Answer to Question #345958 in Statistics and Probability for rhairhai

Question

In a random sample of 300 patients, 21 experienced nausea. A drug manufacturer claims that fewer than 10% of patients who take its new drug for treating Alzheimer’s disease will experience nausea. Test this claim at the 0.05 significance level. What is the appropriate null hypothesis for this situation

Answer 1

The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:p\ge0.10$

$H_a:p<0.10$

This corresponds to a left-tailed test, for which a z-test for one population proportion will be used.

Evidence:

Based on the information provided, the significance level is $\alpha = 0.05 ,$ and the critical value for a left-tailed test is $z_c = -1.6449.$

The rejection region for this left-tailed test is $R = \{z: z < -1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{21/300-0.1}{\sqrt{\dfrac{0.1(1-0.1)}{300}}}=-1.732$$

Since it is observed that $z = -1.732<-1.6449= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z<-1.732)= 0.041637,$ and since $p=0.041637<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is less than 0.10, at the $\alpha = 0.05$ significance level.