Answer:
Step-by-step explanation:
Hello!
Out of 846 patients taking a prescription drug daily, 18 complained of flulike symptoms.
It is known that the population proportion of patients that take the drug of the competition and complain of flulike is 1.8%
Be the variable of interest:
X: number of patients that complained of flulike symptoms after taking the prescription drug, out of 846.
sample proportion p'= 18/846= 0.02
You have to test if the population proportion of patients that experienced flulike symptoms as a side effect is greater than 1.8% (p>0.018)
Assuming that the patients for the clinical trial were randomly selected.
The expected value for this sample is np=846*0.02= 1658 (the expected value of successes is greater than 10) and the sample is less than 10% of the population, you can apply the test for the proportion:
The hypotheses are:
H₀: p ≤ 0.018
H₁: p > 0.018
α: 0.01
[tex]Z= \frac{p'-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]≈N(0;1)
[tex]Z_{H_0}= \frac{0.02-0.018}{\sqrt{\frac{0.018*0.982}{846} } }= 0.437[/tex]
The p-value for this test is 0.331056
The decision rule is
If p-value ≤ α, reject the null hypothesis
If p-value > α, do not reject the null hypothesis
The p-value is greater than α, the decision is to reject the null hypothesis.
So at 1% significance level there is no significant evidence to reject the null hypothesis, you can conclude that the population proportion of patients that took the prescription drug daily and experienced flulike symptoms as a side effect is less or equal to 1.8%
I hope this helps!
