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Testing null hypothesis \({H_0}:\sigma _1^2 = \sigma _2^2\) versus alternative hypothesis\({H_a}\), under the assumption that the two populations are normal and independent, can be performed using test statistic value

\(f = \frac{{s_1^2}}{{s_2^2}}\)

Depending on alternative hypothesis \({H_a}\) the \(P\)value is corresponding area under the \({F_{m - 1,n - 1}}\)curve.

The hypotheses of interest are \({H_0}:\sigma _1^2 = \sigma _2^2\)versus\({H_a}:\sigma _1^2 \ne \sigma _2^2\). The missing values to compute the \(f\) value are the sample standard deviations \({s_1}\)and \[{s_2}\]

The Sample Variance \[{s_2}\] is

\({s^2} = \frac{1}{{n - 1}} \cdot {S_{xx}}\)

where

\({S_{xx}} = \sum {{{\left( {{x_i} - \bar x} \right)}^2}} = \sum {x_i^2} - \frac{1}{n} \cdot {\left( {\sum {{x_i}} } \right)^2}\)

The Sample Standard Deviation\(s\)is

\(s = \sqrt {{s^2}} = \sqrt {\frac{1}{{n - 1}} \cdot {S_{xx}}} \)

Thus, since\(m = 10\)and\[n = 5,\]the sample standard deviations are

\({s_1}{\rm{ }} = \sqrt {\frac{1}{{10 - 1}} \cdot \left[ {\left( {{{29}^2} + {{34}^2} + \ldots + {{27}^2}} \right) - \frac{1}{{10}} \cdot {{(29 + 34 + \ldots + 27)}^2}} \right]} \)

\( = 2.75\)

and

\({s_1}{\rm{ }} = \sqrt {\frac{1}{{5 - 1}} \cdot \left[ {\left( {{{18}^2} + {{15}^2} + \ldots + {{12}^2}} \right) - \frac{1}{5} \cdot {{(18 + 15 + \ldots + 12)}^2}} \right]} \)

\( = 4.44\)

Therefore, the\(f\)statistic value is

\(\begin{array}{c}f = \frac{{{{2.75}^2}}}{{{{4.44}^2}}}\\ = 0.384\end{array}\)

The corresponding \(P\)value, because the alternative hypothesis is two sided, is

\(2 \cdot {\rm{min}}\left( {{A_R},{A_L}} \right)\)

Using software, it is easy to obtain the values to the left of\(f\)value

\({A_L} = P(F \le 0.384) = 0.107.\)

and to the right of\(f\)

\({A_R} = P(F > 0.384) = 0.893\)

Where\(f\)has Fisher's distribution with\(m - 1 = 10 - 1 = 9\)and\(n - 1 = 5 - 1 = 4\)degrees of freedom. The minimum value is\[{A_L},\]thus the\(P\)value is

\(\begin{array}{c}P = 2{A_L}\\ = 2 \times 0.107\\ = 0.214\end{array}\)

The \(P\) value could be obtained using the table in the appendix but in a little harder way. First take reciprocal of \(f\), which is

\(\frac{1}{f} = \frac{1}{{0.384}} = 2.61\)

and the degrees of freedom are now \(4\) and \(9\) (opposite of before). From the table you can notice that the one tailed value is a little bigger than \(0.1\)(thus two tailed is twice as big) and nearly \(0.2\)(it is known that is is equal to \(0.214\)as mentioned before).

However, since

\(P = 0.214 > 0.1 = \alpha \)

Do not reject null hypothesis at the given significance level. There is no statistical significant difference between the two standard deviations.