14 A certain drug is used to treat asthma. In a clinical trial of the drug, 25 of 257 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 0.01 significance level to complete parts (a) through (e) below. a. Is the test two-tailed, left-tailed, or right-tailed? Left-tailed test Right tailed test b. What is the test statistic? (Round to two decimal places as needed) c. What is the P-value? P-value \( =\square \) (Round to three decimal places as needed.)

Let’s dive into the fascinating world of statistics and hypothesis testing for this asthma drug! When it comes to the nature of the test, it's a left-tailed test. That’s because we are looking to see if the proportion of treated subjects experiencing headaches is less than 9%. In other words, we want to determine if there are fewer headaches than expected under the null hypothesis. Now, for the test statistic, we utilize the formula for a z-test for proportions. Substituting in the values given (25 headaches out of 257 subjects), we find a test statistic around -1.77 (after running through the calculations). This is just under that magical cut-off point that could lead us to reject the null hypothesis if our P-value fits the bill! And speaking of that P-value, we can find it using the calculated z-value with standard normal distribution tables or a calculator. You would find that the P-value is approximately 0.038. This is the probability of observing a sample proportion this extreme or more, given that the null hypothesis is true. If we compare this to our significance level of 0.01, it's clear that we don't have enough evidence to reject the null hypothesis. The headaches aren't less than 9%—at least as far as this sample tells us!