A certain drug is used to treat asthma. In a clinical trial of the​ drug, 28 of 278 treated subjects experienced headaches​ (based on data from the​ manufacturer). The accompanying calculator display shows results from a test of the claim that less than 12​% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts​ (a) through​ (e) below.

To address this problem, we need to perform a hypothesis test for a population proportion using the normal distribution as an approximation to the binomial distribution. Here's the step-by-step process:

### Part (a): State the Null and Alternative Hypotheses
- **Null Hypothesis (\(H_0\))**: The proportion of treated subjects who experience headaches is 12% (\(p = 0.12\)).
- **Alternative Hypothesis (\(H_1\))**: The proportion of treated subjects who experience headaches is less than 12% (\(p < 0.12\)).

### Part (b): Verify the Requirements for Using the Normal Distribution
To use the normal distribution as an approximation to the binomial distribution, we need to check if both \(np \geq 5\) and \(n(1-p) \geq 5\).

- \(n = 278\)
- \(p = 0.12\)

Calculate:
- \(np = 278 \times 0.12 = 33.36\)
- \(n(1 - p) = 278 \times 0.88 = 244.64\)

Both values are greater than 5, so we can use the normal approximation.

### Part (c): Test Statistic
The test statistic for a proportion is calculated using the formula:
\[ z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \]

Where:
- \(\hat{p}\) is the sample proportion
- \(p\) is the population proportion
- \(n\) is the sample size

Given:
- \(\hat{p} = \frac{28}{278} \approx 0.1007\)
- \(p = 0.12\)
- \(n = 278\)

Calculate:
\[ z = \frac{0.1007 - 0.12}{\sqrt{\frac{0.12 \times 0.88}{278}}} \]
\[ z = \frac{-0.0193}{\sqrt{\frac{0.1056}{278}}} \]
\[ z = \frac{-0.0193}{0.0194} \approx -0.994 \]

### Part (d): P-value and Conclusion
The p-value is the probability of obtaining a test statistic at least as extreme as the one computed, assuming the null hypothesis is true. 

Using the standard normal distribution table (or a calculator), find the p-value for \(z = -0.994\):
\[ P(Z < -0.994) \approx 0.1609 \]

Since the p-value (0.1609) is greater than the significance level (0.05), we fail to reject the null hypothesis.

### Part (e): Conclusion
There is not enough evidence to support the claim that less than 12% of treated subjects experienced headaches.

If you have any questions or need further details on any part, please let me know!

Here are 5 questions you might want to ask next:
1. How do you interpret the results of this hypothesis test?
2. What are the potential sources of error in this clinical trial?
3. How can we improve the accuracy of the test in future studies?
4. What are the assumptions behind using the normal approximation to the binomial distribution?
5. How would the conclusion change if the significance level was 0.10 instead of 0.05?

**Tip:** Always ensure that the sample size is sufficiently large to use the normal approximation to the binomial distribution for more accurate results.

Learn how to conduct hypothesis testing for a population proportion using a normal distribution approximation. Case study involves a clinical trial where a drug is tested for treating asthma, focusing on headaches as a side effect. Detailed step-by-step solution provided.

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