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The standard deviation of the birth weights for 30 girls is 829.5 hg.

The level of significance is 0.01 to test the claim that both girls and boys have the same standard deviation of birth weights, which is 660.2 hg.

To test the claim that the birth weights of girls have the same standard deviation as the birth weights of boys, the null and alternative hypotheses are formulated as follows.

$$\begin{gathered} H_0:蟽=660.2 \\ {H}_1:蟽鈮�660.2 \end{gathered}$$

Here, $蟽$is the true standard deviation of the birth weights for girls.

The test statistic ${蠂}^2$with degrees of freedom $\left(n-1\right)$is given as follows.

.$$\begin{gathered} {蠂}^2=\frac{\left(n-1\right)s^2}{{蟽}^2} \\ =\frac{\left(30-1\right){\left(829.5\right)}^2}{{\left(660.2\right)}^2} \\ =45.7804 \end{gathered}$$

The degree of freedom is computed as follows.

$$\begin{gathered} df=n-1 \\ =30-1 \\ =29 \end{gathered}$$

Thus, the test statistic is 45.78 with 29 degrees of freedom.

The test is two-tailed.

The critical values are ${{蠂}^2}_L,{{蠂}^2}_R$, such that

$$\begin{gathered} P\left({蠂}^2<{{蠂}^2}_L\right)=\frac{0.01}{2}\left(0.005\right) \\ P\left({蠂}^2>{{蠂}^2}_L\right)=0.995 \\ P\left({蠂}^2>{{蠂}^2}_R\right)=\frac{0.01}{2}\left(0.005\right) \end{gathered}$$

Using the chi-square table for 29 degrees of freedom and a 0.005 level of significance, the right-tailed critical value is 52.336, and the left-tailed one is 13.121.

Thus,

$$\begin{gathered} {蠂}_L^2=13.121 \\ {蠂}_R^2=52.336 \end{gathered}$$

The decision rule states the following:

If the test statistic lies between the critical values, the null hypothesis will fail to be rejected; otherwise, it will be rejected.

In this case, the test statistic lies between the critical values, and hence, the null hypothesis is failed to be rejected at a 0.01 level of significance.

Thus, it can be concluded that there is sufficient evidence to support the claim that the standard deviation of the birth weights of girls is equal to the standard deviation of the birth weights of boys.