Respuesta :
Answer:
The 95% confidence interval for the difference in population proportions who returned home after spring break is (0.31, 0.61).
Step-by-step explanation:
The (1 - α)% confidence interval for the difference between two population proportions is:
[tex]CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}}[/tex]
The information provided is:
Stayed Returned
1st year 19 68
4th year 36 17
Compute the sample proportion of 1st year students who returned home as follows:
[tex]\hat p_{1}=\frac{68}{68+19}=0.782[/tex]
Compute the sample proportion of 4th year students who returned home as follows:
[tex]\hat p_{2}=\frac{17}{17+36}=0.321[/tex]
The critical value of z for 95% confidence level is:
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
Compute the 95% confidence interval for the difference between two proportions as follows:
[tex]CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}}[/tex]
[tex]=(0.782-0.321)\pm 1.96\times \sqrt{\frac{0.782(1-0.782)}{87}+\frac{0.321(1-0.321)}{53}}}[/tex]
[tex]=0.461\pm (1.96\times 0.078)\\=0.461\pm 0.1529\\=(0.3081, 0.6139)\\\approx (0.31, 0.61)[/tex]
Thus, the 95% confidence interval for the difference in population proportions who returned home after spring break is (0.31, 0.61).
The 95% confidence interval for difference between two population proportions, (0.31, 0.61) implies that there is a 0.95 probability that the true difference between the proportions is included in the interval.
