Question

a certain drug is used to treat asthma. in a clinical trial of the drug, 26 of 292 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 8% 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. 1-propztest prop < 0.08 z = 0.569473598 p = 0.7154626085 p = 0.0890410959 n = 292 a. is the test two-tailed, left-tailed, or right-tailed? right tailed test left-tailed test two-tailed test b. what is the test statistic? z = 0.57 (round to two decimal places as needed.) c. what is the p-value? p-value = (round to three decimal places as needed.)

Explanation:

Step1: Identify the type of test

The claim is that less than 8% of treated subjects experienced headaches. In hypothesis testing, when the alternative hypothesis is of the form \(H_1:p < p_0\) (where \(p_0 = 0.08\) in this case), it is a left - tailed test.

Step2: Recall the formula for the P - value in a left - tailed z - test

For a left - tailed z - test, the P - value is \(P(Z<z)\), where \(z\) is the test statistic. Given \(z = 0.57\) (this is incorrect as per the calculator display \(z=0.569473598\approx0.57\)). But using the calculator value \(z = 0.57\), we use the standard normal distribution table or a calculator with a normal distribution function.
Using a standard normal table or a calculator (e.g., in Excel: \(=\text{NORM.S.DIST}(0.57, \text{TRUE})\)), we find the P - value.
\(P(Z < 0.57)=0.7157\) (using a standard normal table: the value corresponding to \(z = 0.57\) is \(0.7157\)).

Answer:

a. Left - tailed test
b. \(z = 0.57\)
c. \(P-\text{value}=0.716\) (rounded to three decimal places)