91影视

The given data,\({y_{ij}},i,j = 1,2,3,4\), should be transformed using

\({x_{ij}} = \sqrt {{y_{ij}}} ,\)\(i,j \in \left\{ {1,2,3,4} \right\},\)

Because the data follow Poisson distribution for each brand.

The hypothesis of interest is

\({H_0}:{\mu _i} = {\mu _j},i \ne j\)

Versus alternative hypothesis

\({H_a}:\)at least two of the \({\mu _i}\)鈥檚 is different,

where\({\mu _i}\)is the true average of brand i, i=1,2,3,4.

The transformed data is given in the table below

This table summarizes everything needed to carry out F test(ANOVA).

Here is explanation how to obtain those values.

Notice first that

I=4, columns 鈥� treatments (usually they are in rows),

And

J=5,rows 鈥� samples of each type,

The following data needs to be filled with corresponding values;

The degrees of freedom are

\(\begin{aligned}{l}I - 1 = 4 - 1 = 3\\I \cdot (J - 1) = 4 \cdot (5 - 1) = 16\\I \cdot J - 1 = 4 \cdot 5 - 1 = 19\end{aligned}\)

Denote with

The total sum of squares

(SST),

Treatment sum of squares

(SSTr),and

Error sum of squares

(SSE) are given by

The mean squares are

\(\begin{aligned}{l}MSTr = \frac{1}{{I - 1}} \cdot SSTr\\MSE = \frac{1}{{I \cdot (J - 1)}} \cdot SSE\end{aligned}\)

F is the ratio of the two mean squares

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

Compute all the values one by one.by summing corresponding columns (usually rows),values of\({x_i}\)are

\(\begin{aligned}{l}{x_1} = 3.16 + 2.24 + \ldots + 2.83 = 15.43;\\{x_2} = 3.74 + 3.46 + \ldots 2.83 = 17.16;\\{x_3} = 3.61 + 4.24 + \ldots + 4.24 = 19.13;\\{x_4} = 4.12 + 4.00 + 3.74 = 20.02;\end{aligned}\)

Values of

Are given by

The grand mean is

And

The total sum of squares is

\(\begin{aligned}{l} = {3.16^2} + {3.74^2} + {...4.24^2} + {3.74^2} - \frac{1}{{4 \cdot 5}} \cdot {71.74^2}\\ = 264 - 257.33\\ = 6.67\end{aligned}\)

The treatment sum of squares is

\(\begin{aligned}{l} = \frac{1}{5} \cdot ({15.43^2} + {17.16^2} + {19.13^2} + {20.02^2}) - \frac{1}{{4 \cdot 5}} \cdot {71.74^2}\\ = 259.82 - 257.33\\ = 2.49\end{aligned}\)

Fundamental identity

SST + SSTr = SSE

Error sum of squares is

SSE = SST 鈥� SSTr = 6.67 - 2.49 = 4.18

The mean squares can be computed now as

\(\begin{aligned}{l}MSTr = \frac{1}{{I - 1}} \cdot SSTr = \frac{1}{3} \cdot 2.49 = 0.83\\MSE = \frac{1}{{I \cdot (J - 1)}} \cdot SSE = \frac{1}{{4 \cdot (5 - 1)}} \cdot 4.18 = 0.26\end{aligned}\)

The value of f static is

\(f = \frac{{MSTr}}{{MSE}} = \frac{{0.83}}{{0.26}} = 3.18\)

The ANOVA table becomes

As for usual tests,you can either make conclusion about the hypothesis look at the F critical value or a P value. Remember that the hypothesis of interest is

\({H_0}:{\mu _i} = {\mu _j},i \ne j\)

Versus alternative hypothesis

\({H_a}:\)at least two of the \({\mu _i}\)鈥檚 are different

The P value is the area to the right of f value under the F curve where F has fisher鈥檚 distribution with degrees of freedom 3 and 16;thus

\(P = P(F > f) = P(F > 3.18) = 0.05\)

Which was computed using software (you could estimate it using the table in the appendix).Because

\(P = 0.05 > 0.01 = \alpha \)

Do not reject null hypothesis

At significance level. There is no statistically significance difference is true averages among the four different brand types

Using the table, you could use. value \({F_{0.01,3,16}}\) for which the area under the F curve to the right of \({F_{0.01,3,16}}\) is 0.01.The value is

\({F_{0.01,3,16}} = 5.29 > 3.18 = f\)

Which indicates to not reject null hypothesis at significance level 0.01