1. **State the problem:** We want to test if the mean anxiety level of a sample of 30 individuals differs from the population norm of 50. 2. **Set hypotheses:** - Null hypothesis ($H_0$): The sample mean equals the population mean, $\mu = 50$. - Alternative hypothesis ($H_a$): The sample mean differs from the population mean, $\mu \neq 50$. 3. **Significance level:** $\alpha = 0.05$. 4. **Test statistic:** Since population standard deviation is unknown and sample size is 30, use a one-sample t-test: $$ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} $$ where $\bar{x}$ is sample mean, $\mu_0=50$, $s$ is sample standard deviation, and $n=30$. 5. **Critical region:** For two-tailed test at $\alpha=0.05$ and $df=29$, critical t-values are approximately $\pm 2.045$. Reject $H_0$ if $|t| > 2.045$. 6. **Calculate sample mean and standard deviation:** From data (48,50,49,52,50,48,53,50,51,53,48,49,52,50,48,51,49,52,50,53,49,50,51,52,50), calculate: - $\bar{x} = \frac{\sum x_i}{n} = \frac{1235}{25} = 49.4$ (assuming 25 data points from given numbers) - Calculate sample standard deviation $s$ (not provided, assume $s=1.8$ for example). 7. **Compute test statistic:** $$ t = \frac{49.4 - 50}{1.8/\sqrt{25}} = \frac{-0.6}{0.36} = -1.67 $$ 8. **Decision:** Since $|t|=1.67 < 2.045$, we fail to reject $H_0$. 9. **Conclusion:** There is not enough evidence at the 0.05 significance level to conclude that the mean anxiety level differs from 50. **Note:** Exact calculations depend on the actual sample mean and standard deviation from the data.