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.

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:

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

The sample size is n=500.

The sample proportion is:

The standard deviation of the proportion is:

The z-statistic is then:

The P-value for this left tailed test is:

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%.

step-by-step explanation:

10.50 + taxes which added up to 11.50

the answer is 31.27mph