Find the test statistic and rejection region

Question

An experimenter has prepared a drug-dose level that he claims will induce sleep for at least 70% of people suffering from insomnia. After examining the dosage we feel that his claims regarding the effectiveness of his dosage are too high. In an attempt to disprove his claim, we administer his prescribed dosage to 70 insomniacs and observe that 44 of them have had sleep induced by the drug dose. Is there enough evidence to refute his claim at the 5% level of significance?

Answer 1

Null and alternate hypotheses

H₀: p<0.70

H_a: p≥0.70

Calculate test statistic

Z = (p̂-p)/√[p̂(1-p̂)/n]

Z = (44/70-0.70)/√[44/70(1-44/70)/70]

Z ≈ -1.237

Our critical value for 95% confidence is Z=1.96. Because -1.237<1.96 (test statistic < critical value), we fail to reject the null hypothesis. This means that we are 95% confident that the true population proportion is actually less than p=0.70

Hope this helped!