Question

Suppose

202202

subjects are treated with a drug that is used to treat pain and

5050

of them developed nausea. Use a

0.050.05

significance level to test the claim that more than

2020​%

of users develop nausea.

Answer 1

Solution: For the given question we construct our null and alternative hypotheses as:

H0: p = 0.20 vs Ha: p > 0.20 [Since it is to be tested whether more than 20% of users develop nausea, so it is a right-tailed test]

[Here p= unknown population proportion of the subjects developing nausea after usage of the drug to reduce pain.]

The test statistic for this is Z = (p_hat - p0)/sqrt(p0*(1-p0)/n) ~ N(0,1) under H0.

where p0 = hypothetical value of the unknown population proportion, n= sample size, p_hat = sample proportion.

Here p0 = 0.2, p_hat = 50/202, n = 202

We reject H0 if Z(observed) > tau(alpha) where tau(alpha) is the upper alpha point of a standard normal distribution.

Here Z(observed) = 1.688634 and tau(alpha) = 1.644854.

So, we see that Z(observed) > tau(alpha), hence we reject H0 and conclude at a 5% level of significance on the basis of the given sample measures that there is enough evidence to support the claim that more than 20% of users develop nausea when treated with the drug to treat pain.