91影视

For given values, the \({\phi ^2}\) value is

\(\phi 2 = \frac{{2\,.\,\left( {{0^2} + {0^2} + {{\left( { - 1} \right)}^2} + {1^2}} \right)}}{1} = 4\)

Where

\({\alpha _1} = {\alpha _2} = 0\,,\,\,{\alpha _3} = - 1\,,\,\,{\alpha _4} = 1\)

Hence, the \(\phi \) value is

\(\phi = 2\)

Degrees of freedom for the F test are \({v_1} = 4 - 1 = 3\) and \({v_2} = 4\,\,.\,\left( {6 - 1} \right) = 20\). The power is approximately 0.9 using the figure

\(\beta \approx 0.9\,\)

Considering the fact that

\({\phi ^2} = \frac{1}{2}\,\,.\,\,J\)

or equally

\(\phi = 0.707\,\,.\,\,\sqrt J ,\)

And the fact that the degrees of freedom \({v_2}\) are

\({v_2} = 4\,\,.\,\left( {J - 1} \right),\)

From the given figure you can conclude that

\(J = 9\)

Four of the \({\mu _1}\)鈥榮 is equal, let

\({\mu _i} = {\mu _j}\,,\,\,\,\,i\,,\,j\,\, \in \,\left\{ {1,2,3,4} \right\}\)

And let

\({\mu _5} = {\mu _1} + 1\)

From this, the value of \({\alpha _i}\) can be computed. Since

\(\mu = {\mu _1} + \frac{1}{5}\)

The following are \({\alpha _i}\) s

\(\begin{aligned}{l}{\alpha _i} = - \frac{1}{5},\,\,\,i = 1,2,3,4,\\{\alpha _5} = \frac{4}{5}\end{aligned}\)

Using the same formula, the value of \({\phi ^2}\) is

\({\phi ^2} = \frac{2}{1}.\frac{{20}}{{25}} = 1.6\)

Hence, the value of \(\phi \) is

\(\phi = 1.26\)

Degrees of freedom for the F test are \({v_1} = 5 - 1 = 4\) and \({v_2} = 5\,\,.\,\left( {10 - 1} \right) = 45\). The power is approximately 0.55 using the figure

\(\beta \approx 0.55\,\)

Here, the final result is,

1. \(\beta \approx 0.9\,\)
2. \(J = 9\)
3. \(\beta \approx 0.55\,\)