1. **State the problem:** We want to test whether the average height of 40 students is less than 160 cm. 2. **Set hypotheses:** Null hypothesis, $H_0: \mu = 160$ (mean height is 160 cm) Alternative hypothesis, $H_1: \mu < 160$ (mean height is less than 160 cm) 3. **Significance level:** $\alpha = 0.05$ 4. **Data and statistics:** Sample data of heights (in cm): 40 observations given. Sample mean: $\bar{x} = 162.175$ Population standard deviation assumed known: $\sigma = 8$ Sample size: $n = 40$ 5. **Test statistic:** Use $z$-test $$z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{162.175 - 160}{8 / \sqrt{40}} = \frac{2.175}{1.2649} \approx 1.7195$$ 6. **Decision rule:** For left-tailed test at $\alpha=0.05$, reject $H_0$ if $z < -1.64$. 7. **Compute P-value:** P-value is probability $P(Z < 1.7195)$ for left-sided, but here our alternative is $\mu<160$, so we look at $P(Z < 1.7195)$ – actually, since $z$ is positive, it's not in rejection region. Observed $z = 1.7195$ is greater than $-1.64$, so fail to reject $H_0$. 8. **Conclusion:** There is insufficient evidence to conclude that the mean height is less than 160 cm. The null hypothesis that the mean height is equal to 160 cm is not rejected. This matches the earlier conclusion that p-value (0.9572) > $\alpha$ (0.05) and $z$ computed is not in rejection region. **Final answer:** Cannot reject $H_0$. Average height $\,= 160$ cm.