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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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