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.

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dfbustos dfbustos · Answer 1 · 2020-07-14T02:57:11+02:00

Answer:

The statistic for this case would be:

[tex] z=\frac{\hat p -p_o}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}[/tex]

And replacing we got:

[tex] z= \frac{0.436-0.4}{\sqrt{\frac{0.436*(1-0.436)}{700}}}= 1.92[/tex]

Step-by-step explanation:

For this case we have the following info:

[tex] n =700[/tex] represent the sample size

[tex] X= 305[/tex] represent the number of employees that earn more than 50000

[tex]\hat p=\frac{305}{700}= 0.436[/tex]

We want to test the following hypothesis:

Nul hyp. [tex] p \leq 0.4[/tex]

Alternative hyp : [tex] p>0.4[/tex]

The statistic for this case would be:

[tex] z=\frac{\hat p -p_o}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}[/tex]

And replacing we got:

[tex] z= \frac{0.436-0.4}{\sqrt{\frac{0.436*(1-0.436)}{700}}}= 1.92[/tex]

And the p value would be given by:

[tex] p_v = P(z>1.922)= 0.0274[/tex]