Answer:
- The lowest value of the confidence interval is 0.5262 or 52.62%
- The highest value of the confidence interval is 0.5538 or 55.38%
Step-by-step explanation:
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;
[tex]C.I=p+-z*\sqrt{\frac{p(1-p)}{n} }[/tex]
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
[tex]\frac{54}{100} *5000 = 2700\\\\p=\frac{2700}{5000} =0.54[/tex]
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
[tex]0.54(1-0.54)=0.2484[/tex]
Divide the value of p(1-p) with the sample size, n
[tex]\frac{0.2448}{5000} =0.00004968[/tex]
- Find the square-root of p(1-p)/n
[tex]=\sqrt{0.00004968} =0.007048[/tex]
Here multiply the square-root of p(1-p)/n by the z*
[tex]=0.007048*1.96=0.0138[/tex]
The 95% confidence interval for the lower end value is p-margin of error
[tex]=0.54-0.0138=0.5262[/tex]
The 95% confidence interval for the upper end value is p+margin of error
[tex]0.54+0.0138=0.5538[/tex]
