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Magazine Chapter 11: Q1E (page 448)
An experiment was carried out to investigate the effect of species (factor A, with I = 4) and grade (factor B, with J = 3) on breaking strength of wood specimens. One observation was made for each species鈥攇rade combination鈥攔esulting in SSA = 442.0, SSB = 428.6, and SSE = 123.4. Assume that an additive model is appropriate.
- Test\({H_0}:{\alpha _1} = {\alpha _2} = {\alpha _3} = {\alpha _4} = 0\)(no differences in true average strength due to species) versus Ha: at least one\({\alpha _i} \ne 0\)using a level .05 test.
- Test\({H_0}:{\beta _1} = {\beta _2} = {\beta _3} = 0\)(no differences in true average strength due to grade) versus Ha: at least one\({\beta _j} \ne 0\)using a level .05 test.
Short Answer
- There is sufficient evidence to support the claim that there is difference in the true average strength due to species.
- There is sufficient evidence to support the claim that there are difference in the true average strength due to grade.
Step by step solution
Determine the mean square values:
A)- Given:
\(\begin{aligned}{*{20}{c}}{SSA = 442.0}\\{SSB = 428.6}\\{SSE = 123.4}\\{I = 4}\\{J = 3}\\{\alpha = 0.05}\\{{H_0}:{\alpha _1} = {\alpha _2} = {\alpha _3} = {\alpha _4} = 0}\end{aligned}\)
The alternative hypothesis states the opposite of the null hypothesis:
\({H_a}:{\rm{\;not all of\;}}{\alpha _i}{\rm{\;are zero\;}}\)
Determine the mean square values:
\(\begin{aligned}{*{20}{c}}{MSA = \frac{{SSA}}{{I - 1}} = \frac{{442.0}}{{4 - 1}} \approx 147.3333}\\{MSE = \frac{{SSE}}{{(I - 1)(J - 1)}} = \frac{{123.4}}{{(4 - 1)(3 - 1)}} \approx 20.5667}\end{aligned}\)
The F-value is the ratio of the mean square values:
\({f_A} = \frac{{MSA}}{{MSE}} = \frac{{147.3333}}{{20.5667}} \approx 7.164\)
The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of the F-distribution table in the appendix containing the F-value in the row
\(\begin{aligned}{*{20}{c}}{dfn = I - 1 = 4 - 1 = 3{\rm{\;and\;}}dfd = (I - 1)(J - 1) = (4 - 1)(3 - 1) = 6{\rm{\;:\;}}}\\{0.010 < P < 0.050}\end{aligned}\)
If the P-value is less than the significance level, then reject the null hypothesis.
\(P < 0.05 \Rightarrow {\rm{\;Reject\;}}{H_0}\)
There is sufficient evidence to support the claim that there are difference in the true average strength due to species.
B)-
Given:
The alternative hypothesis states the opposite of the null hypothesis:
\({H_a}:{\rm{\;not all of\;}}{\beta _i}{\rm{\;are zero\;}}\)
Determine the mean square values:
\(\begin{aligned}{*{20}{c}}{MSB = \frac{{SSB}}{{J - 1}} = \frac{{428.6}}{{3 - 1}} \approx 214.3}\\{MSE = \frac{{SSE}}{{(I - 1)(J - 1)}} = \frac{{123.4}}{{(4 - 1)(3 - 1)}} \approx 20.5667}\end{aligned}\)
The F-value is the ratio of the mean square values:
\({f_A} = \frac{{MSA}}{{MSE}} = \frac{{214.3}}{{20.5667}} \approx 10.420\)
The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of the F-distribution table in the appendix containing the F-value in the row
\(\begin{aligned}{*{20}{c}}{dfn = J - 1 = 3 - 1 = 3{\rm{\;and\;}}dfd = (I - 1)(J - 1) = (4 - 1)(3 - 1) = 6{\rm{\;:\;}}}\\{0.010 < P < 0.050}\end{aligned}\)
If the P-value is less than the significance level, then reject the null hypothesis.
\(P < 0.05 \Rightarrow {\rm{\;Reject\;}}{H_0}\)
There is sufficient evidence to support the claim that there are difference in the true average strength due to species.
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