Question

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 _________.

Answer 1

Answer:

$z=\frac{0.7 -0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}=1.048$  

$p_v =P(z>1.048)=0.147$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.1$ we have $p_v>\alpha$ 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

$\hat p=\frac{70}{100}=0.7$ estimated proportion of people who graduate in 6 years

$p_o=0.65$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

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 is higher than 0.65:  

Null hypothesis:$p \leq 0.65$  

Alternative hypothesis:$p >0.65$  

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:  

$z=\frac{0.7 -0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}=1.048$  

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 $\alpha=0.1$. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

$p_v =P(z>1.048)=0.147$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.1$ we have $p_v>\alpha$ 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 .