Answer:
The decision is to not reject the null hypothesis.
At a significance level of 0.01, there is not enough evidence to support the claim that the proportion of all those using the drug that experience relief is significantly higher than 50% (P-value = 0.1443).
Step-by-step explanation:
This is a hypothesis test for a proportion.
The claim is that the proportion of all those using the drug that experience relief is significantly higher than 50%.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi=0.5\\\\H_a:\pi>0.5[/tex]
The significance level is 0.01.
The sample has a size n=200.
The sample proportion is p=0.54.
[tex]p=X/n=108/200=0.54[/tex]
The standard error of the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.5*0.5}{200}}\\\\\\ \sigma_p=\sqrt{0.00125}=0.035[/tex]
Then, we can calculate the z-statistic as:
[tex]z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.54-0.5-0.5/200}{0.035}=\dfrac{0.038}{0.035}=1.061[/tex]
This test is a right-tailed test, so the P-value for this test is calculated as:
[tex]\text{P-value}=P(z>1.061)=0.1443[/tex]
As the P-value (0.1443) is greater than the significance level (0.01), the effect is not significant.
The null hypothesis failed to be rejected.
At a significance level of 0.01, there is not enough evidence to support the claim that the proportion of all those using the drug that experience relief is significantly higher than 50%.