Question: The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the reduction in systolic blood pressure is $50.5$ for a sample of size $372$ and standard deviation $17.4$. Estimate how much the drug will lower a typical patient's systolic blood pressure (using a 95% confidence level). Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
1. **State the problem:** We want to estimate the population mean reduction $\mu$ in systolic blood pressure using a 95% confidence interval based on the sample mean $\bar{x} = 50.5$, sample size $n = 372$, and sample standard deviation $s = 17.4$.
2. **Formula for confidence interval:** For large samples, the 95% confidence interval for the mean is given by:
$$\bar{x} \pm z_{\alpha/2} \times \frac{s}{\sqrt{n}}$$
where $z_{\alpha/2}$ is the critical value from the standard normal distribution for 95% confidence, which is approximately $1.96$.
3. **Calculate the standard error (SE):**
$$SE = \frac{s}{\sqrt{n}} = \frac{17.4}{\sqrt{372}}$$
Calculate $\sqrt{372}$:
$$\sqrt{372} \approx 19.3132$$
So,
$$SE = \frac{17.4}{19.3132} \approx 0.9009$$
4. **Calculate margin of error (ME):**
$$ME = 1.96 \times SE = 1.96 \times 0.9009 \approx 1.766$$
5. **Construct the confidence interval:**
$$50.5 - 1.766 < \mu < 50.5 + 1.766$$
Calculate the bounds:
$$48.734 < \mu < 52.266$$
6. **Round to one decimal place:**
$$48.7 < \mu < 52.3$$
**Final answer:**
$$48.7 < \mu < 52.3$$