Question

Private and public universities are located in the same city. For the private university, 1042 alumni were surveyed and 655 said that they attended at least one class reunion. For the public university, 804 out of 1317 sampled alumni claimed they have attended at least one class reunion. Is the difference in the sample proportions statistically significant? (Use α=0.05)

Answer 1

Answer:

If we compare the p value with the significance level provided $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the true proportion's for public and private universities are not significantly different at 5% of significance.  

Step-by-step explanation:

1) Data given and notation  

$X_{1}=655$ represent the number of successes for private university

$X_{2}=804$ represent the number of successes for public university

$n_{1}=1042$ sample of 1 selected

$n_{2}=1317$ sample of 2 selected

$\hat p_{1}=\frac{655}{1042}=0.629$ represent the sample proportion for private university

$\hat p_{2}=\frac{804}{1317}=0.610$ represent the sample proportion 2  

z would represent the statistic (variable of interest)  

$p_v$ represent the value for the test (variable of interest)

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the we have significant differences betwen the two proportions, the system of hypothesis would be:  

Null hypothesis:$p_{1} = p_{2}$  

Alternative hypothesis:$p_{1} \neq p_{2}$  

We need to apply a z test to compare proportions, and the statistic is given by:  

$z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$   (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{655+804}{1042+1317}=0.618$

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

$z=\frac{0.629-0.610}{\sqrt{0.618(1-0.618)(\frac{1}{1042}+\frac{1}{1317})}}=0.943$  

4) Statistical decision

Since is a bilateral test the p value would be:  

$p_v =2*P(Z>0.943)=0.346$  

If we compare the p value with the significance level provided $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the true proportion's for public and private universities are not significantly different at 5% of significance.