1. **Problem statement:** We want to test if the normal body temperature is less than 98.6 degrees. 2. **Given data:** Sample size $n=18$, sample mean $\bar{x}=98.217$, sample standard deviation $s=0.684$, significance level $\alpha=0.05$. 3. **Hypotheses:** - Null hypothesis $H_0: \mu = 98.6$ - Alternative hypothesis $H_a: \mu < 98.6$ 4. **Test statistic:** Since population standard deviation is unknown and sample size is small, use the t-test: $$ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} $$ where $\mu_0=98.6$. 5. **Calculate t:** $$ t = \frac{98.217 - 98.6}{0.684/\sqrt{18}} = \frac{-0.383}{0.1613} \approx -2.375 $$ 6. **Degrees of freedom:** $df = n-1 = 17$. 7. **Critical value:** For a left-tailed test at $\alpha=0.05$ and $df=17$, $t_{critical} \approx -1.740$. 8. **Decision rule:** Reject $H_0$ if $t < t_{critical}$. 9. **Conclusion:** Since $-2.375 < -1.740$, reject $H_0$. There is sufficient evidence at the 5% level to conclude the mean body temperature is less than 98.6 degrees. **Final answer:** The data supports the claim that normal body temperature is less than 98.6 degrees.