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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

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 in weight pain for low-dose treatment and variance in weight gain for control condition, respectively.

The corresponding sample standard deviations are \({s_1} = 54\)and \({s_2} = 32.\)The degrees of freedom are \(m - 1 = 20 - 1 = 19\)and \(n - 1 = 23 - 1 = 22.\)

The test statistic is,

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

Thus, the value of the statistic is

\(\begin{array}{c}f = \frac{{{{54}^2}}}{{{{32}^2}}}\\ = 2.85\end{array}\)

At significance level \(\alpha = 0.01,\)and mentioned degrees of freedom, value \({F_{0.01,19,22}}\) is

\({F_{0.01,19,22}} = 2.0837\)

The test is upper sided, and \({F_{0.05,19,22}} = 2.0837 < 2.85 = f\)

Hence reject null hypothesis at given significance level.

The \(P\)value could have been used as well. For the upper sided test, the \(P\) value is

\(\begin{array}{c}P = P(F > 2.85)\\ = 0.010036\end{array}\)

Where \(F\)has Fisher's distribution with degrees of freedom \({\nu _1} = 19\)and\({\nu _2} = 22\). The value was computed using a software (you can use table estimate the value as mentioned above). Since for the \(P\)the following holds

\(P = 0.010036 < 0.05 = \alpha \)

Reject null hypothesis at any given significance level. Hence there is more variability in the low dose weight gains.