1. **State the problem:** Jack wants to test if the true mean pH of river water is $7.4$ using a sample of size $28$, with population standard deviation $\sigma=0.29$ and significance level $\alpha=0.05$. This is a two-tailed test: $$H_0: \mu = 7.4, \quad H_a: \mu \neq 7.4$$ 2. **Calculate the sample mean $\bar{x}$:** Sum all the 28 pH values and divide by 28. $pH$ values: $7.26, 7.54, 7.35, 7.92, 7.11, 7.01, 7.22, 6.96, 7.35, 7.41, 6.92, 7.06, 7.63, 6.92, 6.99, 7.17, 7.31, 6.61, 6.79, 7.05, 7.33, 6.96, 6.93, 7.25, 7.39, 7.46, 7.38, 7.63$ Sum $= 199.83$, so $$\bar{x} = \frac{199.83}{28} = 7.1375$$ 3. **Calculate the $z$ test statistic:** $$z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{7.1375 - 7.4}{0.29 / \sqrt{28}}$$ Calculate denominator: $$0.29 / \sqrt{28} = 0.29 / 5.2915 = 0.0548$$ Calculate $z$: $$z = \frac{-0.2625}{0.0548} = -4.79$$ 4. **Find the $p$-value for two-tailed test:** The $p$-value is $2 \times P(Z < -4.79)$. From standard normal distribution tables or software, $$P(Z < -4.79) \approx 0.0000016$$ So, $$p\text{-value} = 2 \times 0.0000016 = 0.000003$$ 5. **Conclusion:** Since $p$-value $< 0.05$, we reject $H_0$ and conclude the true mean pH differs from $7.4$. Final answers: $z = -4.79$ $p$-value $= 0.000$