Answer:
If we compare the p value and the significance level given we see that
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
estimated proportion of college students expressed an intent to vote
is the value that we want to test
represent the significance level
Confidence=99% or 0.99
z would represent the statistic (variable of interest)
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:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statistic, and the is given by:
(1)
The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value
.
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
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 . The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
If we compare the p value and the significance level given we see that
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
