1. The problem: Perform a t-test to determine if the mean of a sample differs significantly from a known population mean. 2. Formula: The t-test statistic is calculated as $$t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$$ where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $s$ is the sample standard deviation, and $n$ is the sample size. 3. Important rules: Use the t-distribution when the population standard deviation is unknown and the sample size is small (typically $n < 30$). 4. Example: Suppose a sample of size $n=10$ has a mean $\bar{x} = 52$ and standard deviation $s=8$. Test if this differs from a population mean $\mu=50$ at a 5% significance level. 5. Calculate the t-statistic: $$t = \frac{52 - 50}{8 / \sqrt{10}} = \frac{2}{8 / 3.162} = \frac{2}{2.529} \approx 0.79$$ 6. Degrees of freedom: $df = n - 1 = 9$. 7. Compare the calculated t-value to the critical t-value from the t-distribution table for $df=9$ at 5% significance (two-tailed), which is approximately 2.262. 8. Since $0.79 < 2.262$, we fail to reject the null hypothesis; there is no significant difference between the sample mean and population mean. This completes the t-test problem solution.