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The hypothesis of interest are\({H_0}:{\mu _i} = {\mu _j},i \ne j\)versus alternative hypothesis\({H_a}:At{\rm{ }}least{\rm{ }}two{\rm{ }}of{\rm{ }}the\,\,\,\mu \_i's\)are different, where\({\mu _i}\)is the true average of CFF for\(3\)colours,\(i = 1,2,3\)

Let\({J_i},i = 1,2,...,l\)be the sample sizes of\(I\)treatments, and let\(n = \int\limits_{i = 1}^l {{J_i}} \)be the total number of observations.

Let us denote with

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

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,

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

With degrees of freedom, respectively,

\(\begin{aligned}{l}df = n - 1;\\df = I - 1;\\df = \int\limits_{i = 1}^I {\left( {{J_i} - 1} \right) = n - I;} \end{aligned}\)

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

The mean square are,

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

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

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

The P-value is the area under \({F_{I - 1,n - I}}\) curve to the right of f.

The degrees of freedom for \(n = 133\), are,

\(\begin{aligned}{l}df = n - 1 = 19 - 1 = 18;\\df = I - 1 = 3 - 1 = 2;\\df = \int\limits_{i = 1}^I {\left( {{J_i} - 1} \right) = n - I = 19 - 3 = 16.} \end{aligned}\)

The treatment sum of square is,

\(\begin{aligned}{l}SSTr = \int\limits_{i = 1}^I {\int\limits_{j = 1}^{{J_i}} {{{\left( {{{\overline x }_i} - {{\overline x }_{..}}} \right)}^2} = \frac{1}{{{J_i}}} \times \int\limits_{i = 1}^I {{x^2}_i - \frac{1}{n}{x_{..}}^2} } } \\ = \frac{1}{8} \times {204.7^2} + \frac{1}{5} \times {134.6^2} + \frac{1}{6} \times {169.0^2} + \frac{1}{{19}} \times {508.3^2}\\ = 12621.36 - 13598.36\\ = 23\end{aligned}\)

The total sum of square is,

\(\begin{aligned}{l}SST = \int\limits_{i = 1}^I {\int\limits_{j = 1}^{{J_i}} {{{\left( {{x_{ij}} - {{\overline x }_{..}}} \right)}^2} = \int\limits_{i = 1}^I {\int\limits_{j = 1}^{{J_i}} {{x^2}_{ij}} } } } - \frac{1}{n}{x_{..}}^2\\ = 13659.67 - 13598.36\\ = 61.31\end{aligned}\)

Fundamental Identity :

Error sum of square is \(\begin{aligned}{l}SSE = SST - SSTr\\ = 61.31 - 23 = 38.31\end{aligned}\)

The mean square is computed as,

\(\begin{aligned}{l}MSTr = \frac{1}{{I - 1}} \times SSTr = \frac{1}{2} \times 23 = 11.5;\\MSE = \frac{1}{{n - I}} \times SSE = \frac{1}{{16}} \times 38.31 = 2.39\end{aligned}\)

The value of \(F\)statistics is,

\(f = \frac{{MSTr}}{{MSE}} = \frac{{11.5}}{{2.3}} = 4.803\)

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 \(2and16\),

\(P = P\left( {F > f} \right) = P\left( {F > 4.803} \right) = 0.02\)which was computed using the software.

Because, \(P = 0.02 < 0.05 = \alpha \)

So reject null hypotheses at significance level.

\({F_{0.05,2,16}} = 3.63 < f = 4.803 < 6.23 = {F_{0.01,2,16}}\)

Which indicates to reject the null hypothesis at significance level. The conclusion is that there is significant difference in true average for the three treatment.

For the problem \(\left( b \right)\), denote with,

\({w_{ij}} = {Q_{\alpha ,I,n - I}} \times \sqrt {\frac{{MSE}}{2} \times \frac{1}{{{J_i}}} + \frac{1}{{{J_j}}}} \)

Then the probability that

\({\overline x _i} - {\overline x _j} - {w_{ij}} < {\mu _i} - {\mu _j} < {\overline x _i} - {\overline x _j} + {w_{ij}}\)

Is approximately \(1 - \alpha \)for every \(i = 1,2,...,I\) and every \(j = 1,2,...,I\) where \(i \ne j\)

From the table in the appendix for \(\alpha = 0.05,I = 3,n - I = 16\), the following holds,

\({Q_{0.05,3,16}} = 3.65\)

Using the modified Turkey鈥檚 method to obtain the confidence interval for each pair you can conclude difference between iris colors.

Given the average \({\overline x _i}and{Q_{\alpha ,I,n - I}}\) the confidence interval can be computed using the below mentioned formula,

\(\)\({\overline x _i} - {\overline x _j} - {w_{ij}} < {\mu _i} - {\mu _j} < {\overline x _i} - {\overline x _j} + {w_{ij}}\)

Compute \({w_{ij}}\),

\(\begin{aligned}{l}{w_{12}} = {Q_{0.05,3,16}} \times \sqrt {\frac{{MSE}}{2} \times \frac{1}{8} + \frac{1}{5}} = 2.27;\\{w_{13}} = {Q_{0.05,3,16}} \times \sqrt {\frac{{MSE}}{2} \times \frac{1}{8} + \frac{1}{6}} = 2.15;\\{w_{23}} = {Q_{0.05,3,16}} \times \sqrt {\frac{{MSE}}{2} \times \frac{1}{5} + \frac{1}{6}} = 2.42\end{aligned}\)

The MSE computed in the part \(\left( a \right)\) is \(MSE = 2.39\)

Hence the intervals are,

\(\begin{aligned}{l}\left( {{{\overline x }_{1.}} - {{\overline x }_{2.}} - {w_{12}},{{\overline x }_{1.}} - {{\overline x }_{2.}} + {w_{12}}} \right) = \left( { - 3.6,0.94} \right);\\\left( {{{\overline x }_{1.}} - {{\overline x }_{3.}} - {w_{13}},{{\overline x }_{1.}} - {{\overline x }_{3.}} + {w_{13}}} \right) = \left( { - 4.73, - 0.43} \right);\\\left( {{{\overline x }_{2.}} - {{\overline x }_{3.}} - {w_{23}},{{\overline x }_{2.}} - {{\overline x }_{3.}} + {w_{23}}} \right) = \left( { - 3.67,1.17} \right);\end{aligned}\)

Since only for the interval of pair \(\left( {2,3} \right)\)does not include zero, so that you can conclude that there is only statistical difference between CFF for brown and blue color.