227k views
1 vote
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?

User GlinesMome
by
6.8k points

1 Answer

4 votes

Answer:


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

User Bids
by
7.0k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.