91影视

The hypotheses of interest are

\({H_0}:\,\,\,{\mu _1} = {\mu _j},\,i \ne j\,\)

Versus alternative hypothesis

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

Where \({\mu _i}\)are the true averages yield due to different salinity levels.

The sample sizes of a treatments are unequal; thus, the ANOVA for unequal sample sizes should be used. The following table summarizes quantities

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

Let \({J_{i,\,\,}}i = 1,2, \ldots ,l\,\,\,\,\,\,\,I\,\,\)be the sample sizes of 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

total sum of squares

(SST),

treatment sum of squares

(SSTr), and

error sum of squares

(SSE) are given by

\(\begin{aligned}{l}\,\,\,\,\,SST = {\sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^{{J_i}} \, \left( {\,{x_{i\,j}} - \mathop x\limits^\_ ..} \right)} ^2} = \sum\limits_{i = 1}^I \, \sum\limits_{j = 1}^{{J_i}} \, x_{_{i\,j}}^2 - \frac{1}{n}x_{ \cdot \cdot }^2\,;\\SSTr = {\sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^{{J_i}} \, \left( {\,{{\mathop x\limits^\_ }_{i\,.}} - \mathop x\limits^\_ ..} \right)} ^2} = \frac{1}{J} \cdot \sum\limits_{i = 1}^I {\frac{1}{{{J_i}}}\,} x_{_{i\,.}}^2 - \frac{1}{n}x_{ \cdot \cdot }^2\,;\\\,\,\,\,\,SST = {\sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^{{J_i}} \, \left( {\,{x_{i\,j}} - {{\mathop x\limits^\_ }_{i.}}} \right)} ^2} = \sum\limits_{i = 1}^I {\left( {{J_i} - 1} \right)\,} s_i^2,\end{aligned}\)

with degrees of freedom, respectively

\(\begin{aligned}{l}df = n - 1;\\df = I - 1;\\df = \sum\limits_{i = 1}^I \, \left( {{J_i} - 1} \right) = n - I;\end{aligned}\)

\(SST = SSTr + SSE\).

The mean squares are

\(\begin{aligned}{l}MSTr = \frac{1}{{I - 1}}.\,SSTr;\\\,\,\,\,\,MSE = \frac{1}{{n - I\,}}.\,SSE\,.\end{aligned}\)

F is ratio of the two mean squares

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

The P-value is the area under the \({F_{I - 1,\,n - I}}\)curve to the right of f.

Compute everything one by one. Starting with 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

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

The grand sum is

\({x_{ \cdot \cdot }} = \sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^{{J_i}} {\,{x_{i\,j}}} \, = 59.5 + 55.2 + \ldots + 58.7 + 46.1 = 943.9,} \)

And the grand mean is

\(\mathop x\limits^\_ ..\frac{1}{n}.\,{x_{ \cdot \cdot }} = \frac{1}{{18}}.943.9 = 52.44\)

Degrees of freedom are, for \(n = 5 + 4 + 4 + 5 = 19\)

\(\begin{aligned}{l}df = n - 1 = 19 - 1 = 18;\\df = I - 1 = 4 - 1 = 3;\\df = \sum\limits_{i = 1}^I \, \left( {{J_i} - 1} \right) = n - I = 19 - 4 = 15;\end{aligned}\)

The total sum of squares is

\(\begin{aligned}{l}SST = {\sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^{{J_i}} \, \left( {\,{x_{i\,j}} - \mathop x\limits^\_ ..} \right)} ^2} = \sum\limits_{i = 1}^I \, \sum\limits_{j = 1}^{{J_i}} \, x_{_{i\,j}}^2 - \frac{1}{n}x_{ \cdot \cdot }^2\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {{{59.5}^2} + {{55.2}^2} + \ldots + {{58.7}^2} + {{46.1}^2}} \right) - \frac{1}{{18}}{.52.44^2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 50,078.07 - 49,497.06722\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 581.0028\end{aligned}\)

The treatment sum of squares is

\(\begin{aligned}{l}SSTr = {\sum\limits_{i = 1}^I {\,\sum\limits_{j = 1}^{{J_i}} \, \left( {\,{{\mathop x\limits^\_ }_{i\,.}} - \mathop x\limits^\_ ..} \right)} ^2} = \frac{1}{J} \cdot \sum\limits_{i = 1}^I {\frac{1}{{{J_i}}}\,} x_{_{i\,.}}^2 - \frac{1}{n}x_{ \cdot \cdot }^2\,\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{5}{.291.4^2} + \frac{1}{4}{.221.6^2} + \frac{1}{4}{.203.4^2} + \frac{1}{5}{.225.4^2} - \frac{1}{{18}}{.52.44^2}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 49,953.572 - 49,497.07\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 456.5048.\end{aligned}\)

From the fundamental Identity

\(SST = SSTr + SSE\)

The error sum of squares is

\(SSE = SST - SSTr = 581.0028 - 456.5048 = 124.498\)

The mean squares are

\(\begin{aligned}{l}MSTr = \frac{1}{{I - 1}}.\,SSTr = \frac{1}{3}.456.5048 = 152.1683;\\\,\,\,\,\,MSE = \frac{1}{{n - I}}.\,SSE\, = \frac{1}{{14}}.124.498 = 8.89\end{aligned}\)

F is ratio of the two mean squares

\(f = \frac{{MSTr}}{{MSE}} = \frac{{152.1683}}{{8.89}} = 17.12\)

The P-value is the area under the \({F_{3,14}}\)curve to the right of f

\(P\left( {F > f} \right) = P\left( {F > 17.12} \right) \approx 0\)

Which was computed using a software; you can estimate the value using the table in the appendix as

\({F_{0.05,3,14}} = 3.34\)

Where the area to the right of \({F_{0.05,3,14}} = 3.34\) under the \({F_{3,14}}\) curve is 0.05, and because

\(f = 17.12 > 3.34 = \,{F_{0.05,3,14}}\)

reject null hypothesis

At given significant level. Also, if you used the P value, because

\(P = 0 < 0.05 = \alpha \)

reject null hypothesis

At given significant level. The conclusion is that there is a significant difference in yield of tomatoes for the different salinity levels.

Here, the final result is Reject null hypothesis.