Answer in Statistics and Probability for Jin #204937

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?

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

“H_0: pleq0.6”

“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.

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