Question

An e-commerce research company claims that 60% or more graduate students have bought merchandise on-line. A consumer group is suspicious of the claim and thinks that the proportion is lower than 60%. A random sample of 80 graduate students show that only 22 students have ever done so.

Is there enough evidence to show that the true proportion is lower than 60%? Conduct the test at 10% Type I error rate, and use both p-value and rejection region approaches.

**Make sure to include your null and alternative hypotheses, whether it is a left-tailed, right-tailed, or two-tailed test, whether the conditions are met for a one-proportion z-test, the rejection region and p-value, and your conclusion.

Answer 1

here as np=80*0.6 =48 and n(1-p) =80*.4 =32 both are greater than or equal to 10 ; therefore  conditions are met for a one-proportion z-test

as p value is less than 0.10 level of significance , we reject null hypothesis and conclude that  true proportion is lower than 60%