SOLUTION: A random sample of 800 home owners in a particular city found 232 home owners who had a swimming pool in their backyard. Find a 95% confidence interval for the true percent of home

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Question 1197599: A random sample of 800 home owners in a particular city found 232 home owners who had a swimming pool in their backyard. Find a 95% confidence interval for the true percent of home owners in this city who have a swimming pool in their backyard. Express your results to the nearest hundredth of a percent.
Answer: to %
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi 
  p(had pool) = 232/800 = .29
95% CI   ⇒   z = 1.96   |  2-sided  Invnorm( .95 + .05/2) = Invnorm(.975)
ME = z%2Asqrt%28%28p%281-p%29%29%2Fn%29
Plug and Play
ME = 1.96%2Asqrt%28%28.29%28.71%29%29%2F800%29 
.29 - ME < mu < .29 + ME
Will let You finish it Up
Wish You the Best in your Studies.