91影视

Using evidence, it is necessary to address whether Design 1 is superior to Design 2.

\(\begin{aligned}{l}n1 &= 8\\n2 &= 8\end{aligned}\)

The variance measures variability. The amount of scattered is indicated by variance.

The variance is more significant with the mean and the more dispersed the data.

\(\begin{aligned}{l}{\sigma ^2}1 &= Variance\,of\,design\,1\\{\sigma ^2}2 &= \,Variance\,of\,design\,2\end{aligned}\)

The test hypothesis is as follows.

\(\begin{aligned}{l}H0:\,{\sigma ^2}1 < {\sigma ^2}2\\Ha:\,{\sigma ^2}1 &= {\sigma ^2}2\end{aligned}\)

Denominator\({S^2}2\) and numerator\({S^2}1\) both

\(df = n1 - 1\)

\(v1 = n1 - 1\)

\(\begin{aligned}{l}df &= 8 - 1\\df &= 7\end{aligned}\)

\(df = n2 - 1\)

\(v2 = n2 - 1\)

\(\begin{aligned}{l}df &= 8 - 1\\df &= 7\end{aligned}\)

\(\begin{aligned}{l}\bar x &= \frac{{\sum xi}}{n}\\\bar x &= \frac{{9,439}}{8}\\\bar x &= 1,179.875\end{aligned}\)

\(\begin{aligned}{l}S{1^2} &= \frac{{\sum {{\left( {xi - \bar x} \right)}^2}}}{{n - 1}}\\S{1^2} = \frac{{41,047.83}}{7}\\S{1^2} &= 5,863.97\end{aligned}\)

\(\begin{aligned}{l}\bar y &= \frac{{\sum yi}}{n}\\\bar y &= \frac{{10,836}}{8}\\\bar y = 1,354.5\end{aligned}\)

\(\begin{aligned}{l}S{2^2} &= \frac{{\sum {{\left( {yi - \bar y} \right)}^2}}}{{n - 1}}\\S{2^2} &= \frac{{42,487.879}}{7}\\S{2^2} = 6,069.697\end{aligned}\)

The test statistics will therefore be.

\(\begin{aligned}{l}F &= {\textstyle{{Larger\,\,sample\,\,variance} \over {Smaller\,\,sample\,\,variance}}}\\F &= \frac{{{S^2}1}}{{{S^2}2}}\\F &= \frac{{41,047.83}}{{42,487.879}}\\F &= 0.96610\end{aligned}\)

Reject\(H0:\,{\sigma ^2}1 = {\sigma ^2}2\)for F=0.966 When the calculated design-2 exceeds the design1 .So, reject\(H0:\,{\sigma ^2}1 = {\sigma ^2}2\)

Therefore, this test indicates that both variances will differ. The probability of seeing a value of F at least as similar to \(Ha:{\sigma ^2}1 < {\sigma ^2}2\) as F=0.96610 .So, \(Ha\) is true would be design-2 is greater than design1.