Question 1165189: A psychiatrist is testing a new antianxiety drug, which seems to have the potentially harmful side effect of lowering the heart rate. For a sample of 50 medical students whose pulse was measured after 6 weeks of taking the drug, the mean heart rate was 70 beats per minute (bpm). If the mean heart rate for the population is 72 bpm with a standard deviation of 12, can the psychiatrist conclude that the new drug lowers heart rate significantly?
Answer by CPhill(2138)   (Show Source):
You can put this solution on YOUR website! This is a hypothesis testing problem. We will use a one-sample $Z$-test for the mean, as the population standard deviation ($\sigma$) is known and the sample size ($n=50$) is large.
## 1. State the Hypotheses
The psychiatrist is concerned that the drug **lowers** the heart rate.
* **Null Hypothesis ($H_0$):** The drug has no effect; the mean heart rate ($\mu$) is still $72$ bpm.
$$H_0: \mu = 72$$
* **Alternative Hypothesis ($H_1$):** The drug lowers the mean heart rate ($\mu$). (This is a left-tailed test.)
$$H_1: \mu < 72$$
## 2. Identify the Given Data
| Variable | Symbol | Value |
| :---: | :---: | :---: |
| Population Mean (Hypothesized) | $\mu_0$ | $72 \text{ bpm}$ |
| Population Standard Deviation | $\sigma$ | $12 \text{ bpm}$ |
| Sample Size | $n$ | $50$ |
| Sample Mean | $\bar{x}$ | $70 \text{ bpm}$ |
| Significance Level | $\alpha$ | $0.05$ (Standard assumption since none is given) |
## 3. Calculate the Test Statistic ($Z_{\text{test}}$)
We use the $Z$-test formula:
$$Z_{\text{test}} = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}$$
### a. Calculate the Standard Error ($\text{SE}$)
$$\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{50}} \approx \frac{12}{7.071}$$
$$\text{SE} \approx 1.697 \text{ bpm}$$
### b. Calculate the $Z$-score
$$Z_{\text{test}} = \frac{70 - 72}{1.697} = \frac{-2}{1.697}$$
$$Z_{\text{test}} \approx -1.1786$$
## 4. Determine the P-value
Since this is a left-tailed test, the $P$-value is the probability of observing a $Z$-score of $-1.1786$ or lower.
$$P\text{-value} = P(Z \le -1.1786)$$
Using a standard normal distribution table or calculator:
$$P\text{-value} \approx 0.1192$$
## 5. Make a Decision and Conclusion
We compare the $P$-value to the assumed significance level, $\alpha = 0.05$.
$$P\text{-value} (0.1192) > \alpha (0.05)$$
**Decision:** Since the $P$-value is greater than $\alpha$, we **fail to reject the null hypothesis ($H_0$)**.
**Conclusion:**
The psychiatrist **cannot conclude** that the new drug significantly lowers the heart rate. The sample mean of $70 \text{ bpm}$ is not far enough below the population mean of $72 \text{ bpm}$ to be considered statistically significant at the $0.05$ level. The observed difference could easily be due to random sampling variation.
|
|
|
