Question

An acne specialist has found that 60% of acne sufferers are helped by existing treatments. He decides to test a new treatment on a random sample of 40 acne patients. If 28 of these respond to this treatment, can he conclude that it is significantly better at the 0.05 level?

Answer 1

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

$H_0: p\leq0.6$

$H_1:p>0.6$

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

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

The rejection region for this right-tailed test is $R=\{z: z>1.6449\}$

The z-statistic is computed as follows:

$$\hat{p}=\dfrac{x}{n}=\dfrac{28}{40}=0.7$$

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.7-0.6}{\sqrt{\dfrac{0.6(1-0.6)}{40}}}\approx1.291$$

Since it is observed that $z=1.291<1.6449=z_c,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is greater than 0.6, at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value is $p=P(z>1.291)=0.09835,$ and since $p=0.09835>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is greater than 0.6, at the $\alpha=0.05$ significance level.

Therefore he cannot conclude that it is significantly better at the 0.05 level.