In a survey of 703 randomly selected workers 1593 got their

In a survey of 703 randomly selected workers, 15.93% got their jobs through newspaper ads. Consider a hypothesis test that uses a 0.05 significance level to test the claim that less than 20% of workers get their jobs through newspaper ads.

What is the P-Value?

Solution

The hypothesis test in this question is:

H0: p 0.2 vs. H1: p < 0.2

The test statistic is:
z = ( 0.1593 - 0.2 ) / ( ( 0.2 * (1 - 0.2 ) / 703 )
z = -2.697814

The p-value = P( Z < z )
= P( Z < -2.697814 )
= 0.003489816

Since the p-value is less than the significance level of 0.05 we reject the null hypothesis and conclude the alternate hypothesis p < 0.2 is true.