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There is no prior information about the proportion of Americans who support free trade in 2018. If we want to estimate a 97.5% confidence interval
There is no prior information about the proportion of Americans who support free trade in
2018. If we want to estimate a 97.5% confidence interval for the true proportion of Americans who
support free trade in 2018 with a 0.16 margin of error, how many randomly selected Americans must be
surveyed?
I don't understand how to get p and q from this question.
my critical value (Z) is: 1.959963985
my margin of error: .16
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