1. **State the problem:** A researcher wants to test if the fat content of a processed food exceeds 30% based on a sample of size $n=7$ with values 31.5, 30.3, 31.1, 30.7, 29.9, 29.6, and 31.8. 2. **Set hypotheses:** - Null hypothesis $H_0$: $\mu \leq 30$ (fat content does not exceed 30%) - Alternative hypothesis $H_a$: $\mu > 30$ (fat content exceeds 30%) 3. **Calculate sample mean $\bar{x}$:** $$\bar{x} = \frac{31.5 + 30.3 + 31.1 + 30.7 + 29.9 + 29.6 + 31.8}{7} = \frac{215.9}{7} = 30.8429$$ 4. **Calculate sample standard deviation $s$:** First find squared deviations: $$(31.5 - 30.8429)^2 = 0.4316$$ $$(30.3 - 30.8429)^2 = 0.2956$$ $$(31.1 - 30.8429)^2 = 0.0663$$ $$(30.7 - 30.8429)^2 = 0.0205$$ $$(29.9 - 30.8429)^2 = 0.8897$$ $$(29.6 - 30.8429)^2 = 1.5473$$ $$(31.8 - 30.8429)^2 = 0.9153$$ Sum of squared deviations = 4.1663 Sample variance: $$s^2 = \frac{4.1663}{7-1} = \frac{4.1663}{6} = 0.6944$$ Sample standard deviation: $$s = \sqrt{0.6944} = 0.8331$$ 5. **Calculate test statistic $t$:** $$t = \frac{\bar{x} - 30}{s/\sqrt{n}} = \frac{30.8429 - 30}{0.8331/\sqrt{7}} = \frac{0.8429}{0.315} = 2.676$$ 6. **Determine critical value:** At significance level $\alpha = 0.01$ and degrees of freedom $df = 6$, the critical $t$ value for a one-tailed test is approximately 3.143. 7. **Make decision:** Since calculated $t = 2.676 < 3.143$, we fail to reject the null hypothesis. 8. **Conclusion:** At the 0.01 significance level, there is not enough evidence to conclude that the fat content exceeds 30%. Therefore, the researcher would conclude (a) Fat contents do not exceed 30%.