Answer: (0.565, 0.635)
Step-by-step explanation:
Given : Sample size : n= 750
The number of voters favored the school bond initiative =450
Then , the proportion of voters favored the school bond initiative :
[tex]p=\dfrac{450}{750}=0.6[/tex]
Significance level : [tex]1-0.95=0.05[/tex]
Critical value : [tex]z_{\alpha/2}=1.96[/tex]
The confidence interval for population proportion is given by :-
[tex]p\ \pm\ z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}\\\\=0.6\pm (1.96)\sqrt{\dfrac{(0.6)(0.4)}{750}\approx0.6\pm0.035\\\\=(0.6-0.035,0.6+0.035)\\\\=(0.565, 0.635)[/tex]
Hence, the 95% confidence interval for the true proportion of voters favoring the initiative is (0.565, 0.635).
