Question

Suppose 243 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea. Use a 0.05 significance level to test the claim that more than 20%​ of users develop nausea.

Answer 1

The following information is provided: The sample size is N=243, the number of favorable cases is X=54, and the sample proportion is , and the significance level is

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: p = 0.20

Ha: p>0.20

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

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is Zc​=1.64.

The rejection region for this right-tailed test is R={z:z>1.64}

(3) Test Statistics

The z-statistic is computed as follows:

(4) The decision about the null hypothesis

Since it is observed that , it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.1932, and since p=0.1932≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion p is greater than p0​, at the α=0.05 significance level.

Graphically

Let me know in the comments if anything is not clear. I will reply ASAP! Please do upvote if satisfied!