A recent article in USA Today 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.
(A) At the .01 significance level, can we conclude that a larger proportion of students at your school have jobs?

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Answer:

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

[tex]p_v =P(z>2.105)=0.0176[/tex]  

We see that the p value is [tex]p_v>\alpha[/tex] 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 with jobs is not significantly higher than 1/3 or 0.33

Step-by-step explanation:

Data given and notation

n=200 represent the random sample taken

X=80 represent the number of students with jobs

[tex]\hat p=\frac{80}{200}=0.4[/tex] estimated proportion of students with jobs

[tex]p_o=0.33[/tex] is the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Hypothesis test

We need to conduct a hypothesis in order to test the claim that the true proportion of students with job is higher than 0.33 or 1/3.:  

Null hypothesis:[tex]p \leq 0.33[/tex]  

Alternative hypothesis:[tex]p > 0.33[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

Calculate the statistic  

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

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

Statistical decision  

The significance level provided [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.  

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

[tex]p_v =P(z>2.105)=0.0176[/tex]  

We see that the p value is [tex]p_v>\alpha[/tex] 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 with jobs is not significantly higher than 1/3 or 0.33

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