There is no prior information about the proportion of Americans who support gun control in 2019....
There is no prior information about the proportion of Americans who support gun control in 2019. If we want to estimate 93% confidence interval for the true proportion of Americans who support gun control in 2019 with a 0.18 margin of error, how many randomly selected Americans must be surveyed?
Homework Answers
Given : Margin of error = E=0.18
Significance level=
Since , there is no prior information about the proportion.
Therefore , assume that p=q=0.5
Therefore , the required sample size is ,
Where , 
; From the standard normal probability table
Hence , there are 25 randomly selected Americans must be surveyed.
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There is no prior information about the proportion of Americans
who support gun control in 2019....
There is no prior information about the proportion of Americans who support gun control in 2019....
There is no prior information about the proportion of Americans who support gun control in 2019. If we want to estimate 97% confidence interval for the true proportion of Americans who support gun control in 2019 with a 0.27 margin of error, how many randomly selected Americans must be surveyed? Answer:
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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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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)
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