91影视

Let

Where the following holds

And where the are independent normally distributed random variable with mean 0 and variance \({\sigma ^2}\). The hypotheses interest are

\({H_{0A}}:{\alpha _1} = {\alpha _2} = \ldots = {\alpha _I} = 0\)versus \({H_{aA}}:{\rm{\;at least one\;}}{\alpha _ - }i \ne 0\)

And, for the factor B

\({H_{0B}}:{\beta _1} = {\beta _2} = \ldots = {\beta _J} = 0\) versus \({H_{aB}}:{\rm{\;at least one\;}}{\beta _ - }j \ne 0\)

The following table summarizes values which will be required to compute the test statistic value.

Values of I and J are

I=4-number of rows;

J=4 - number of columns.

Before continuing, denote with \({\bar X_i}\) the average of measurements obtained when factor A is held at level i

With \({\bar X_{ \cdot j}}\)the average of measurements obtained when factor B is held at level j

And with \({\bar X_{..}}\), the grand mean

Observed values are denoted with small x instead of big \(X\). The notation without line over\(X\)are just the sums.

\({x_{2.}} = 123 + 406 + \ldots + 416 = 1509\)

\({x_{3.}} = 68 + 416 + \ldots + 379 = 1525\)

\({x_4} = 62 + 75 + \ldots + 564 = 1063\).

The sum of measurements for factor B at corresponding level

\({x_{.1}} = 1155 + 123 + 68 + 62 = 1408\)

\({x_{.2}} = 2255 + 406 + 416 + + 75 = 3152\)

\({x_{.3}} = 3505 + 564 + 662 + 362 = 5093\)

\({x_{.4}} = 4632 + 416 + 379 + 564 = 5991\)

The grand sum is

The sum of squares are given by

With degrees of freedom respectively

\(d{f_T} = IJ - 1\)

\(d{f_A} = I - 1\)

\(d{f_B} = J - 1\)

\(d{f_E} = (I - 1)(J - 1)\)

SST can be computed as

\( = \left( {{{1155}^2} + {{2255}^2} + \ldots + {{362}^2} + {{564}^2}} \right) - \frac{1}{{4 \cdot 4}} \cdot 15,{644^2}\)

\( = 42,048,790 - 15,295,921\)

\( = 26,752,869\)

SSA can be computed as

\( = \frac{1}{4} \cdot \left( {11,{{547}^2} + {{1509}^2} + + {{1525}^2} + {{1063}^2}} \right) - \frac{1}{{4 \cdot 4}} \cdot 15,{644^2}\)

\( = 34,766,471 - 15,295,921\)

\( = 19,470,550\)

SSB can be computed as

\( = \frac{1}{4} \cdot \left( {{{1408}^2} + {{1509}^2} + + {{3152}^2} + {{5093}^2} + {{5991}^2}} \right) - \frac{1}{{4 \cdot 4}} \cdot 15,{644^2}\)

\( = 18,437,074.5 - 15,295,921\)

\( = 3,141,153.5\)

Fundamental Idendity:

\(SST = SSA + SSB + SSE\)

By the fundamental identity, the SSE can be computed as

\(SSE = SST - SSA - SSB = 26,752,869 - 19,470,550 - 3,141,153.5\)

=4,141,165.5

When testing hypotheses \({H_{0A}}\) versus \({H_{aB}}\),the test static value is

\({f_A} = \frac{{MSA}}{{MSE}}\)

And the P-value is the area under the \({F_{I - 1,(I - 1)(J - 1)}}\)curve to the right to the test static value\({f_A}\)

When testing hypotheses\({H_{0B}}\)versus \({H_{aB}}\)the tst sstatic value is

\({f_B} = \frac{{MSB}}{{MSE}}\),

And the P-value is the area under the \({F_{J - 1,(I - 1)(J - 1)}}\)curve to the right of the test static value\({f_B}\).

The degrees of freeom are

\(d{f_T} = IJ - 1 = 4 \cdot 4 - 1 = 15\)

\(d{f_A} = I - 1 = 4 - 1 = 3\)

\(d{f_B} = J - 1 = 4 - 1 = 3\)

\(d{f_E} = (I - 1)(J - 1) = (4 - 1) \cdot (4 - 1) = 9\)

The mean squares are

\(MSA = \frac{1}{{I - 1}} \cdot SSA = \frac{1}{3} \cdot 19,470,550 = 6,490,183.33\)

\(MSB = \frac{1}{{J - 1}} \cdot SSB = \frac{1}{3} \cdot 3,141,153.5 = 1,047,051.17\)

\(MSE = \frac{1}{{(I - 1)(J - 1)}} \cdot SSE = \frac{1}{9} \cdot 4,141,165.5 = 460,129.50\)

The value of test statics are

\({f_A} = \frac{{MSA}}{{MSE}} = \frac{{6,490,183.33}}{{460,129.50}} = 14.11\)

\({f_B} = \frac{{MSB}}{{MSE}} = \frac{{1,047,051.17}}{{460,129.50}} = 2.280\)

\({\rm{\;There are two ways to make conclusion}}{\rm{. Using the table in the appendix or compute\;}}P{\rm{\;value using a software}}{\rm{.\;}}\)

The \({P_A}\)value when testing the hypotheses\({H_{0A}}\) versus\({H_{aA}}\)is \({P_A} = P\left( {F > {f_A}} \right) = P(F > 14.11) = 0.00\)

Which was computed using the software, and random variables \(F\) has Fisher鈥檚 distribution with degree of freedom,\(I - 1 = 3\)and \((I - 1)(J - 1) = 9\),because \({P_A} = 0.00 < 0.05 = \alpha \)

Reject hypotheses \({H_{0A}}\) at given significance level.

The \({P_B}\)value when testing hypothesis\({H_{0B}}\)versus \({H_{aB}}\)is \({P_B} = P\left( {F > {f_B}} \right) = P(F > 2.28) = 0.15\)

which was computed using software, and random variable \(F\)has Fisher鈥檚 distribution with degree of value \(J - 1 = 3\)and \((I - 1)(J - 1) = 9\)because,

\({P_B} = 0.15 > 0.05 = \alpha \)

Do not reject hypotheses \({H_{0B}}\)

At given significance level.

Using the fact that \({F_{\alpha ,I - 1,(I - 1)(J - 1)}} = {F_{0.05,3,9}} = 3.86\)

which was obtained from the table in the appendix, and from the fact that

\({F_{0.05,3,9}} = 3.86 < 14.11 = {f_A}\)

Do not reject hypotheses \({H_{0A}}\)at given significance level.

And also using the fact that \({F_{\alpha ,J - 1,(I - 1)(J - 1)}} = {F_{0.05,3,9}} = 3.86\)

which was obtained from the table in the appendix, and the fact that

\({F_{0.05,3,9}} = 3.86 > 2.28 = {f_B}\)

Do not reject hypotheses \({H_{0B}}\)

At the significance level.

Finally, the result can be represented in a table

It appears that the accumulation does not depend on the sowing rate.