There is no prior information about the proportion of Americans who support the gun control in 2018. If we want to estimate 95% confidence interval for the true proportion of Americans who support the gun control in 2018 with a 0.2 margin of error, how many randomly selected Americans must be surveyed?
(Round up your answer to nearest whole number)
Sample size formula is
n = 0.25*(z/E)^2
where z = 1.96 [From z table for 95% confidence level]
E = margin of error = 0.2
this gives us
n = 0.25*(1.96/0.2)^2
= 0.25*96.04
= 24.01
=25 (rounded up to nearest integer)
Required sample size is 25
There is no prior information about the proportion of Americans who support the gun control in...
There is no prior information about the proportion of Americans who support the gun control in 2018. If we want to estimate 92% confidence interval for the true proportion of Americans who support the gun control in 2018 with a 0.24 margin of error, how many randomly selected Americans must be surveyed?
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