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

`z=\frac{0.7 -0.65}{\sqrt{(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 .

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=(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{(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{(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 .