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Hypotheses are typically written in the form of if/then statements, such as if someone consumes a lot of sugar, they will develop cavities in their teeth.

Given,

The data is given table.

This table summarizes everything needed to carry out F teat. Here is explanation How to obtain those values.

\(I = 4\) Column-treatments

And

\(J = 6\)Row,

The following table needs to be filled with corresponding values:

The degrees of freedom are

\(\begin{aligned}{l}I - 1 = 4 - 1 = 3\\I.\left( {J - 1} \right)4.\left( {6.1} \right) = 36\\I\,\,.\,J - 1 = 4.6 - 1 = 23\end{aligned}\)

Denote with

\(\begin{aligned}{l}{x_{i.}}\sum\limits_{j = 1}^J {{x_{ij.}}} \\{x_{..}}\sum\limits_{i = 1}^J {\sum\limits_{j = 1}^J {{x_{ij.}}} } \end{aligned}\)

The total sum of squares

\(\left( {SST} \right),\)

And treatment sum of squares

\(\left( {SSTr} \right)\),

And Error sum of squares

\({\rm{ }}\left( {SSE} \right)\)are given by

\(SST = \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{{\left( {{x_{ij}} - \overline x ..} \right)}^2}} = } \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{x^2}_{ij} - \frac{1}{{I\,\,.\,J}}} {x^2}_{..;}} \)

\(SSTr = \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{{\left( {\overline {{x_{i.}}} - \overline x ..} \right)}^2}} = \frac{1}{J}.} \sum\limits_{j = 1}^J {{x^2}_{i.} - \frac{1}{{I\,\,.\,J}}{x^2}_{..;}} \)

\(SSE = \sum\limits_{i = 1}^I {{{\sum\limits_{j = 1}^J {\left( {{x_{ij}} - \overline {{x_{i.}}} } \right)} }^2}} .\)

The mean squares are

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

F is ratio of the two mean

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

Compute all value:

\(\begin{aligned}{l}{X_{1.}} = 5.2 + 4.5 + ... + 5.8 = 34.3\\{X_{2.}} = 6.5 + 8 + ... + 5.6 = 39.6\\{X_{3.}} = 5.8 + 4.7 + ... + 5.2 = 33\\{X_{4.}} = 8.3 + 6.1 + ... + 7.2 = 41.9\end{aligned}\)

Values of

\(\overline {{x_{i.}}} = \frac{1}{J}.{x_{i.}}\)

Are given by

\(\begin{aligned}{l}\overline {{x_{1.}}} = \frac{1}{6}.34.3 = 5.72\\\overline {{x_{2.}}} = \frac{1}{6}.39.6 = 6.06\\\overline {{x_{3.}}} = \frac{1}{6}.33 = 5.50\\\overline {{x_{4.}}} = \frac{1}{6}.41.9 = 6.98\end{aligned}\)

The grand mean is

\(\overline {x..} = \frac{1}{{I\,\,.\,\,J}}.x.. = \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{x_{ij}}} = \frac{1}{{4.6}}.\left( {5.2 + 6.5 + ... + 5.2 + 7.2} \right)} \)

And

\(x.. = \sum\limits_{j = 1}^J {{x_{ij}} = } \left( {5.2 + 6.5 + ... + 5.2 + 7.2} \right)\)\( = 148.8\)

Total sum square

\(SST = \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{{\left( {\overline {{x_{ij}}} . - \overline {x..} } \right)}^2}} = } \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{x^2}_{ij}} } - \frac{1}{{I\,\,.\,J}}{x^2}..\)

\(\begin{aligned}{l} = \left( {{{5.2}^2} + {{6.5}^2} + ... + {{5.2}^2} + {{7.2}^2}} \right) - \frac{1}{{4\,.\,6}}{.148.8^2}\\ = 946.68 - 922.56\\ = 24.12\end{aligned}\)

The treatment sum of square is

\(SSTr = \sum\limits_{i = 1}^I {\sum\limits_{j = 1}^J {{{\left( {\overline {{x_i}} . - \overline {x..} } \right)}^2}} = \frac{1}{J}.} {\rm{ }}\sum\limits_{i = 1}^I {{x_{i.}}^2 - \frac{1}{{I - J}}{x^2}..} \)

\(\begin{aligned}{l} = \frac{1}{6}.\left( {{{34.3}^2} + {{39.6}^2} + {{33}^2} + {{41.9}^2}} \right) - \frac{1}{{4.\,6}}{.148.8^2}\\ = 931.54 - 922.56\\ = 8.98\end{aligned}\)

Fundamental Identify

SST = SSTr + SSE.

Error sum of squares is

\(SSE = SST - SSTr = 24.12 - 8.98 = 15.14\)

The mean computed

\(\begin{aligned}{l}MSTr = \frac{1}{{I - 1}}.SSTr = \frac{1}{3}.8.98 = 2.99\\MSE = \frac{1}{{I\,.\left( {J - 1} \right)}}.SSE = \frac{1}{{4.\left( {6 - 1} \right)}}.15.14 = 0.757\end{aligned}\)

The value of F statistic is

\(f = \frac{{MSTr}}{{MSE}} = \frac{{2.99}}{{0.757}} = 3.95\)

ANOVA table now

As for usual tests, you can either make conclusion about the hypotheses look at the F critical value or a p value. Remember that the hypotheses of interest are

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

versus alternative hypothesis

\({H_a}:\)at least two of the are different

The p value is the area to the right of f value under the F curve where F has Fisher's distribution with degrees of freedom \(4\) and \(30\) ; thus

\(P = P\left( {F > f} \right) = P\left( {33.95} \right) = 0.023\)

which was computed using software

\({\rm{P = 0}}{\rm{.023 < 0}}{\rm{.05 = }}\alpha \)

Reject null hypothesis at given significance level. There is no statistically significance difference in true averages among the four types of iron formation.

Using the table, you could use e.g. \({F_{0,05,3,20}}\,\)value for which the area under the curve to the right \({F_{0,05,3,20}}\,\,is\,0.05\)of. The value is

\({F_{0,05,3,30}} = 3.1 < 3.95 = f\)

And value \({F_{0,05,3,20}}\) for which the area under the F curve to the right

\({F_{0,05,3,20}} = 4.94 < 3.95 = f\)

Following holds

The p value is between \(0.01\)and\(0.05\)which indicates to reject null hypothesis

Significance level\(0.05\)

Hence,

Reject null hypothesis at any reasonable significance level.