SOLUTION: In a survey of 225 randomly selected gun owners, it was found that 69 of them said they owned a gun primarily for protection.
Find the margin of error and 95% confidence interva
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Find the margin of error and 95% confidence interva
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Question 1199178: In a survey of 225 randomly selected gun owners, it was found that 69 of them said they owned a gun primarily for protection.
Find the margin of error and 95% confidence interval for the percentage of all gun owners who would say that they own a gun primarily for protection. Report answers to at least 2 decimal places. Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! sample size = 225
p = 69/225
q = (1 - 69) / 225 = 225/225 - 69/225 = (225 - 69) / 225 = 156/225
mean = 69/225
standard error = sqrt(p*q/n) = sqrt((69/225*156/225)/225) = .0307406515.
z = (x-m)/s
z is the z-score
(x-m) is the margin of error.
s is the standard error.
your confidence interval is 95%.
at 95% confidence interval, the critical z-score is plus or minus 1.96
to find the margin of error, you only need to use the hi side z-score because the confidence interval is symmetric about the mean and the margin of error above the mean is the same as the margin of error below the mean.
formula becomes 1.96 = (x-m) / standard error which becomes 1.96 = (x-m) / .0307406515.
solve for (x-m) to get (x-m) = 1.96 * .0307406515 = .0602516769.
that should be the margin of error on either side of the mean at 95% confidence interval.
the mean is equal to 69/225
that minus the margin of error yields a low side value of .2464149897.
that plus the margin of error yields a high side value of .3669183436.
