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The data on oxidation-induction time (min) for various commercial oils is provided

Let x represent the sample values.

The sample mean is computed as,

\(\begin{array}{c}\bar x &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{87 + 103 + 130 + ... + 119 + 129}}{{18}}\\ \approx 134.89\end{array}\)

Thus, the sample mean is 134.89 min.

The sample variance is given as,

\(\begin{array}{c}{s^2} &=& \frac{{\sum {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}\\ &=& \frac{{{S_{xx}}}}{{n - 1}}\end{array}\)

The calculations are represented as,

Substituting the values, the variance is given as,

\(\begin{array}{c}{s^2} &=& \frac{{\sum {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}\\ &=& \frac{{22765.7899}}{{19 - 1}}\\ &=& 1264.766\end{array}\)

Therefore, the sample variance is 1264.77 min square.

The standard deviation is computed as,

\(\begin{array}{c}s &=& \sqrt {{s^2}} \\ &=& \sqrt {1264.766} \\ &=& 35.564\end{array}\)

Therefore, the standard deviation is 35.564 min.

b.

Referring to part a, the sample variance is 1264.766 min and the sample standard deviation is 35.564 min.

If the observations wereexpressed in hours,the measures of variability change as per the constant change in each of the sample observations.

In this case, each observation would be divided by 60 to transform in terms of hours.

The variance measure for the observations expressed in hours is computed as follows,

\(\begin{array}{c}{\left( {{s^{'}}} \right)^2} &=& \frac{{{s^2}}}{{{{60}^2}}}\\ &=& \frac{{1264.766}}{{{{60}^2}}}\\ &=& 0.351\end{array}\)

The standard deviation is computed as,

\(\begin{array}{c}{s^{'}} &=& \frac{s}{{60}}\\ &=& \frac{{35.564}}{{60}}\\ &=& 0.593\end{array}\)

Therefore, if the data is re expressed in hours, the variance and standard deviation is 0.351 hour square and 0.593 hour respectively.