1. **Problem Statement:** We want to find the critical value $c$ such that the null hypothesis $H_0: \mu = 20$ is rejected if the sample mean $\overline{X} > c$ at a 5% level of significance. 2. **Given Data:** - Population mean under $H_0$: $\mu = 20$ - Population standard deviation: $\sigma = 6$ - Sample size: $n = 36$ - Significance level: $\alpha = 0.05$ 3. **Test Type:** This is a right-tailed z-test because $H_A: \mu > 20$. 4. **Formula for the critical value $c$:** $$ c = \mu + z_{\alpha} \times \frac{\sigma}{\sqrt{n}} $$ where $z_{\alpha}$ is the z-score corresponding to the upper $\alpha$ percentile of the standard normal distribution. 5. **Find $z_{\alpha}$:** At 5% significance level for a right-tailed test, $z_{0.05} = 1.645$ (from z-tables). 6. **Calculate the standard error:** $$ \frac{\sigma}{\sqrt{n}} = \frac{6}{\sqrt{36}} = \frac{6}{6} = 1 $$ 7. **Calculate $c$:** $$ c = 20 + 1.645 \times 1 = 20 + 1.645 = 21.645 $$ 8. **Final answer:** The critical value $c$ at 5% significance level is **21.65** (rounded to two decimal places).