Respuesta :
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
