Question

To plan the budget for next year a college must update its estimate of the proportion of next year's freshmen class that will need financial aid. Historically 35% of freshmen at this college have needed financial aid. In a random sample of 150 freshman applications received thus far, 67 of the applicants request financial aid. Is there evidence that the proportion of next year's freshmen class needing financial aid has increased

Answer 1

`z=\frac{0.447 -0.35}{\sqrt{(0.35(1-0.35))/(150)}}=2.49`

`p_v =P(z>2.49)=0.0064`

So the p value obtained was a very low value and using the significance level assumed
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants request financial aid is significantly higher than 0.35

Explanation:

Data given and notation

n=150 represent the random sample taken

X=67 represent the applicants who request financial aid

`\hat p=(67)/(150)=0.447` estimated proportion of applicants request financial aid

`p_o=0.35` is the value that we want to test

`\alpha` represent the significance level

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 of applicants request financial aid is higher than 0.35:

Null hypothesis:
`p \leq 0.35`

Alternative hypothesis:
`p > 0.35`

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{0.447 -0.35}{\sqrt{(0.35(1-0.35))/(150)}}=2.49`

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 assumed is
`\alpha=0.05`. The next step would be calculate the p value for this test.

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

`p_v =P(z>2.49)=0.0064`

So the p value obtained was a very low value and using the significance level assumed
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants request financial aid is significantly higher than 0.35