91Ó°ÊÓ

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

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 \(H:0:{\sigma _1} = \sigma _2^2\) versus\({H_a}:\sigma _1^2 > \sigma _2^2\), where \({\sigma _1},{\sigma _2}\) denote the variance of the not fused population and fused population, respectively. The test is one sided (upper).

From the exercise 9.7, page 376, the corresponding sample standard deviations are \({s_1} = 277.3\) and\({s_2} = 205.9\). The degrees of freedom are \(m - 1 = 10 - 1 = 9\)and\(n - 1 = 8 - 1 = 7\).

Thus, the value of the statistic is

\(\begin{array}{c}f = \frac{{s_1^2}}{{s_2^2}}\\ = \frac{{{{277.3}^2}}}{{{{205.9}^2}}}\\ = 1.814\end{array}\)

\(\)

At significance level\(\alpha = 0.01\), and mentioned degrees of freedom, value \({F_{0.01,9,7}}\) is

\({F_{0.01,9,7}} = 6.72\)

From the table in the appendix (or you could use a software which is better).

The test is upper sided, and

\({F_{0.01,9,7}} = 6.72 > 1.814 = f\)

Hence do not reject null hypothesis at given significance level.

P-value approach:

For the upper sided test, the\(P\)value is

\(\begin{array}{c}P = P(F > 1.814)\\ = 0.2223\end{array}\)

Where \(F\) has Fisher's distribution with degrees of freedom \({\nu _1} = 9\)and \({\nu _2} = 7.\)The value was computed using a software. The \(P\) value is large, and

\(P = 0.2223 > 0.05 = \alpha \)

Thus. do not reject null hypothesis at any reasonable significance level. The variances are equal.