Question

Women athletes at the a certain university have a long-term graduation rate of 67%. Over the past several years, a random sample of 38 women athletes at the school showed that 22 eventually graduated. Does this indicate that the population proportion of women athletes who graduate from the university is now less than 67%? Use a 10% level of significance.

Answer 1

Answer:

$z=\frac{0.579 -0.67}{\sqrt{\frac{0.67(1-0.67)}{38}}}=-1.193$  

$p_v =P(z$  

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 to reject the null hypothesis, and we can conclude that  the proportion of women athletes graduated is not significantly lower than 0.67 or 67% at 10% of significance

Step-by-step explanation:

Data given and notation

n=38 represent the random sample taken

X=22 represent the number of women athletes graduated

$\hat p=\frac{22}{38}=0.579$ estimated proportion of women athletes graduated

$p_o=0.67$ 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 lower than 0.67 or no:  

Null hypothesis:$p \geq 0.67$  

Alternative hypothesis:$p < 0.67$  

When we conduct a proportion test we need to use the z statisitc, 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:  

$z=\frac{0.579 -0.67}{\sqrt{\frac{0.67(1-0.67)}{38}}}=-1.193$  

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:  

$p_v =P(z$  

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 to reject the null hypothesis, and we can conclude that  the proportion of women athletes graduated is not significantly lower than 0.67 or 67% at 10% of significance