Question

A recent article reported that a job awaits only one in three new college graduates. The major reasons given were an overabundance of college graduates and a weak economy. A survey of 200 recent graduates from your school revealed that 80 students had jobs. At the 0.01 significance level, can we conclude that a larger proportion of students at your school have jobs?

Answer 1

`z=\frac{0.4 -0.33}{\sqrt{(0.33(1-0.33))/(200)}}=2.105`

`p_v =P(z>2.105)=0.018`

If we compare the p value obtained and the significance level given
`\alpha=0.01` we have
`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 of students that had job at the school mentioned is not significantly higher then 0.33 .

Explanation:

1) Data given and notation

n=200 represent the random sample taken

X=80 represent the number of students that had jobs

`\hat p=(80)/(200)=0.4` estimated proportion of students that had jobs

`p_o=0.3333` 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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion of students that have jobs at the school mentioned is higher than the reported value at the article:

Null hypothesis:
`p\leq 0.33`

Alternative hypothesis:
`p > 0.33`

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`.

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.4 -0.33}{\sqrt{(0.33(1-0.33))/(200)}}=2.105`

4) 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 right tailed test the p value would be:

`p_v =P(z>2.105)=0.018`

If we compare the p value obtained and the significance level given
`\alpha=0.01` we have
`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 of students that had job at the school mentioned is not significantly higher then 0.33 .