1. **State the problem:** We are testing the claim about the population variance $\sigma^2$ with the hypotheses: $$H_0: \sigma^2 = 30.9$$ $$H_a: \sigma^2 \neq 30.9$$ at the significance level $\alpha = 0.10$. 2. **Formula used:** The test statistic for variance is given by $$\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}$$ where $n$ is the sample size, $s^2$ is the sample variance, and $\sigma_0^2$ is the claimed population variance under $H_0$. 3. **Given values:** - $n = 81$ - $s^2 = 39.9$ - $\sigma_0^2 = 30.9$ 4. **Calculate the test statistic:** $$\chi^2 = \frac{(81-1) \times 39.9}{30.9} = \frac{80 \times 39.9}{30.9}$$ 5. **Intermediate step with cancellation:** $$\chi^2 = \frac{\cancel{80} \times 39.9}{\cancel{30.9}} \approx 103.24$$ 6. **Interpretation:** The calculated test statistic is $\chi^2 = 103.24$ (rounded to two decimal places). This value will be compared to the critical values from the chi-square distribution with $n-1=80$ degrees of freedom to decide whether to reject $H_0$. **Final answer:** $$\chi^2 = 103.24$$