1. The problem involves hypothesis testing for a population mean $\mu$ with null hypothesis $H_0: \mu = \mu_0$ and alternative hypothesis $H_1: \mu < \mu_0$. 2. The test statistic critical value $x_c$ is given by the formula: $$x_c = \mu_0 - t_{n-1,1-\alpha} \cdot \frac{s}{\sqrt{n}}$$ where: - $\mu_0$ is the hypothesized mean under $H_0$, - $t_{n-1,1-\alpha}$ is the critical value from the t-distribution with $n-1$ degrees of freedom at significance level $\alpha$, - $s$ is the sample standard deviation, - $n$ is the sample size. 3. We reject the null hypothesis $H_0$ if the sample mean $\bar{x} < x_c$. 4. This is a left-tailed t-test because the alternative hypothesis is $\mu < \mu_0$. 5. To perform the test: - Calculate the critical value $x_c$ using the formula. - Compare the sample mean $\bar{x}$ to $x_c$. - If $\bar{x} < x_c$, reject $H_0$; otherwise, do not reject $H_0$. This method helps determine if there is sufficient evidence to conclude the population mean is less than $\mu_0$ at the chosen significance level.