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The hypothesis of interest are\({H_0}:{\sigma _A} = 0\)versus alternative hypothesis\({H_a}:{\sigma _A} > 0\)

The summarized data given in the table is,

\(\begin{aligned}{l}\begin{aligned}{*{20}{c}}{No}&{1:}&{2:}&{3:}&{4:}\\1&{85.06}&{84.99}&{84.48}&{84.1}\\2&{85.25}&{84.28}&{84.72}&{84.55}\\3&{84.87}&{84.88}&{85.1}&{84.05}\\{{x_i}}&{255.18}&{254.15}&{254.30}&{252.7}\end{aligned}\\\begin{aligned}{*{20}{c}}{\overline {{x_i}} \,\,\,\,\,\,}&{85.06}&{84.72}&{84.77}\end{aligned}\,\,\,\,\,\,84.23\\{\overline x _{..}} = 1016.33\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x_{...}} = 84.69\end{aligned}\)

The above table gives us everything needed to carry out\(F\)test.

Let us notice that in first place is,

\(I = 4\), columns- treatment and

\(J = 3\), rows-sample of each type.

The following table needs to be filled with the corresponding values,

The degrees of freedom are,

\(\begin{aligned}{l}I - 1 = 4 - 1 = 3;\\I \times \left( {J - 1} \right) = 4 \times \left( {3 - 1} \right) = 8;\\I \times J - 1 = 4 \times 3 - 1 = 11.\end{aligned}\)

Let us denote with

\(\begin{aligned}{l}{x_i} = \int\limits_{j = 1}^J {{x_{ij}}} \\{x_{..}} = \int\limits_{i = 1}^I {\int\limits_{j = 1}^J {{x_{ij}}} } \end{aligned}\)

The total sum of squares\(\left( {SST} \right)\), the treatment sum of squares\(\left( {SSTr} \right)\), the error sum of squares\(\left( {SSE} \right)\)are given by,

\(\begin{aligned}{l}SST = {\int\limits_{i = 1}^I {\int\limits_{j = 1}^J {\left( {{x_{ij}} - \overline x ..} \right)} } ^2} = \int\limits_{i = 1}^I {\int\limits_{j = 1}^J \begin{aligned}{l}{x_{ij}}^2 - \frac{1}{{I \times J}}{x_{..}}^2;\\\end{aligned} } \\SSTr = \int\limits_{i = 1}^I {\int\limits_{j = 1}^J {{{\left( {{{\overline x }_i} - {{\overline x }_{..}}} \right)}^2} = \frac{1}{J} \times \int\limits_{i = 1}^I {{x^2}_i - \frac{1}{{I \times J}}{x_{..}}^2;} } } \\SSE = \int\limits_{i = 1}^I {\int\limits_{j = 1}^J {{{\left( {{x_{ij}} - {{\overline x }_{i.}}} \right)}^2}} } \end{aligned}\)

The mean square are,

\(\begin{aligned}{l}MSTr = \frac{1}{{I - 1}} \times SSTr;\\MSE = \frac{1}{{I \times \left( {J - 1} \right)}} \times SSE\end{aligned}\)

\(F\)is the ratio of the two mean square,

\(F = \frac{{MSTr}}{{MSE}}\)

Calculate all the values one by one. By summing corresponding column, value of \({x_i}\)are,

\(\begin{aligned}{l}{x_1} = 85.06 + 85.25 + 84.87 = 255.18;\\{x_2} = 84.99 + 84.28 + 84.88 = 254.15;\\{x_3} = 84.48 + 84.72 + 85.1 = 254.30;\\{x_4} = 84.1 + 84.55 + 84.05 = 252.70.\end{aligned}\)

The value of \({\overline x _i} = \frac{1}{J} \times {x_i}\) are given by,

\(\begin{aligned}{l}{\overline x _1} = \frac{1}{3} \times 255.18 = 85.06;\\{\overline x _2} = \frac{1}{3} \times 254.15 = 84.72;\\{\overline x _3} = \frac{1}{3} \times 254.30 = 84.77;\\{\overline x _4} = \frac{1}{3} \times 252.70 = 84.23.\end{aligned}\)

The grand mean \({\overline x _{..}}\)is,

\(\begin{aligned}{l}{\overline x _{..}} = \frac{1}{{I \times J}} \times {x_{..}} = \frac{1}{{I \times J}}\int\limits_{i = 1}^I {\int\limits_{j = 1}^J \begin{aligned}{l}{x_{ij}} = \frac{1}{{4 \times 3}} \times \left( {85.06 + 84.99 + .. + 85.1 + 84.05} \right) = 84.69,\\\end{aligned} } \\{x_{..}} = \int\limits_{j = 1}^J {{x_{ij}} = 85.06 + 84.99 + ... + 85.1 + 84.05 = 1016.33} \end{aligned}\)

The total sum of square is,

\(\begin{aligned}{l}SST = \int\limits_{i = 1}^I {\int\limits_{j = 1}^J {{{\left( {{x_{ij}} - {{\overline x }_{..}}} \right)}^2} = \int\limits_{i = 1}^I {\int\limits_{j = 1}^J {{x^2}_{ij}} } } } - \frac{1}{{I \times J}}{x_{..}}^2\\ = {85.06^2} + {84.99^2} + .. + {85.1^2} + {84.05^2} - \frac{1}{{4 \times 3}} \times {1016.33^2}\\ = 86078.99 - 86077.22\\ = 1.77\end{aligned}\)

The treatment sum of square is,

\(\begin{aligned}{l}SSTr = \int\limits_{i = 1}^I {\int\limits_{j = 1}^J {{{\left( {{{\overline x }_i} - {{\overline x }_{..}}} \right)}^2} = \frac{1}{J} \times \int\limits_{i = 1}^I {{x^2}_i - \frac{1}{{I \times J}}{x_{..}}^2} } } \\ = \frac{1}{3} \times {255.18^2} + {254.15^2} + {254.30^2} + {252.70^2} - \frac{1}{{4 \times 5}} \times {1016.33^3}\\ = 86078.28 - 86077.22\\ = 1.06\end{aligned}\)

Fundamental Identity : \(SST = SSTr + SSE\)

Error sum of square is \(\begin{aligned}{l}SSE = SST - SSTr\\ = 1.77 - 1.06 = 0.71\end{aligned}\)

The mean square is computed as,

\(\begin{aligned}{l}MSTr = \frac{1}{{I - 1}} \times SSTr = \frac{1}{3} \times 1.06 = 0.352;\\MSE = \frac{1}{{I \times \left( {J - 1} \right)}} \times SSE = \frac{1}{{4 \times \left( {3 - 1} \right)}} \times 0.71 = 0.089\end{aligned}\)

The value of \(F\)statistics is,

\(f = \frac{{MSTr}}{{MSE}} = \frac{{0.352}}{{0.089}} = 3.958\)

The ANOVA table now becomes,

For the usual test, you can either make conclusion about the hypothesis look at the \(F\)critical value or a \(P\)value.

The interest of hypothesis are \({H_0}:{\sigma _A} = 0\)versus alternative hypothesis\({H_a}:{\sigma _A} > 0\)

The\(P\)value is the area to the right of \(f\)value under \(F\)curve where \(F\)has Fisher鈥檚 distribution with degree of freedom \(3and8\),

\(P = P\left( {F > f} \right) = P\left( {F > 3.958} \right) = 0.053\)which was computed using the software.

Because, \(P = 0.053 > 0.05 = \alpha \)

So donot reject null hypotheses at significance level. There is no statistically difference in true averages among the four different laboratories.

Using the table, we could use e.g. value \({F_{0.05,3,8}}\)for which the area under the \(F\) curve to the right of \({F_{0.05,3,8}}\)is \(0.05\). The value is

\({F_{0.05,3,8}} = 4.07 > 3.958 = f\)

Which indicates not to reject the null hypothesis at significance level \(0.05\).