Question

The National Collegiate Athletic Association​ (NCAA) measures the Graduation Success Rate​ (GSR), which is the percentage of eligible athletes who graduate within six years of entering college. According to the​ NCAA, the GSR for all scholarship athletes in a particular division is 57​%. The GSR for all students in this division is 62​%. Suppose the NCAA report was based on a sample of 500 ​student-athletes, of which 285 graduated within six years. Is this sufficient information to conclude that the GSR for all scholarship athletes in this division differs from 62​%? Carry out the test using a Type I error rate of 0.05.

Answer 1

Answer:

Yes. There is enough evidence to support the claim that the GSR for all scholarship athletes in this division differs from 62%.

Step-by-step explanation:

We have to perform a hypothesis test on a proportion.

The claim is that the GSR for all scholarship athletes in this division differs from 62%. Then, the null and alternative hypothesis are:

$H_0: \pi=0.62\\\\H_a:\pi$

The significance level, named here as Type I error rate, is 0.05.

The sample size is n=500.

The sample proportion is:

$p=X/n=285/500=0.57$

The standard deviation of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.62(1-0.62)}{500}}=\sqrt{0.0004712}=0.022$

The z-statistic is then:

$z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.57-0.62+0.5/500}{0.022}=\dfrac{-0.049}{0.022} = -2.227$

The P-value for this left tailed test is:

$P-value=P(z$

The P-value is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough evidence to support the claim that the GSR for all scholarship athletes in this division differs from 62%.

Answer 2

Answer:

$z=\frac{0.57 -0.62}{\sqrt{\frac{0.62(1-0.62)}{500}}}=-2.303$  

$p_v =2*P(z$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of student athletes who graduate within 6 years is significantly different from 0.62 or 62%

Step-by-step explanation:

Data given and notation

n=500 represent the random sample taken

X=285 represent the student athletes who graduate within 6 years

$\hat p=\frac{285}{500}=0.57$ estimated proportion of student athletes who graduate within 6 years

$p_o=0.62$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that true proportion differs from 0.62.:  

Null hypothesis:$p=0.62$  

Alternative hypothesis:$p \neq 0.62$  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

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

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.57 -0.62}{\sqrt{\frac{0.62(1-0.62)}{500}}}=-2.303$  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided $\alpha=0.05$. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

$p_v =2*P(z$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of student athletes who graduate within 6 years is significantly different from 0.62 or 62%