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;
![C.I=p+-z*\sqrt{\frac{p(1-p)}{n} }](https://tex.z-dn.net/?f=C.I%3Dp%2B-z%2A%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%20%7D)
where ;
p=sample proportion
n=sample size
z*=value of z* from the standard normal distribution for 95% confidence level
Given;
n=5000
<u>Find p</u>
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](https://tex.z-dn.net/?f=%5Cfrac%7B54%7D%7B100%7D%20%2A5000%20%3D%202700%5C%5C%5C%5Cp%3D%5Cfrac%7B2700%7D%7B5000%7D%20%3D0.54)
<u>To calculate the 95% confidence interval, follow the steps below;</u>
- 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
![0.54(1-0.54)=0.2484](https://tex.z-dn.net/?f=0.54%281-0.54%29%3D0.2484)
Divide the value of p(1-p) with the sample size, n
![\frac{0.2448}{5000} =0.00004968](https://tex.z-dn.net/?f=%5Cfrac%7B0.2448%7D%7B5000%7D%20%3D0.00004968)
- Find the square-root of p(1-p)/n
![=\sqrt{0.00004968} =0.007048](https://tex.z-dn.net/?f=%3D%5Csqrt%7B0.00004968%7D%20%3D0.007048)
Here multiply the square-root of p(1-p)/n by the z*
![=0.007048*1.96=0.0138](https://tex.z-dn.net/?f=%3D0.007048%2A1.96%3D0.0138)
The 95% confidence interval for the lower end value is p-margin of error
![=0.54-0.0138=0.5262](https://tex.z-dn.net/?f=%3D0.54-0.0138%3D0.5262)
The 95% confidence interval for the upper end value is p+margin of error
![0.54+0.0138=0.5538](https://tex.z-dn.net/?f=0.54%2B0.0138%3D0.5538)
Yea I’m gonna go watch my internet and then go to go home to go to
The second question is the first choice