Here you estimate the proportion of people in the population that said did not have children under 18 living at home.It can also be given as a percentage.

The general expression to apply here is;

$C.I=p+-z*\sqrt{\frac{p(1-p)}{n} }$

where ;

p=sample proportion

n=sample size

z*=value of z* from the standard normal distribution for 95% confidence level

Given;

n=5000

Find p

From the question 54% of people chosen said they did not have children under 18 living at home

$\frac{54}{100} *5000 = 2700\\\\p=\frac{2700}{5000} =0.54$

To calculate the 95% confidence interval, follow the steps below;

Find the value of z* from the z*-value table

The value of z* from the table is 1.96

calculate the sample proportion p

The value of p=0.54 as calculated above

Find p(1-p)

$0.54(1-0.54)=0.2484$

Find p(1-p)/n

Divide the value of p(1-p) with the sample size, n

$\frac{0.2448}{5000} =0.00004968$

Find the square-root of p(1-p)/n

$=\sqrt{0.00004968} =0.007048$

Find the margin of error

Here multiply the square-root of p(1-p)/n by the z*

$=0.007048*1.96=0.0138$

The 95% confidence interval for the lower end value is p-margin of error

$=0.54-0.0138=0.5262$

The 95% confidence interval for the upper end value is p+margin of error

$0.54+0.0138=0.5538$

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