Question

The University of Arkansas recently approved out of state tuition discounts for high school students from any state. The students must qualify by meeting certain standards in terms of GPA and standardized test scores. The goal of this new policy is to increase the geographic diversity of students from states beyond Arkansas and its border states. Historically, 90% of all new students came from Arkansas or a bordering state. Ginger, a student at the U of A, sampled 180 new students the following year and found that 157 of the new students came from Arkansas or a bordering state. Does Ginger’s study provide enough evidence to indicate that this new policy is effective with a level of significance 10%? What would be the correct decision?
a. Reject H0; conclude that the new policy does not increase the percentage of students from states that don’t border Arkansasb. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
c. Reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
d. Fail to reject H0; conclude that the new policy does not increase the percentage of students from states that don’t border Arkansas

Answer 1

Answer:

$z=\frac{0.872 -0.9}{\sqrt{\frac{0.9(1-0.9)}{180}}}=-1.252$  

$p_v =P(z<-1.252)=0.105$  

So the p value obtained was a very low 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 to reject the null hypothesis, and we can said that at 10% the proportion of students who came from Arkansas or a bordering state is not significantly lower than 0.9

b. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas

Step-by-step explanation:

Data given and notation n  

n=180 represent the random sample taken

X=157 represent the students who came from Arkansas or a bordering state

$\hat p=\frac{157}{180}=0.872$ estimated proportion of students who came from Arkansas or a bordering state

$p_o=0.9$ 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 the true proportion is higher or not than 0.9.:  

Null hypothesis:$p\geq 0.9$  

Alternative hypothesis:$p < 0.9$  

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:  

$z=\frac{0.872 -0.9}{\sqrt{\frac{0.9(1-0.9)}{180}}}=-1.252$  

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 left tailed test the p value would be:  

$p_v =P(z<-1.252)=0.105$  

So the p value obtained was a very low 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 to reject the null hypothesis, and we can said that at 10% the proportion of students who came from Arkansas or a bordering state is not significantly lower than 0.9

b. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas