91影视

a)-

As suggested, subtract 400 from all observations to obtain new observations. This does not affect the result - it has been proved in Chapter 10.

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

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

The following table summarizes values which will be required to compute the test statistic value. NOTE: from all values you should subtract 400 to obtain the following data!

Values of I and J are

\(I = 4\)- number of rows;

J = 3 鈥� number of columns.

Before continuing, denote withthe average of measurements obtained when factor A is held at level i.

Withthe average of measurements obtained when factor B is held at level j

And with.. the grand mean

Observed values are denoted with small instead of big X. The notations without line over X are just the sums.

Compute them one by one. Starting with average of measurements for factor A - Paint brand at level corresponding level

\(\begin{aligned}{*{20}{c}}{{x_{1.}} = 54 + 46 + 51 = 151;}\\{{x_{2.}} = 46 + 44 + 47 = 137;}\\{{x_{3.}} = 39 + 42 + 44 = 125;}\\{{x_{4.}} = 44 + 37 + 43 = 124.}\end{aligned}\)

Values of

\({\bar x_{i.}} = \frac{1}{J} \cdot {x_i}\)

Are given by

\(\begin{aligned}{*{20}{c}}{{{\bar x}_{1.}} = \frac{1}{3} \cdot 151 = 50.33}\\{{{\bar x}_{2.}} = \frac{1}{3} \cdot 137 = 45.67}\\{{{\bar x}_{3.}} = \frac{1}{3} \cdot 125 = 41.67}\\{{{\bar x}_{4.}} = \frac{1}{3} \cdot 124 = 41.33.}\end{aligned}\)

The average of measurements for factor B- Roller at level corresponding level

\(\begin{aligned}{*{20}{c}}{{x_{.1}} = 54 + 46 + 39 + 44 = 183}\\{{x_{.2}} = 46 + 44 + 42 + 37 = 169}\\{{x_{.3}} = 5147 + 44 + 43 = 185}\end{aligned}\)

Values of

\({\bar x_{ \cdot j}} = \frac{1}{I} \cdot {x_{ \cdot j}}\)

Are given by

\(\begin{aligned}{*{20}{c}}{{{\bar x}_{.1}} = \frac{1}{4} \cdot 183 = 45.75}\\{{{\bar x}_{.2}} = \frac{1}{4} \cdot 169 = 42.25}\\{{{\bar x}_{.3}} = \frac{1}{4} \cdot 185 = 46.25.}\end{aligned}\)

The grand sum is

The grand mean is

The sum of squares are given by

with degrees of freedom respectively,

\(\begin{aligned}{*{20}{c}}{d{f_T} = IJ - 1}\\{d{f_A} = I - 1}\\{d{f_B} = J - 1}\\{d{f_E} = (I - 1)(J - 1).}\end{aligned}\)

SST can be computed as

SSA can be computed as

SSB can be computed as

\(SST = SSA + SSB + SSE\)

By the fundamental identity, the SSE can be computed as

\(SSE = SST - SSA - SSB = 238.25 - 159.58 - 38 = 40.67.\)

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 statistic value f_B.

The degrees of freedom are

\(\begin{aligned}{*{20}{c}}{d{f_T} = IJ - 1 = 4 \cdot 3 - 1 = 11}\\{d{f_A} = I - 1 = 4 - 1 = 3}\\{d{f_B} = J - 1 = 3 - 1 = 2}\\{d{f_E} = (I - 1)(J - 1) = (4 - 1) \cdot (3 - 1) = 6.}\end{aligned}\)

The mean squares are

\(\begin{aligned}{*{20}{c}}{MSA = \frac{1}{{I - 1}} \cdot SSA = \frac{1}{3} \cdot 159.58 = 53.19}\\{MSB = \frac{1}{{J - 1}} \cdot SSB = \frac{1}{2} \cdot 38 = 19}\\{MSE = \frac{1}{{(I - 1)(J - 1)}} \cdot SSE = \frac{1}{6} \cdot 40.67 = 6.78}\end{aligned}\)

The value of test statistics are

\(\begin{aligned}{*{20}{c}}{{f_A} = \frac{{MSA}}{{MSE}} = \frac{{53.19}}{{6.78}} = 7.85}\\{{f_B} = \frac{{MSB}}{{MSE}} = \frac{{19}}{{6.78}} = 2.80}\end{aligned}\)

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

\({P_A} = P\left( {F > {f_A}} \right) = P(F > 7.85) = 0.02\)

which was computed using software, and random variable F has Fisher's distribution with degrees of freedom\(I - 1 = 3{\rm{\;and\;}}(I - 1)(J - 1) = 6\)because

\({P_A} = 0.02 < 0.05 = \alpha \)

reject hypothesis H_0A

at given significance level.

The P_B value when testing hypotheses H_OB versus H_aB is

\({P_B} = P\left( {F > {f_B}} \right) = P(F > 2.8) = 0.14\)

which was computed using software, and random variable F has Fisher's distribution with degrees of freedom\(J - 1 = 2{\rm{\;and\;}}(I - 1)(J - 1) = 6.{\rm{ Because\;}}\)

\({P_B} = 0.14 > 0.05 = \alpha \)

do not reject hypothesis H_0B

at given significance level.

using the fact that

\({F_{\alpha ,I - 1,(I - 1)(J - 1)}} = {F_{0.05,3,6}} = 4.76,\)

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

\({F_{0.05,3,6}} = 4.76 < 7.85 = {f_A}\)

reject hypothesis H_0A

at given significance level.

Also, using the fact that

\({F_{\alpha ,J - 1,(I - 1)(J - 1)}} = {F_{0.05,2,6}} = 5.14,\)

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

\({F_{0.05,2,6}} = 5.14 > 2.8 = {f_B}\)

do not reject hypothesis H_0B

at given significance level.

Finally, the result can be represented in a table

b)-

As mentioned, and as you can see from the computed table, the P_A value when testing hypotheses H_0A versus H_aA is

\({P_A} = P\left( {F > {f_A}} \right) = P(F > 7.85) = 0.02\)

which was computed using software, and random variable F has Fisher's distribution with degrees of freedom\(J - 1 = 3{\rm{\;and\;}}(I - 1)(J - 1) = 6.{\rm{ Because\;}}\)

\({P_A} = 0.02 < 0.05 = \alpha \)

reject hypothesis H_0A

at given significance level.

The amount of coverage do depend on factor A - Paint Brand.

c)-

The P_B value when testing hypotheses H_OB versus H_aB is

\({P_B} = P\left( {F > {f_B}} \right) = P(F > 2.8) = 0.14\)

which was computed using software, and random variable F has Fisher's distribution with degrees of freedom\(J - 1 = 2{\rm{\;and\;}}(I - 1)(J - 1) = 6.{\rm{ Because\;}}\)

\({P_B} = 0.14 > 0.05 = \alpha \)

do not reject hypothesis H_0B

at given significance level.

The amount of coverage do depend on factor B - Paint Brand.

d)-

The hypothesis that was reject is the null hypothesis for factor A - Paint Brand, so it makes sense to use the Turkey's method to identify differences in levels of mentioned factor.

In order to compare levels of factor B, first obtain\({Q_{\alpha ,J,(I - 1)(.J - 1)}}\). In second step compute value

\(w = {Q_{\alpha ,J,(I - 1)(J - 1)}} \cdot \sqrt {\frac{{MSE}}{I}} \)

In the last step, arrange the sample means in increasing order and underscore pairs which differs less than w, and identify pairs not underscored by the same line as corresponding to significantly different levels of the given factor.

From the table in the appendix the value of Q is

\({Q_{\alpha ,I,(I - 1)(J - 1)}} = {Q_{0.05,4,6}} = 4.9\)

The w can be computed as

\(\begin{aligned}{*{20}{c}}{w = {Q_{\alpha ,I,(I - 1)(J - 1)}} \cdot \sqrt {\frac{{MSE}}{J}} = {Q_{0.05,4,6}} \cdot \sqrt {\frac{{MSE}}{3}} }\\{ = 4.9 \cdot \sqrt {\frac{{6.78}}{3}} = 7.37}\end{aligned}\)

From the table, notice that the ordered means are

\(x.1 < x.2 < x.3 < x.4\)

thus; compute all differences and see which are smaller than w.

The following are the differences

\(\begin{aligned}{*{20}{c}}{{x_{.3}} - {x_{.4}} = 41.67 - 41.33 = 0.33 < w = 7.37}\\{{x_{.2}} - {x_{.4}} = 45.67 - 41.33 = 4.33 < w = 7.37}\\{{x_{.1}} - {x_{.4}} = 50.33 - 41.33 = 9.00 > w = 7.37}\\{{x_{.2}} - {x_{.3}} = 45.67 - 41.67 = 4.00 < w = 7.37}\\{{x_{.1}} - {x_{.3}} = 50.33 - 41.67 = 8.67 > w = 7.37}\\{{x_{.1}} - {x_{.2}} = 50.33 - 45.67 = 4.67 > w = 7.37}\end{aligned}\)

The differences are the ones smaller than w; thus, do not differ a significantly. The other pair do differ significantly. The pair which differ are

(4,1), (3,1), (2,1).

It is easy to notice that the first brand significantly differs from other brands. This can be represented using line as