Chapter 13.2, Problem 1E

The following data are from a completely randomized design.

	Treatment
	A	B	C
	162	142	126
	142	156	122
	165	124	138
	145	142	140
	148	136	150
	174	152	128
Sample mean	156	142	134
Sample variance	164.4	131.2	110.4

1. a. Compute the sum of squares between treatments.
2. b. Compute the mean square between treatments.
3. c. Compute the sum of squares due to error.
4. d. Compute the mean square due to error.
5. e. Set up the ANOVA table for this problem.
6. f. At the α = .05 level of significance, test whether the means for the three treatments
7. g. are equal.

To determine

Compute the sum of squares between the treatments.

Answer to Problem 1E

The sum of squares between the treatments is 1,488.

Explanation of Solution

Calculation:

The data represents the completely randomized design for three treatments A, B and C.

The value of $\overline{\overline{x}}$ is,

$\begin{array}{c}\overline{\overline{x}}=\frac{156+142+134}{3}\\ =144\end{array}$

The sum of squares between the treatments is,

$\begin{array}{c}SSTR={\displaystyle \sum _{j=1}^k{n}_j}{\left({\overline{x}}_j-\overline{\overline{x}}\right)}^2\\ =6{\left(156-144\right)}^2+6{\left(142-144\right)}^2+6{\left(134-144\right)}^2\\ =1,488\end{array}$

Thus, the sum of squares between the treatments is 1,488.

To determine

Compute the mean square between the treatments.

Answer to Problem 1E

The mean square between the treatments is 744.

Explanation of Solution

Calculation:

The mean square between the treatments is,

$\begin{array}{c}MSTR=\frac{SSTR}{k-1}\\ =\frac{1488}{3-1}\\ =\frac{1488}{2}\\ =744\end{array}$

Thus, the mean square between the treatments is 744.

To determine

Compute the sum of squares due to error.

Answer to Problem 1E

The sum of squares due to error is 2,030.

Explanation of Solution

Calculation:

The sum of squares due to error is,

$SSE={\displaystyle \sum _{j=1}^k\left(n_j-1\right)}{\left(s_j\right)}^2$

Substitute the values $s_1^2=164.4,s_2^2=131.2s_3^2=110.4$.

$\begin{array}{c}SSE={\displaystyle \sum _{j=1}^k\left(n_j-1\right)}{\left(s_j\right)}^2\\ =5\left(164.4\right)+5\left(131.2\right)+5\left(110.4\right)\\ =2,030\end{array}$

Thus, the sum of squares due to error is 2,030.

To determine

Compute the mean square due to error.

Answer to Problem 1E

The mean square due to error is 135.3.

Explanation of Solution

Calculation:

The mean square due to error is,

$\begin{array}{c}MSE=\frac{SSE}{n_T-k}\\ =\frac{2,030}{\left(12-3\right)}\\ =135.3\end{array}$

Thus, the mean square due to error is 135.3.

To determine

Develop ANOVA table for the problem.

Explanation of Solution

Calculation:

From parts (a), (b), (c) and (d) the values required for the ANOVA table are,

$\text{SSTR}=1,488,\text{SSE}=2,030,\text{SST}=3,518,$$\text{df}\left(\text{treatments}\right)=2,\text{df}\left(\text{error}\right)=15$$\text{df}\left(\text{total}\right)=17,\text{MSTR}=744,\text{MSE}=135.3$.

The F-ratio is obtained as,

$\begin{array}{c}F=\frac{MSTR}{MSE}\\ =\frac{744}{135.3}\\ =5.50\end{array}$

Thus, the ANOVA table is,

Source	df	SS	MS	F
Treatments	2	1,488	744	5.50
Error	15	2,030	135.3
Total	17	3,518

To determine

Check whether there is a significant difference between the treatment means at $\alpha =0.05$ level of significance.

Answer to Problem 1E

There is no evidence that the means for the three treatments are equal.

Explanation of Solution

Calculation:

State the hypotheses:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3$

Alternative hypothesis:

$H_a:$ Not all the population means are equal.

Critical value:

The critical value is obtained from the F table with 2 degrees of freedom for the numerator and 15 for the denominator.

Step by step procedure to obtain F-value using Table 4 of Appendix B is given below:

• Locate the value 2 and $\alpha =0.05$ in the right column of the Table 4 of Appendix B.
• Go through the row corresponding to the value 2 and column corresponding to the value 15 of the Table 4.
• Locate the value corresponding to (2, 15).

The required critical value for the F-distribution with level of significance 0.05, $d{f}_1=2$ and $d{f}_2=15$ is 3.68.

Rejection rule:

If $F\ge 3.68$, reject the null hypothesis $H_0$.

Conclusion:

Here, the F-test statistic is 5.50.

The test statistic is greater than the critical value.

Therefore, the null hypothesis is rejected.

Thus, there is no enough evidence that the means for the three treatments are equal.