In 2009, the Southeastern Conference (SEC) commissioner set a goal to have greater than 65% of athletes that are entering freshmen graduate in 6 years. In 2015, a sample of 100 entering freshmen from 2009 was taken and it was found that 70 had graduated in 6 years. Does this data provide evidence that the commissioner’s graduation goal was met (α = .10)? The value of the test statistic is and the critical value is .

So the p value obtained was a very high value and using the significance level given we have so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 5% of significance the proportion of people who graduate in 6 years is not significantly higher than 0.65 .  

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

X=70 represent the number of people who graduate in 6 years

estimated proportion of people who graduate in 6 years

is the value that we want to test

represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

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 is higher than 0.65:  

Null hypothesis:  

Alternative hypothesis:  

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

(1)  

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

Calculate the statistic  

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

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 . The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

So the p value obtained was a very high value and using the significance level given we have so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 5% of significance the proportion of people who graduate in 6 years is not significantly higher than 0.65 .