1. **State the problem:** We want to test if the standard deviation $\sigma$ of the golf professional's scores on his home course is equal to 1.20 (null hypothesis) against the alternative that it is not equal to 1.20 (two-tailed test). 2. **Given data:** - Sample size $n = 10$ - Sample standard deviation $s = 1.32$ - Hypothesized population standard deviation $\sigma_0 = 1.20$ 3. **Hypotheses:** - Null hypothesis $H_0: \sigma = 1.20$ - Alternative hypothesis $H_a: \sigma \neq 1.20$ 4. **Test statistic:** For testing variance or standard deviation, we use the chi-square test: $$\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}$$ 5. **Calculate the test statistic:** $$\chi^2 = \frac{(10-1) \times (1.32)^2}{(1.20)^2} = \frac{9 \times 1.7424}{1.44} = \frac{15.6816}{1.44} = 10.89$$ 6. **Degrees of freedom:** $df = n - 1 = 9$ 7. **Decision rule:** At a chosen significance level (commonly $\alpha = 0.05$), find critical values from chi-square distribution with 9 degrees of freedom. The null hypothesis is rejected if $\chi^2$ is less than the lower critical value or greater than the upper critical value. 8. **Interpretation:** Compare $\chi^2 = 10.89$ with critical values (approximately 2.7 and 19.0 for $\alpha=0.05$ two-tailed). Since $10.89$ lies between these values, we do not reject the null hypothesis. **Final conclusion:** There is not enough evidence to conclude that the consistency (standard deviation) of his home course scores differs from 1.20.