1. **State the problem:** We want to test the claim that protein bars contain on average 20g of protein. We have a sample of 10 bars with protein amounts: 19.1, 18.7, 21.0, 19.6, 20.3, 20.1, 18.9, 19.8, 20.4, 19.5. 2. **Set up hypotheses:** - Null hypothesis $H_0$: $\mu = 20$ (mean protein is 20g) - Alternative hypothesis $H_a$: $\mu \neq 20$ (mean protein is not 20g) 3. **Calculate sample mean $\bar{x}$:** $$\bar{x} = \frac{19.1 + 18.7 + 21.0 + 19.6 + 20.3 + 20.1 + 18.9 + 19.8 + 20.4 + 19.5}{10} = \frac{197.4}{10} = 19.74$$ 4. **Calculate sample standard deviation $s$:** First find squared deviations: $$(19.1 - 19.74)^2 = 0.4096$$ $$(18.7 - 19.74)^2 = 1.0816$$ $$(21.0 - 19.74)^2 = 1.5876$$ $$(19.6 - 19.74)^2 = 0.0196$$ $$(20.3 - 19.74)^2 = 0.3136$$ $$(20.1 - 19.74)^2 = 0.1296$$ $$(18.9 - 19.74)^2 = 0.7056$$ $$(19.8 - 19.74)^2 = 0.0036$$ $$(20.4 - 19.74)^2 = 0.4356$$ $$(19.5 - 19.74)^2 = 0.0576$$ Sum of squared deviations = $0.4096 + 1.0816 + 1.5876 + 0.0196 + 0.3136 + 0.1296 + 0.7056 + 0.0036 + 0.4356 + 0.0576 = 4.744$. Sample variance $s^2 = \frac{4.744}{10 - 1} = \frac{4.744}{9} = 0.5271$. Sample standard deviation: $$s = \sqrt{0.5271} = 0.7263$$ 5. **Test statistic (t):** $$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{19.74 - 20}{0.7263 / \sqrt{10}} = \frac{-0.26}{0.2295} = -1.133$$ 6. **Decision rule:** Degrees of freedom $df = n - 1 = 9$. At $\alpha = 0.05$ two-tailed test, critical t-value $\approx \pm 2.262$. 7. **Conclusion:** Since $-2.262 < -1.133 < 2.262$, we fail to reject $H_0$. There is not enough evidence to reject the claim that the mean protein content is 20g. **Final answer:** The sample standard deviation is approximately $0.7263$ grams.