Question

A random sample of 250 observations selected from a population produced a sample proportion equal to 0.92.Round your answers to 3 decimal places.a. Make a 90% confidence interval for p.? to ?b. Construct a 95% confidence interval for p.? to ?c. Make a 99% confidence interval for p.? to ?

Answer 1

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

`\text{margin of error = z}*\sqrt[]{(pq)/(n)}`

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,

`\begin{gathered} \text{margin of error = 1.645}*\sqrt[]{(0.92*0.08)/(250)} \\ \text{margin of error = 0.02}8 \end{gathered}`

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

`\begin{gathered} \text{margin of error = 1.96}*\sqrt[]{(0.92*0.08)/(250)} \\ \text{margin of error = 0.034} \end{gathered}`

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

`\begin{gathered} \text{margin of error = 2.576}*\sqrt[]{(0.92*0.08)/(250)} \\ \text{margin of error = 0.044} \end{gathered}`

The 99% confidence interval for p is

0.92 ± 0.044

it is from

0.876 to 0.964