Question

According to a recent report, 46% of college student internships are unpaid. A recent survey of 100 college interns at a local university found that 43 had unpaid internships. Use the five-step p-value approach to hypothesis testing and a 0.05 level of significance to determine whether the proportion of college interns that had unpaid internships is different from 0.46. Let pi be the population proportion. Determine the null hypothesis, H_0, and the alternative hypothesis, H_1. H_0: pi 0.46 H_1: pi 0.46 What is the test statistic?

Answer 1

`z=\frac{0.43 -0.46}{\sqrt{(0.46(1-0.46))/(100)}}=-0.602`

`p_v =2*P(z<-0.602)=0.547`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 5% of significance the proportion of people with unpaid internships is not significantly different from 0.46

Explanation:

Data given and notation

n=100 represent the random sample taken

X=43 represent the people with unpaid internships

`\hat p=(43)/(100)=0.785` estimated proportion of people with unpaid internships

`p_o=0.46` is the value that we want to test

`\alpha=0.05` represent the significance level

Confidence=95% or 0.95

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 ture proportion is 0.46.:

Null hypothesis:
`p=0.46`

Alternative hypothesis:
`p \\eq 0.46`

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{0.43 -0.46}{\sqrt{(0.46(1-0.46))/(100)}}=-0.602`

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

Since is a bilateral test the p value would be:

`p_v =2*P(z<-0.602)=0.547`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 5% of significance the proportion of people with unpaid internships is not significantly different from 0.46