Question

A human resources representative claims that the proportion of employees earning more than $50,000 is less than 40%. To test this claim, a random sample of 700 employees is taken and 305 employees are determined to earn more than $50,000.The following is the setup for this hypothesis test:{H0:p=0.40Ha:p<0.40Find the test statistic for this hypothesis test for a proportion. Round your answer to 2 decimal places.

Answer 1

Answer:

The statistic for this case would be:

$z=\frac{\hat p -p_o}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}$

And replacing we got:

$z= \frac{0.436-0.4}{\sqrt{\frac{0.436*(1-0.436)}{700}}}= 1.92$

Step-by-step explanation:

For this case we have the following info:

$n =700$ represent the sample size

$X= 305$ represent the number of employees that earn more than 50000

$\hat p=\frac{305}{700}= 0.436$

We want to test the following hypothesis:

Nul hyp. $p \leq 0.4$

Alternative hyp : $p>0.4$

The statistic for this case would be:

$z=\frac{\hat p -p_o}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}$

And replacing we got:

$z= \frac{0.436-0.4}{\sqrt{\frac{0.436*(1-0.436)}{700}}}= 1.92$

And the p value would be given by:

$p_v = P(z>1.922)= 0.0274$