Respuesta :
Answer:
The 99% of confidence limits for the proportion that plan to vote for the incumbent.
(0.6473 ,0.7527)
Step-by-step explanation:
Explanation:-
Given data the election of a local construction union involves 2,000 union members. Among them, 500 members are randomly selected.
Given large sample size 'N' = 2000
Given sample size 'n' = 500
Given data Of the 500 surveyed, 350 said they would vote for the incumbent.
The sample Proportion
[tex]p = \frac{x}{n} = \frac{350}{500} =0.7[/tex]
q = 1-p = 1 - 0.7 = 0.3
Confidence intervals:-
The 99% of confidence intervals are determined by
[tex](p-Z_{\alpha } \sqrt{\frac{pq}{n} } , p+Z_{\alpha }\sqrt{\frac{pq}{n} } )[/tex]
The z- score of 0.99 level of significance =2.576
[tex](0.7-2.576\sqrt{\frac{0.7X0.3}{500} } , 0.7+2.576\sqrt{\frac{0.7X0.3}{500} } )[/tex]
on using calculator, we get
(0.7 - 0.0527 ,0.7+0.0527)
(0.6473 ,0.7527)
Conclusion:-
The 99% of confidence limits for the proportion that plan to vote for the incumbent.
(0.6473 ,0.7527)
