Answer:
The 90% level of confidence interval For the Population proportion
(0.6569 ,0.7031)
Step-by-step explanation:
Explanation:-
Given sample size 'n' = 1100
Given sample proportion
[tex]p^{-} =\frac{x}{n} = \frac{750}{1100} = 0.68[/tex]
The 90% level of confidence interval For the Population proportion is determined by
[tex](p^{-} - Z_{\frac{\alpha }{2} } \sqrt{\frac{p(1-p)}{n} } ,p^{-} + Z_{\frac{\alpha }{2} } \sqrt{\frac{p(1-p)}{n} })[/tex]
90% of level of significance Z-value
[tex]Z_{\frac{0.10}{2} } = Z_{0.05} =1.645[/tex]
[tex](0.68 - 1.645 \sqrt{\frac{0.68(1-0.68)}{1100} } ,(0.68 + 1.645 \sqrt{\frac{0.68(1-0.68)}{1100}[/tex]
On calculation , we get
(0.68 -0.0231 ,0.68 +0.0231)
(0.6569 ,0.7031)
Final answer:-
The 90% level of confidence interval For the Population proportion
(0.6569 ,0.7031)
