Question

A random sample found that forty-two percent of 100 Americans were satisfied with the gun control laws in 2017. Compute a 90% confidence interval for the true proportion of Americans who were satisifed with the gun control laws in 2017. Fill in the blanks appropriately. A 90% for the true proportion of Americans who were satisifed with the gun control laws in 2017 is (?, ? ) (Keep 3 decimal places)

Answer #1

x=42

n=100

Estimate for sample proportion

Level of significance is =1-0.90=0.1

Z critical value(using Z table)=1.645

Confidence interval formula is

=(0.339,0.501)

A random sample found that forty-three percent of 100 Americans
were satisfied with the gun control laws in 2017. Compute a 97%
confidence interval for the true proportion of Americans who were
satisifed with the gun control laws in 2017. Fill in the blanks
appropriately.
A 97% for the true proportion of Americans who were satisifed
with the gun control laws in 2017 is ( , )
(Keep 3 decimal places).
Can you please in include how the formula to solve?
n=...

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