1. **Problem:** For test $H_0: \mu=100$ vs $H_1: \mu>100$, sample $z=2.15$. Find the p-value. 2. This is a right-tailed test so p-value = $P(Z \geq 2.15)$. 3. Using standard normal tables or calculator, $P(Z \geq 2.15) = 1 - P(Z \leq 2.15) = 1 - 0.9842 = 0.0158$. 4. **Problem:** For test $H_0: \mu=100$ vs $H_1: \mu \neq 100$, sample $z=2.15$. Find p-value. 5. This is a two-tailed test, so p-value = $2P(Z \geq 2.15) = 2(0.0158) = 0.0316$. 6. **Problem:** For test $H_0: \mu=80$ vs $H_1: \mu < 80$, sample $z=1.63$. Find p-value. 7. This is a left-tailed test, so p-value = $P(Z \leq 1.63)$. 8. From tables, $P(Z \leq 1.63) = 0.9484$. But since alternative is $\mu < 80$, p-value = $1 - P(Z \leq 1.63) = 1 - 0.9484 = 0.0516$. 9. **Problem:** For test $H_0: \mu=72$ vs $H_1: \mu \neq 72$, sample $z=1.63$. Find p-value. 10. Two-tailed test, p-value = $2 P(Z \geq 1.63) = 2(1 - 0.9484) = 2(0.0516) = 0.1032$. 11. **Problem:** A sample of $n=200$ school managers had mean $\bar{x}=78$, $s=4.2$. Population mean $\mu=73$, $\sigma=8$. Test for significant difference at $\alpha=0.05$ using p-value method. 12. Calculate test statistic: $$z=\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} = \frac{78 - 73}{\frac{4.2}{\sqrt{200}}} = \frac{5}{4.2 / 14.142} = \frac{5}{0.296} = 16.89$$ 13. This is a two-tailed test (difference test), find p-value: $p = 2 P(Z \geq 16.89)$. 14. Since 16.89 is extremely high, p-value is nearly zero, much less than 0.05. 15. Conclusion: Reject $H_0$, significant difference exists. **Final answers:** 1) p-value = 0.0158 2) p-value = 0.0316 3) p-value = 0.0516 4) p-value = 0.1032 5) p-value $\approx 0$, reject $H_0$.