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Magazine Chapter 11: Q23E (page 459)
The accompanying data was obtained in an experiment to investigate whether compressive strength of concrete cylinders depends on the type of capping material used or variability in different batches (" The Effect of Type of Capping Material on the Compressive Strength of Concrete Cylinders, Proceedings ASTM, 1958: 11661186). Each number is a cell total based on K=3 observations.
Short Answer
The ANOVA table is
Step by step solution
Step 1:
Let
where the are independent normally distributed random variable with mean 0 and variance \({\sigma ^2}\). The hypotheses of interest are
\({H_{0A}}:{\alpha _1} = {\alpha _2} = \ldots = {\alpha _I} = 0{\rm{\;versus\;}}{H_{aA}}:{\rm{\;at least one\;}}{\alpha _ - }i \ne 0{\rm{,\;}}\)
For the factor B
\({H_{0B}}:\sigma _B^2 = 0\)versus \({H_{aB}}:\sigma _B^2 > 0\)
\({H_{0G}}:\sigma _G^2 = 0\)versus \({H_{aG}}:\sigma _G^2 > 0\)
Sum of squares are given by
With degrees of freedom respectively,
\(\begin{aligned}{*{20}{c}}{d{f_T} = IJK - 1}\\{d{f_E} = IJ(K - 1)}\\{d{f_A} = I - 1}\\{d{f_B} = J - 1}\\{d{f_{AB}} = (I - 1)(J - 1).}\end{aligned}\)
The given data can be represented in a following table
The sum of measurements obtained when factor B is held at level j are
The grand sum is
\({x_ \ldots } = 1847 + 1942 + \ldots + 1891 + 1756 = 27,479\)
The sum of squares can now be computed. This is a little bit trickier because not all the data points are given; however, it is possible to compute using the given values.
The SST is
\(\begin{aligned}{*{20}{c}}{ = \frac{1}{{5 \cdot 3}} \cdot \left( {{{9410}^2} + {{8835}^2} + {{9234}^2}} \right) - \frac{1}{{3 \cdot 5 \cdot 3}} \cdot 27,{{479}^2}}\\{ = 16,791,472.067 - 16,779,898.689}\\{ = 11,573.378}\end{aligned}\)
The SSA is
\(\begin{aligned}{*{20}{c}}{ = \frac{1}{{5 \cdot 3}} \cdot \left( {{{9410}^2} + {{8835}^2} + {{9234}^2}} \right) - \frac{1}{{3 \cdot 5 \cdot 3}} \cdot 27,{{479}^2}}\\{ = 16,791,472.067 - 16,779,898.689}\\{ = 11,573.378.}\end{aligned}\)
The SSB is
\(\begin{aligned}{*{20}{c}}{ = \frac{1}{{3 \cdot 3}} \cdot \left( {{{5432}^2} + {{5684}^2} + {{5619}^2} + {{5567}^2} + {{5177}^2}} \right) - \frac{1}{{3 \cdot 5 \cdot 3}} \cdot 27,{{479}^2}}\\{ = 16,797,828.778 - 16,779,898.689}\\{ = 17,930.089}\end{aligned}\)
The SSE is
\(\begin{aligned}{*{20}{c}}{ = 16,815,853 - \frac{1}{3} \cdot 50,443,409}\\{ = 1383.333}\end{aligned}\)
By the fundamental identity the SSAB
\(\begin{aligned}{*{20}{c}}{SSAB = SST - SSA - SSB - SSE}\\{ = 35,954.311 - 11,573.378 - 17,930.089 - 1383.333}\\{ = 5067.511.}\end{aligned}\)
The degrees of freedom are
\(\begin{aligned}{*{20}{c}}{d{f_T} = IJK - 1 = 3 \cdot 5 \cdot 2 - 1 = 44}\\{d{f_E} = IJ(K - 1) = 3 \cdot 5 \cdot (3 - 1) = 30}\end{aligned}\)
\(\begin{aligned}{*{20}{c}}{d{f_A} = I - 1 = 3 - 1 = 2}\\{d{f_B} = J - 1 = 5 - 1 = 4}\\{d{f_{AB}} = (I - 1)(J - 1) = (3 - 1) \cdot (5 - 1) = 8}\end{aligned}\)
The mean squares are
\(\begin{aligned}{*{20}{c}}{MSA = \frac{1}{{I - 1}} \cdot SSA = \frac{1}{2} \cdot 11,573.378 = 5786.689}\\{MSB = \frac{1}{{J - 1}} \cdot SSB = \frac{1}{4} \cdot 17,930.089 = 4482.522}\end{aligned}\)
\(\begin{aligned}{*{20}{c}}{MSAB = \frac{1}{{(I - 1)(J - 1)}} \cdot SSAB = \frac{1}{8} \cdot 5067.511 = 633.439}\\{MSE = \frac{1}{{IJ(K - 1)}} \cdot SSE = \frac{1}{{30}} \cdot 1383.333 = 46.111}\end{aligned}\)
When testing hypotheses \({H_{0A}}\)versus \({H_{aB}}\)the test statistic value is
\({f_A} = \frac{{MSA}}{{MSAB}}\)
and the P-value is the area under the \({F_{I - 1,(I - 1)(J - 1)}}\)curve to the right of the test statistic value \({f_A}\)
When testing hypotheses \({H_{0B}}\)versus \({H_{aB}}\)the test statistic value is
\({f_B} = \frac{{MSB}}{{MSAB}}\)
and the P-value is the area under the \({F_{I - 1,(I - 1)(J - 1)}}\)curve to the right of the test statistic value \({f_B}\)
When testing hypotheses \({H_{0G}}\)versus \({H_{aG}}\)the test statistic value is
\({f_G} = \frac{{MSAB}}{{MSE}}\)
and the P-value is the area under the \({F_{I - 1,(I - 1)(J - 1)}}\)curve to the right of the test statistic value \({f_B}\)
Hence, the values f values are
\(\begin{aligned}{*{20}{c}}{{f_A} = \frac{{MSA}}{{MSAB}} = \frac{{5786.689}}{{633.439}} = 9.141}\\{{f_B} = \frac{{MSB}}{{MSAB}} = \frac{{4482.522}}{{633.439}} = 7.076}\\{{f_G} = \frac{{MSAB}}{{MSE}} = \frac{{633.439}}{{46.111}} = 13.737}\end{aligned}\)
The critical values are, respectively,
\(\begin{aligned}{*{20}{c}}{{F_{\alpha ,I - 1,(I - 1)(J - 1)}} = {F_{0.01,2,8}} = 8.65}\\{{F_{\alpha ,J - 1,(I - 1)(J - 1)}} = {F_{0.01,4,8}} = 7.01}\\{{F_{\alpha ,(I - 1)(J - 1),IJ(K - 1)}} = {F_{0.01,8,30}} = 3.17,}\end{aligned}\)
which were computed from the table in the appendix.
The P values are the mentioned areas under corresponding F curve, their values are, respectively,
\(\begin{aligned}{*{20}{c}}{{P_A} = 0.009}\\{{P_B} = 0.0097}\\{{P_G} = 0.00}\end{aligned}\)
which were computed using a software.
The ANOVO table is
Always test hypothesis\({H_{0G}}\). Since the ANOVA table has been already computed, there is no need for that because the values have been computed already.
When testing\({H_{0A}}\)versus\({H_{aA}}\)because
\(\begin{aligned}{*{20}{c}}{{F_{0.01,2,8}} = 8.65 < 9.141 = {f_A};}\\{{P_A} = 0.009 < 0.01 = \alpha ;}\end{aligned}\)
reject null hypothesis $H_{0 A}$
at given significance level.
When testing\({H_{0A}}\)versus\({H_{0B}}\)because
\(\begin{aligned}{*{20}{c}}{{F_{0.01,4,8}} = 7.01 < 7.076 = {f_B}}\\{{P_B} = 0.0097 < 0.01 = \alpha }\end{aligned}\)
reject null hypothesis $H_{0 B}$
at given significance level.
When testing \({H_{0G}}\) versus \({H_{aG}}\) because
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