SOLUTION: If 2,000 card-carrying members of a political party were randomly sampled, and 1,600 said they wanted a change in leadership, what is the 95% confidence interval for the true popul

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Question 1155816: If 2,000 card-carrying members of a political party were randomly sampled, and 1,600 said they wanted a change in leadership, what is the 95% confidence interval for the true population percentage for all of the card-carrying members of the party who wanted a change in leadership?
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
mean is 0.80
SE is z0.975,* sqrt(p*(1-p)/n)
=1.96*sqrt (.8*.2/2000)
=0.0175
interval is (0.7825, 0.8175)