A certain drug is used to treat asthma. In a clinical trial of the drug, 15 of 295 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 $9 \%$ 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. c. What is the P-value? $P$-value $=$ (Round to four decimal places as needed.) d. What is the null hypothesis, and what do you conclude about it? Identify the null hypothesis. A. $\mathrm{H}_{0}: \mathrm{p} \neq 0.09$ B. $H_{0}: p=0.09$

Step 1 ：We are testing the claim that less than 9% of treated subjects experienced headaches. The null hypothesis would be that the proportion of subjects who experienced headaches is equal to 9%.

Step 2 ：To calculate the P-value, we can use the normal approximation to the binomial distribution. The formula for the z-score is: \[ z = \frac{p - \hat{p}}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}} \] where p is the proportion under the null hypothesis (0.09 in this case), \(\hat{p}\) is the observed proportion (15/295), and n is the sample size (295).

Step 3 ：Substituting the given values into the formula, we get: p = 0.09, \(\hat{p}\) = 0.05084745762711865, n = 295, z = -2.349789859949005.

Step 4 ：Using a standard normal distribution table, we find the P-value corresponding to the calculated z-score. The P-value is 0.009392006201522616.

Step 5 ：Final Answer: The P-value is \(\boxed{0.0094}\). The null hypothesis is \(\boxed{H_{0}: p=0.09}\). Since the P-value is less than the significance level of 0.05, we reject the null hypothesis. This suggests that the proportion of subjects who experienced headaches is significantly less than 9%.