91影视

The following table summarizes quantities

The following is required to understand how to compute mentioned and required quantities.

Unequal Sample Sizes

Let \({J_{i,\,\,}}i = 1,2, \ldots ,l\) be the sample sizes of I treatments, and let \(n = \sum\nolimits_{i = 1}^l {\,\,{J_i}} \) be the total number of observations. Denote with

\(\begin{aligned}{l}{x_i} = \sum\limits_{j = 1}^{{J_i}} {{x_{i\,j}}} \,;\\{x_{ \cdot \cdot }} = \sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^{{J_i}} {\,{x_{i\,j}}} \,} \end{aligned}\)

The sum of observations in the \({i^{th}}\) treatment

\(\begin{aligned}{l}{x_1} = 59.5 + 53.3 + \ldots + 58.7 = 291.4;\\{x_2} = 55.2 + 59.1 + 52.8 + 54.5 = 221.6;\\{x_3} = 51.7 + 48.8 + \ldots + 46.1 = 203.4;\\{x_4} = 44.6 + 48.5 + 41 + 47.3 = 227.5;\end{aligned}\)

and the averages of the \({i^{th}}\) treatment are

\(\begin{aligned}{l}{\mathop x\limits^\_ _{1.}} = \frac{1}{{{J_1}}}.{x_{1.}} = \frac{1}{5}.291.4 = 58.28;\\{\mathop x\limits^\_ _{2.}} = \frac{1}{{{J_2}}}.{x_{2.}} = \frac{1}{4}.221.6 = 55.4;\\{\mathop x\limits^\_ _{3.}} = \frac{1}{{{J_3}}}.{x_{3.}} = \frac{1}{4}.203.4 = 50.85;\\{\mathop x\limits^\_ _{4.}} = \frac{1}{{{J_4}}}.{x_{4.}} = \frac{1}{5}.225.4 = 45.5;\end{aligned}\)

Denote with

\(\,{w_{ij}} = {Q_{\alpha ,I,n - I}} \cdot \sqrt {\frac{{MSE}}{2}.\left( {\frac{1}{{{J_i}}} + \frac{1}{{{J_j}}}} \right)} ,\)

Then the probability that is

\({\mathop X\limits^\_ _{i.}} - {\mathop X\limits^\_ _{j.}} - {w_{ij}} < {\mu _i} - {\mu _j} < {\mathop X\limits^\_ _{i.}} - {\mathop X\limits^\_ _{j.}} + {w_{ij}}\)

Approximately \(1 - \alpha \) for every \(i = 1,2, \ldots ,l\,\) and every \(j = 1,2,....I\,\)where \(i \ne j\).

The MSE was computed in the exercise 22, and it is

\(MSE = 8.89\)

The quantity \({Q_{\alpha ,I,n - I}}\) can be found in the appendix, and in this case, it is

\({Q_{\alpha ,I,n - I}} = {Q_{0.05,4,18 - 4}} = {Q_{0.05,4,18 - 4}} = 4.11\,\)

The goal is to compute the confidence intervals for every, \(i = 1,2, \ldots ,l\,\) \(j = 1,2,....I\,\) where\(i \ne j\,\), and notice if zero belong to the intervals. If it does, the \(\mu \)鈥榮 are not significantly different.

Compute first \({w_{1j}}\,,\,\,J = 2,3,4\) required for the confidence intervals as

\(\begin{aligned}{l}{w_{12}} = 4.11\,\,\sqrt {\frac{{8.89}}{2}.\left( {\frac{1}{5} + \frac{1}{4}} \right)\,} \, = 5.81;\\{w_{13}} = 4.11\,\,\sqrt {\frac{{8.89}}{2}.\left( {\frac{1}{5} + \frac{1}{4}} \right)\,} \, = 5.81;\\{w_{14}} = 4.11\,\,\sqrt {\frac{{8.89}}{2}.\left( {\frac{1}{5} + \frac{1}{5}} \right)\,} \, = 5.48;\end{aligned}\)

Now compute \({w_{1j}}\,,\,\,J = 3,4\)

\(\begin{aligned}{l}{w_{23}} = 4.11\,\,\sqrt {\frac{{8.89}}{2}.\left( {\frac{1}{4} + \frac{1}{4}} \right)\,} \, = 6.13;\\{w_{24}} = 4.11\,\,\sqrt {\frac{{8.89}}{2}.\left( {\frac{1}{4} + \frac{1}{5}} \right)\,} \, = 5.81;\end{aligned}\)

and the last is \({w_{34}}\)

\({w_{34}} = 4.11\,\,\sqrt {\frac{{8.89}}{2}.\left( {\frac{1}{4} + \frac{1}{5}} \right)\,} \, = 5.81;\)

The corresponding differences (respectively to given above) are

\(\begin{aligned}{l}{\mathop x\limits^\_ _{1.}} - {\mathop x\limits^\_ _{2.}} = 2.88;\\{\mathop x\limits^\_ _{1.}} - {\mathop x\limits^\_ _{3.}} = 7.43;\\{\mathop x\limits^\_ _{1.}} - {\mathop x\limits^\_ _{4.}} = 12.78;\\{\mathop x\limits^\_ _{2.}} - {\mathop x\limits^\_ _{3.}} = 4.55;\\{\mathop x\limits^\_ _{3.}} - {\mathop x\limits^\_ _{4.}} = 9.9;\\{\mathop x\limits^\_ _{3.}} - {\mathop x\limits^\_ _{4.}} = 5.35;\end{aligned}\)

The confidence intervals

\({\mathop X\limits^\_ _{i.}} - {\mathop X\limits^\_ _{j.}} - {w_{ij}} < {\mu _i} - {\mu _j} < {\mathop X\limits^\_ _{i.}} - {\mathop X\limits^\_ _{j.}} + {w_{ij}}\)

Can be computed now using

\(\begin{aligned}{l}{\mathop x\limits^\_ _{i.}} - {\mathop x\limits^\_ _{j.}} - {w_{ij}} < {\mu _i} - {\mu _j} < {\mathop x\limits^\_ _{i.}} - {\mathop x\limits^\_ _{j.}} + {w_{ij}}\\\end{aligned}\)

Where everything has been calculated; thus, the intervals are, respectively to the differences given above,

\(\begin{aligned}{l}2.88 - 5.81 < {\mu _1} - {\mu _2} < 2.88 + 5.81;\\7.43 - 5.81 < {\mu _1} - {\mu _3} < 7.43 + 5.81;\\12.78 - 5.48 < {\mu _1} - {\mu _4} < 12.78 + 5.48;\\4.55 - 6.13 < {\mu _2} - {\mu _3} < 4.55 + 6.13;\\9.9 - 5.81 < {\mu _2} - {\mu _4} < 9.9 + 5.81;\\5.35 - 5.81 < {\mu _3} - {\mu _4} < 5.35 + 5.81\,,\end{aligned}\)

or equally

\(\begin{aligned}{l} - 2.93 < {\mu _1} - {\mu _2} < 8.69;\\1.62 < {\mu _1} - {\mu _3} < 13.24;\\7.3 < {\mu _1} - {\mu _4} < 18.26;\\ - 1.58 < {\mu _2} - {\mu _3} < 10.68;\\4.09 < {\mu _2} - {\mu _4} < 15.71;\\ - 0.46 < {\mu _3} - {\mu _4} < 11.16.\end{aligned}\)

The confidence intervals of the following differences include zero

\(\begin{aligned}{l}{\mu _1} - {\mu _2}\\{\mu _2} - {\mu _3}\\{\mu _3} - {\mu _4}.\end{aligned}\)

Hence, the conclusion is that there is no significant differences between true means of treatments 1 and 2; 2 and 3; 3 and 4. However, there is significant difference in true means of other combinations.

Here, the result is There is no significant differences between true means of treatments 1 and 2; 2 and 3; 3 and 4.