Question 1157866: A poll was done where 450 adults were randomly surveyed. 35 % said they were working hard at their jobs and hoped to get a promotion. Calculate and interpret a 95 % confidence interval for the proportion of adults who said they were working hard and hoped for a promotion.
Compute the margin of error for a 95 % interval
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The half-interval for CI=95% is z(1.96)*sqrt(p(1-p)/n)
that is added and subtracted to the point estimate of 35%
SE=sqrt(.35*.65/450)=0.0005056=0.0225
1.96*0.0225=0.0440
95% confidence interval for the true proportion is (0.306, 0.394) This means that we don't know the exact proportion of people who are hoping. That is a parameter and will never be known, but we can be highly confident that the interval will contain that value.
Note: this will be different on a calculator, because 35% of 450 is not an integer, so one has to use 157/450 or 158/450, giving a slightly different interval from 157.5/450
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