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The Friedman鈥檚 test for rendomized block experiment

Test for teststatistic is

\(\begin{array}{l}{F_r} = \frac{{12J}}{{I\left( {I + 1} \right)}}\sum\limits_{i = 1}^I {{{\left( {{{\bar R}_i} - \frac{{I + 1}}{2}} \right)}^2}} \\ = \frac{{12J}}{{IJ\left( {I + 1} \right)}}\sum\limits_{i = 1}^I {\bar R_i^2 - 3J} \left( {I + 1} \right)\end{array}\)

When the null hypothesis is true test statistic \({F_r}\), has approximately chi-squared distribution with I-1 degrees of freedom.

Two table goes together- point\({x_{ij}}\)in the first table corresponds to the\({r_{ij}}\)rank in the second table.

The test statistic value is,

\(\begin{array}{l}{f_r} = \frac{{12J}}{{I\left( {I + 1} \right)}}{\sum\limits_{i = 1}^I {\left( {{{\bar R}_i} - \frac{{I - 1}}{2}} \right)} ^2}\\ = \frac{{12}}{{IJ\left( {I + 1} \right)}}\sum\limits_{i = 1}^I {\bar R_i^2} - 3J\left( {I + 1} \right)\\ = \frac{{12}}{{10 \cdot 3 \cdot 4}} \cdot \left( {361 + 289 + 576} \right) - 3 \cdot 10 \cdot 4\\ = 2.60\end{array}\)

Degrees of freedom are

\(df = I - 1 = 3 - 1 = 2\)

Critical value at significant level 0.05 is

\(\begin{array}{l}X_{0.05,2}^2 = 5.99\\X_{0.05,2}^2 = 5.99 > 2.60 = {f_r}\end{array}\)

Do not reject null hypothesis

Hence, do not reject null hypothesis.