91影视

a)-

Let

Where the following holds

and where the are independent normally distributed random variable with mean 0 and varianceThe hypotheses of interest are

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

And for the factor B

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

With given information it is first required to construct ANOVA table and than make conclusion about appropriate hypotheses.

With given information

\(\begin{aligned}{*{20}{c}}{SST = 396.13,}\\{SSA = 58.16,}\\{SSB = 246.97,}\end{aligned}\)

it is easier to compute corresponding mean squares are the F statistic values.

\(SST = SSA + SSB + SSE\)

By the fundamental identity, the SSE can be computed as

\(SSE = SST - SSA - SSB = 396.13 - 58.16 - 246.97 = 91\)

When testing hypotheses H_0A versus H_aA, the test statistic 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 of the test statistic value f_A.

When testing hypotheses\({H_{0B}}{\rm{\;versus\;}}{H_{aB}}\), the test statistic 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 statisticvalue f_B.

The degrees of freedom are

\(\begin{aligned}{*{20}{c}}{d{f_T} = IJ - 1 = 4 \cdot 5 - 1 = 19}\\{d{f_A} = I - 1 = 4 - 1 = 3}\\{d{f_B} = J - 1 = 5 - 1 = 4}\\{d{f_E} = (I - 1)(J - 1) = (4 - 1) \cdot (5 - 1) = 12}\end{aligned}\)

The mean square are

\(\begin{aligned}{*{20}{c}}{MSA = \frac{1}{{I - 1}} \cdot SSA = \frac{1}{3} \cdot 58.16 = 11.4444{\rm{;\;}}}\\{MSB = \frac{1}{{J - 1}} \cdot SSB = \frac{1}{4} \cdot 246.97 = 61.7425{\rm{;\;}}}\\{MSE = \frac{1}{{(I - 1)(J - 1)}} \cdot SSE = \frac{1}{{12}} \cdot 91 = 7.5833.}\end{aligned}\)

The value of test statistics are

\(\begin{aligned}{*{20}{c}}{{f_A} = \frac{{MSA}}{{MSE}} = \frac{{58.16}}{{7.5833}} = 2.5565}\\{{f_B} = \frac{{MSB}}{{MSE}} = \frac{{246.97}}{{7.5833}} = 8.1419}\end{aligned}\)

There are two ways to make conclusion. Using the table in the appendix or compute P value using a software.

The P_A value when testing hypotheses H_OA versus H_aA is

\({P_A} = P\left( {F > {f_A}} \right) = P(F > 2.5565) = 0.104\)

which was computed using software, and random variable F has Fisher's distribution with degrees of freedom\(I - 1 = 3{\rm{\;and\;}}(I - 1)(J - 1) = 12\)because

\({P_A} = 0.104 > 0.01 = \alpha \)

Do not reject hypothesis H_0A

at given significance level. The true separation force is not affected by the angle of pull (which was required in the exercise). The P_B value when testing hypotheses H_0B versus H_aB is

\({P_B} = P\left( {F > {f_B}} \right) = P(F > 8.1419) = 0.002\)

which was computed using software, and random variable F has Fisher's distribution with degrees of freedom\(J - 1 = 4{\rm{\;and\;}}(I - 1)(J - 1) = 12\)Because

\({P_B} = 0.002 < 0.01 = \alpha \)

Reject hypothesis H_0B

at given significance level.

Using the fact

\({F_{\alpha ,I - 1,(I - 1)(J - 1)}} = {F_{0.01,3,12}} = 5.95,\)

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

\({F_{0.01,3,12}} = 5.95 > 2.5565 = {f_A}\)

Reject hypothesis H_0A

at given significance level.

Finally, the result can be represented in a table