogue box that appears select ‘View Probability’ option and click OK.

Step 3: Next select the distribution as F and enter the corresponding numerator and denominator degrees of freedom in the respective fields and click on OK to obtain the required graph.

The graph obtained is attached below,

(c)

To determine

The p-value.

Answer to Problem 11E

Solution: The p-value will lie in between $\underset{\_}{0.001<P<0.01}$.

Explanation of Solution

From part (a) it is observed that the F-statistic value will lie in between 4.18 and 6.49. From table E provided in Appendix of the textbook the p- value corresponding to $4.18<F<6.49$ will fall in the range $0.001<P<0.01$. Thus, corresponding to the provided test statistic the p-value will lie in between 0.001 and 0.01.

(d)

To determine

Weather all pairs of group means differ significantly or not.

Answer to Problem 11E

Solution: It is not necessary that all the group means are different. There may be a single mean that differs.

Explanation of Solution

The p-value obtained in part (c) is very small to reject the null hypothesis. The null hypothesis assumes all the group means to be equal. If the null hypothesis is rejected then it does not implies that all the group means are unequal, there may be only one or more unequal means.