1. **State the problem:** We want to test if the average weight difference (weight after - weight before) is less than 0, supporting Susie's claim of guaranteed weight loss. 2. **Set hypotheses:** - Null hypothesis $H_0$: $\mu = 0$ (no weight loss on average) - Alternative hypothesis $H_a$: $\mu < 0$ (weight loss on average) 3. **Given data:** - Sample mean difference $\bar{x} = -3.40$ - Sample standard deviation $s = 3.51$ - Sample size $n = 5$ - Significance level $\alpha = 0.05$ 4. **Test statistic:** Use the one-sample t-test since population standard deviation is unknown and sample size is small. $$ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{-3.40 - 0}{3.51/\sqrt{5}} = \frac{-3.40}{1.57} \approx -2.17 $$ 5. **Degrees of freedom:** $df = n - 1 = 4$ 6. **Critical value:** For a left-tailed test at $\alpha=0.05$ and $df=4$, from t-tables or calculator: $$ t_{critical} \approx -2.132 $$ 7. **Decision rule:** Reject $H_0$ if $t < t_{critical}$. 8. **Compare:** $t = -2.17 < -2.132 = t_{critical}$, so we reject the null hypothesis. 9. **Conclusion:** There is sufficient evidence at the 0.05 level to support that the average weight difference is less than zero, supporting Susie's claim. **Answer:** I only (Reject the null hypothesis)