Chapter 13, Problem 1RE

(a)

To determine

The critical value for $\alpha =0.05,\nu =16,k=7$.

Answer to Problem 1RE

Solution: The required value is 4.741.

Explanation of Solution

Given: The significance level is 0.05. The total number of treatments is $k=7$ and the error degrees of freedom are $\nu =16$.

The tabulated value is obtained from Table X in the appendix of the textbook. The row entries correspond to $\nu$ and the column entries correspond to $k$. The value at the intersection of $\left(16,7\right)$ is 4.741, which is the required critical value.

(b)

To determine

The critical value for $\alpha =0.05,\nu =30,k=6$.

Answer to Problem 1RE

Solution: The required value is 4.302.

Explanation of Solution

Given: The total number of treatments is $k=6$ and the error degrees of freedom are $\nu =30$. The significance level is $\alpha =0.05$.

The value corresponding to row 30 and column 6 in Table X of the textbook is 4.302. The rows contain the values of error degrees of freedom and the column represents the values for the number of treatments.

(c)

To determine

The critical value for $\alpha =0.01,\nu =42,k=4$.

Answer to Problem 1RE

Solution: The value is 4.696.

Explanation of Solution

Given: There are a total of four treatments with the provided error degrees of freedom as 42. The confidence level is 1%.

The table for the critical values of Tukey’s test for 99% confidence level is provided in the appendix of the textbook. The critical value corresponding to row 42 and column 4 is 4.696, which is the required critical value.

(d)

To determine

The critical value for $H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4={\mu }_5={\mu }_6,$ with $n=46,\alpha =0.05$.

Answer to Problem 1RE

Solution: The required value is 4.232.

Explanation of Solution

Given: It is provided that a total of six treatment means are to be compared for a sample of size 46 at 5% level of significance.

According to the provided information, $k=6\text{ and }n=46$. The error degrees of freedom are calculated as follows:

$\begin{array}{c}\nu =n-k\\ =46-6\\ =40\end{array}$

The required critical value is the value corresponding to row 40 and column 6, and 4.232 is the required value.