
Respuesta :
Answer:
The difference in the sample proportions is not statistically significant at 0.05 significance level.
Step-by-step explanation:
Significance level is missing, it is  α=0.05
Let p(public) be the proportion of alumni of the public university who attended at least one class reunion Â
p(private) be the proportion of alumni of the private university who attended at least one class reunion Â
Hypotheses are:
[tex]H_{0}[/tex]: p(public) = p(private)
[tex]H_{a}[/tex]: p(public) ≠p(private)
The formula for the test statistic is given as:
z=[tex]\frac{p1-p2}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}[/tex] where
- p1 is the sample proportion of  public university students who attended at least one class reunion  ([tex]\frac{808}{1311}=0.616[/tex])
- p2 is the sample proportion of private university students who attended at least one class reunion  ([tex]\frac{647}{1038}=0.623[/tex])
- p is the pool proportion of p1 and p2 ([tex]\frac{808+647}{1311+1038}=0.619[/tex])
- n1 is the sample size of the alumni from public university (1311)
- n2 is the sample size of the students from private university (1038)
Then z=[tex]\frac{0.616-0.623}{\sqrt{{0.619*0.381*(\frac{1}{1311} +\frac{1}{1038}) }}}[/tex] =-0.207
Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis. Â