a.
Construct an ANOVA table and give a
a.
Answer to Problem 3E
The ANOVA table is,
Source | DF | SS | MS | F | P |
Lighting | 3 | 9,943 | 3,314.33 | 3.3329 | |
Block | 2 | 11,432 | 5,716.00 | 5.7481 | |
Interaction | 6 | 6,135 | 1,022.50 | 1.0282 | |
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
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
The number of degrees of freedom for the lighting methods is,
Thus, the number of degrees of freedom for the lighting methods is 3.
The number of levels for blocks is denoted as J.
Here
The number of degrees of freedom for blocks is,
Thus, the number of degrees of freedom for blocks is 2.
The number of degrees of freedom for interactions is,
Thus, the number of degrees of freedom for interactions is 6.
The number of replication is denoted as K.
Here
The number of degrees of freedom for error is,
Thus, the number of degrees of freedom for error is 24.
The number of degrees of freedom for total is,
The mean square of A can be obtained as follows:
Substitute
Thus, the value of
The mean square of B can be obtained as follows:
Substitute
Thus, the value of
The mean square of AB can be obtained as follows:
Substitute
Thus, the value of
The sum of squares of error can be obtained as follows:
Substitute
Thus, the sum of squares of error is
The mean square of error can be obtained as follows:
Substitute
Thus, the value of
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,
Substitute
Thus, the F-value is 3.3329.
The F-value corresponding to block is,
Substitute
Thus, the F-value is 5.7481.
The F-value corresponding to interaction is,
Substitute
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
Therefore, the P-value is
P-value for block:
From Appendix A table A.8, the upper 10% point of the
Therefore, the P-value is
P-value for interaction:
From Appendix A table A.8, the upper 10% point of the
Therefore, the P-value is
b.
Explain whether the assumptions for a randomized complete block design is satisfied or not.
b.
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:
Alternative hypothesis:
For interaction, the F-test statistic is 1.0282 and
Decision:
If
If
Conclusion:
Interaction:
Here, the P-value is greater than the level of significance.
That is,
Therefore, the null hypothesis is not rejected.
Thus, the interaction is not significant at
Therefore, there is no interaction between treatment and blocking factors.
Thus, all the assumptions for a randomized complete block design is satisfied.
c.
Check whether the ANOVA table provide evidence that the lighting type affects illuminance or not.
c.
Answer to Problem 3E
The ANOVA table provide evidence that the lighting type affects the illuminance
at
Explanation of Solution
Calculation:
Factor A is lighting.
Main effect of factor A:
Null hypothesis:
Alternative hypothesis:
For Factor A, the F-test statistic is 3.3329 and
Decision:
If
If
Conclusion:
Factor A:
Here, the P-value is less than the level of significance.
That is,
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
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Chapter 9 Solutions
Statistics for Engineers and Scientists
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