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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.

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

where \({\rm{ }}{\mu _1}\,and\,\,{\mu _2}\) are true averages lumen output for brand 1 and 2 bulbs, respectively.

The mean squares are is ratio of the two mean squares Using given values in the exercise, the mean squares are

\(\begin{aligned}{l}MS{T_r} = \frac{1}{{1 - 1}}SS{T_r}:\\MSE = \frac{1}{{I.\left( {J - 1} \right)}}.SSE.\end{aligned}\)

F is ratio of the two mean squares

\(F = \frac{{MS{T_r}}}{{MSE}}\)

Using given values in the exercise, the mean squares are

\(MS{T_r} = \frac{1}{{1 - 1}}.SS{T_r} = \frac{{591.2}}{2} = 295.6;\)

\(MSE = \frac{1}{{I.\left( {J - 1} \right)}}.SSE.\frac{{477.3}}{{21}} = 227.3\)

Hence, the value of statistic F is

\(F = \frac{{MS{T_r}}}{{MSE}} = \frac{{295.6}}{{227.3}} = 1.3.\)

The exact p value can be computed using software - the area under the F curve with degrees of freedom \(2\)and \(21\) to the right of value

\(P = P\left( {F > f} \right) = P\left( {F > 1.3} \right) = 0.294.{\rm{ }}\)

Because \(p > 0.05\)

do not reject null hypothesis \({F_{0.1,2,21}} = \,\,\,2.57 > 1.3 = f,\)

there are not statistically significant difference in the true averages

. Using the table, you could use e.g. value \({F_{0.1,2,21}}\) for which the area under the F curve to the right of \({F_{0.1,2,21}}\)is \(0.1\)The value is

which indicates not to reject null hypothesis

Hence,

Do not reject null hypothesis