Answer:
The null and alternative hypothesis are:
![H_0: \pi=0.6\\\\H_a:\pi<0.6](https://img.qammunity.org/2021/formulas/mathematics/college/w4uhd6deq97b7pnqwutbl41gsmxeqwd9bw.png)
The value of the test statistic is z=-2.617.
The critical value for a one-tailed test with significance level α=0.05 is zc=--1.645.
The P-value is P=0.0044.
To calculate this P-value in excel, we use the function NORM.S.DIST(-2.617;TRUE).
Explanation:
This is a hypothesis test for a proportion.
The claim is that the proportion of undergraduate students enrolled at a State University in NY earned their degrees within four years of matriculation has fallen below 60%.
Then, the null and alternative hypothesis are:
![H_0: \pi=0.6\\\\H_a:\pi<0.6](https://img.qammunity.org/2021/formulas/mathematics/college/w4uhd6deq97b7pnqwutbl41gsmxeqwd9bw.png)
The significance level is 0.05.
The sample has a size n=95.
The sample proportion is p=0.4632.
![p=X/n=44/95=0.4632](https://img.qammunity.org/2021/formulas/mathematics/college/1pg1gvr07nf7dqfyaxqjqbfguey9l03f54.png)
The standard error of the proportion is:
![\sigma_p=\sqrt{(\pi(1-\pi))/(n)}=\sqrt{(0.6*0.4)/(95)}\\\\\\ \sigma_p=√(0.00253)=0.0503](https://img.qammunity.org/2021/formulas/mathematics/college/eyihx529xgz1yhx14s4tc6c44j85w05c3o.png)
Then, we can calculate the z-statistic as:
![z=(p-\pi-0.5/n)/(\sigma_p)=(0.4632-0.6+0.5/95)/(0.0503)=(-0.1315)/(0.0503)=-2.617](https://img.qammunity.org/2021/formulas/mathematics/college/lyv29n3fb7j61fnoyh8243sjm4svryoctj.png)
This test is a left-tailed test, so the P-value for this test is calculated as:
![P-value=P(z<-2.617)=0.0044](https://img.qammunity.org/2021/formulas/mathematics/college/2e4nuvjqbodhyyfkniag4apjvnjcnrpzy3.png)
As the P-value (0.0044) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the proportion of undergraduate students enrolled at a State University in NY earned their degrees within four years of matriculation has fallen below 60%.
The critical value for a one-tailed test with significance level α=0.05 is zc=--1.645.
To calculate this P-value in excel, we use the function NORM.S.DIST(-2.617;TRUE).