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

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

`p_v =P(z<-2.24)=0.0125`

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

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=(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{(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{(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<-2.24)=0.0125`

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