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

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) (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 = 
Plug and Play
ME =  
.29 - ME <  < .29 + ME
Will let You finish it Up
Wish You the Best in your Studies.

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