Chapter 9.4, Problem 3E

a.

To determine

Construct an ANOVA table and give a range for the P-value.

Answer to Problem 3E

The ANOVA table is,

Source	DF	SS	MS	F	P
Lighting	3	9,943	3,314.33	3.3329	$0.01<P\text{-value <}0.05$
Block	2	11,432	5,716.00	5.7481	$0.001<P\text{-value <}0.010$
Interaction	6	6,135	1,022.50	1.0282	$P\text{-value >0}\text{.100}$
Error	24	23,866	994.417
Total	35	51,376

Explanation of Solution

Calculation:

It is given that there are four lighting methods, levels for block is three and the number of replication is three. The observed sum of squares are $\text{SSB}=11,432,\text{SSA}=9,943,\text{SSAB}=6,135\text{and SST}=51,376$.

Factor A is treatment and B is block.

The ANOVA table can be obtained as follows:

The level of significance is 0.05.

The number of levels of the lighting methods is denoted as I.

Here $I=4$.

The number of degrees of freedom for the lighting methods is,

$\begin{array}{c}I-1=4-1\\ =3\end{array}$

Thus, the number of degrees of freedom for the lighting methods is 3.

The number of levels for blocks is denoted as J.

Here $J=3$.

The number of degrees of freedom for blocks is,

$\begin{array}{c}J-1=3-1\\ =2\end{array}$

Thus, the number of degrees of freedom for blocks is 2.

The number of degrees of freedom for interactions is,

$\begin{array}{c}\left(I-1\right)\left(J-1\right)=\left(4-1\right)\left(3-1\right)\\ =\left(3\right)\left(2\right)\\ =6\end{array}$

Thus, the number of degrees of freedom for interactions is 6.

The number of replication is denoted as K.

Here $K=3$.

The number of degrees of freedom for error is,

$\begin{array}{c}IJ\left(K-1\right)=\left(4\right)\left(3\right)\left(3-1\right)\\ =\left(12\right)\left(2\right)\\ =24\end{array}$

Thus, the number of degrees of freedom for error is 24.

The number of degrees of freedom for total is, $3+2+6+24=35$.

The mean square of A can be obtained as follows:

$\text{MSA}=\frac{\text{SSA}}{I-1}$

Substitute $\text{SSA}=9,943\text{and I=4}$ in the above equation.

$\begin{array}{c}\text{MSA}=\frac{9,943}{4-1}\\ =\frac{9,943}{3}\\ =3,314.33\end{array}$

Thus, the value of $\text{MSA}=3,314.33$.

The mean square of B can be obtained as follows:

$\text{MSB}=\frac{\text{SSB}}{J-I}$

Substitute $\text{SSB}=11,432\text{and J=3}$ in the above equation.

$\begin{array}{c}\text{MSB}=\frac{11,432}{3-1}\\ =\frac{11,432}{2}\\ =5,716\end{array}$

Thus, the value of $\text{MSB}=5,716$.

The mean square of AB can be obtained as follows:

$\text{MSAB}=\frac{\text{SSAB}}{\left(I-1\right)\left(J-I\right)}$

Substitute $\text{SSAB}=6,135,I=4\text{and J=3}$ in the above equation.

$\begin{array}{c}\text{MSAB}=\frac{6,135}{\left(4-1\right)\left(3-1\right)}\\ =\frac{6,135}{\left(3\right)\left(2\right)}\\ =\frac{6,135}{6}\\ =1,022.5\end{array}$

Thus, the value of $\text{MSAB}=1,022.5$.

The sum of squares of error can be obtained as follows:

$\text{SSE}=\text{SST}-\text{SSA}-\text{SSB}-\text{SSAB}$

Substitute $\text{SSB}=11,432,\text{SSA}=9,943,\text{SSAB}=6,135\text{and SST}=51,376$ in the above equation.

$\begin{array}{c}SSE=51,376-9,943-11,432-6,135\\ =23,866\end{array}$

Thus, the sum of squares of error is $\text{SSE}=23,866$.

The mean square of error can be obtained as follows:

$\text{MSE}=\frac{\text{SSE}}{IJ\left(K-I\right)}$

Substitute $\text{SSE}=23,866,I=4,J=3\text{and K=3}$ in the above equation.

$\begin{array}{c}\text{MSE}=\frac{23,866}{\left(4\right)\left(3\right)\left(3-1\right)}\\ =\frac{23,866}{\left(12\right)\left(2\right)}\\ =\frac{23,866}{24}\\ =994.417\end{array}$

Thus, the value of $\text{MSE}=994.417$.

The F-value can be obtained by dividing each mean square by the mean square for error:

The F-value corresponding to mold temperature is,

$F=\frac{\text{MSA}}{\text{MSE}}$

Substitute $\text{MSA}=3,314.33\text{and MSE}=994.417$ in the above equation.

$\begin{array}{c}F=\frac{3,314.33}{994.417}\\ =3.3329\end{array}$

Thus, the F-value is 3.3329.

The F-value corresponding to block is,

$F=\frac{\text{MSB}}{\text{MSE}}$

Substitute $\text{MSB}=5,716\text{and }MSE=994.417$ in the above equation.

$\begin{array}{c}F=\frac{5,716}{994.417}\\ =5.7481\end{array}$

Thus, the F-value is 5.7481.

The F-value corresponding to interaction is,

$F=\frac{\text{MSAB}}{\text{MSE}}$

Substitute $\text{MSAB}=1,022.5\text{and MSE}=994.417$ in the above equation.

$\begin{array}{c}F=\frac{1,022.5}{994.417}\\ =1.0282\end{array}$

Thus, the F-value is 1.0282.

The number of degrees of freedom for the numerator and denominator of an F statistics is the number of degrees of freedom of its effect and the number of degrees of freedom for error respectively.

P-value for lighting:

From Appendix A table A.8, the upper 5% point of the $F_{3,24}$ distribution is 3.01. The upper 10% point of the $F_{3,24}$ distribution is 4.72. Thus, the P-value corresponding to lighting is between 0.010 and 0.05.

Therefore, the P-value is $0.01<P\text{-value <}0.05$.

P-value for block:

From Appendix A table A.8, the upper 10% point of the $F_{2,24}$ distribution is 5.61. The upper 1% point of the $F_{2,24}$ distribution is 9.34. Thus, the P-value corresponding to block is between 0.010 and 0.05.

Therefore, the P-value is $0.001<P\text{-value <}0.010$.

P-value for interaction:

From Appendix A table A.8, the upper 10% point of the $F_{6,24}$ distribution is 2.04. Thus, the P-value corresponding to interaction is greater than 0.100.

Therefore, the P-value is $P\text{-value >0}\text{.100}$.

b.

To determine

Explain whether the assumptions for a randomized complete block design is satisfied or not.

Answer to Problem 3E

All the assumptions for a randomized complete block design is satisfied.

Explanation of Solution

Calculation:

Assumptions of randomized complete block design:

• There must be no interaction between treatment and blocking factors.

Interaction:

Null hypothesis:

$H_0:$ There is no interaction between treatment and blocking factors.

Alternative hypothesis:

$H_1:$ There is interaction between treatment and blocking factors.

For interaction, the F-test statistic is 1.0282 and $P\text{-value >}0.10$.

Decision:

If $P\text{-value}\le \alpha ,$ reject the null hypothesis $H_0$.

If $P\text{-value}>\alpha ,$ fail to reject the null hypothesis $H_0$.

Conclusion:

Interaction:

Here, the P-value is greater than the level of significance.

That is, $P\text{-value}\left(>0.10\right)>\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is not rejected.

Thus, the interaction is not significant at $\alpha =0.05$ level of significance.

Therefore, there is no interaction between treatment and blocking factors.

Thus, all the assumptions for a randomized complete block design is satisfied.

c.

To determine

Check whether the ANOVA table provide evidence that the lighting type affects illuminance or not.

Answer to Problem 3E

The ANOVA table provide evidence that the lighting type affects the illuminance

at $\alpha =0.05$ level of significance.

Explanation of Solution

Calculation:

Factor A is lighting.

Main effect of factor A:

Null hypothesis:

$H_0:$ Lighting type not affect the illuminance.

Alternative hypothesis:

$H_1:$ Lighting type affects the illuminance.

For Factor A, the F-test statistic is 3.3329 and $0.01<P\text{-value <}0.05$.

Decision:

If $P\text{-value}\le \alpha ,$ reject the null hypothesis $H_0$.

If $P\text{-value}>\alpha ,$ fail to reject the null hypothesis $H_0$.

Conclusion:

Factor A:

Here, the P-value is less than the level of significance.

That is, $P\text{-value}\left(<0.05\right)\le \alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

Thus, lighting type affects the illuminance.

Hence, the ANOVA table provide evidence that the lighting type affects the illuminance

at $\alpha =0.05$ level of significance.