Respuesta :
Confidence intreval for population proportion is written as
sample proportion ± margin of error
The formula for calculating margin of error for a population proportion is expressed as
[tex]\text{margin of error = z}\times\sqrt[]{\frac{pq}{n}}[/tex]where
z is the z score of the confidence level
p = population proportion
q = 1 = p
From the information given,
p = 0.98
q = 1 - 0.92 = 0.08
n = 250
a) For a 90% confidence interval, z = 1.645
By substituting these values into the formula,
[tex]\begin{gathered} \text{margin of error = 1.645}\times\sqrt[]{\frac{0.92\times0.08}{250}} \\ \text{margin of error = 0.02}8 \end{gathered}[/tex]Thus,
The 90% confidence interval for p is
0.92 ± 0.028
it is from
0.892 to 0.948
b)For a 95% confidence interval, z = 1.96
By substituting these values into the formula
[tex]\begin{gathered} \text{margin of error = 1.96}\times\sqrt[]{\frac{0.92\times0.08}{250}} \\ \text{margin of error = 0.034} \end{gathered}[/tex]Thus,
The 95% confidence interval for p is
0.92 ± 0.034
It is from
0.886 to 0.954
c) For a 99% confidence interval, z = 2.576
By substituting these values into the formula
[tex]\begin{gathered} \text{margin of error = 2.576}\times\sqrt[]{\frac{0.92\times0.08}{250}} \\ \text{margin of error = 0.044} \end{gathered}[/tex]The 99% confidence interval for p is
0.92 ± 0.044
it is from
0.876 to 0.964
