Question

A presidential candidate's aide estimates that, among all college students, the proportion who intend to vote in the upcoming election is at least . If out of a random sample of college students expressed an intent to vote, can we reject the aide's estimate at the level of significance?

Answer 1

Answer:

$z=\frac{0.529 -0.6}{\sqrt{\frac{0.6(1-0.6)}{240}}}=-2.24$  

$p_v =P(z$  

If we compare  the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion college students expressed an intent to vote is not higher than 0.6

Step-by-step explanation:

Assuming the following question: A presidential candidate's aide estimates that, among all college students, the proportion p who intend to vote in the upcoming election is at least 60% . If 127 out of a random sample of 240 college students expressed an intent to vote, can we reject the aide's estimate at the 0.1 level of significance?

Data given and notation

n=240 represent the random sample taken

X=127 represent the college students expressed an intent to vote

$\hat p=\frac{127}{240}=0.529$ estimated proportion of college students expressed an intent to vote

$p_o=0.6$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

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 at least 60% of students are intented to vote .:  

Null hypothesis:$p \geq 0.6$  

Alternative hypothesis:$p < 0.6$  

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:  

$z=\frac{0.529 -0.6}{\sqrt{\frac{0.6(1-0.6)}{240}}}=-2.24$  

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 $\alpha=0.01$. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

$p_v =P(z$  

If we compare  the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion college students expressed an intent to vote is not higher than 0.6