Respuesta :
The sample size needed to obtain a margin of error of 0.05 for the estimation of a population proportion is 384.
Given margin of error of 0.05 and confidence interval of 95%.
We have to find the sample size needed to obtain a margin of error of 0.5 with confidence level of 95%.
Margin of error is the difference between real values and calculated values.
The formula of margin of error is as under:
Margin of error=z*σ/[tex]\sqrt{n}[/tex]
where
z is the critical value of z for given confidence level
n is sample size
σ is population standard deviation
We have not given population standard deviation so we will use the following formula:
Margin of error=z*[tex]\sqrt{p(1-p)}[/tex]/[tex]\sqrt{n}[/tex]
We have to find z value for 95% confidence level.
z value=1.96
We know that [tex]\sqrt{p(1-p)}[/tex]<=1/2
put [tex]\sqrt{p(1-p)}[/tex]=1/2
Margin of error=1.96*1/2/[tex]\sqrt{n}[/tex]
0.05=1.96*0.5/[tex]\sqrt{n}[/tex]
[tex]\sqrt{n}[/tex]=0.98/0.05
[tex]\sqrt{n}[/tex]=19.6
squaring both sides
n=384.16
After rounding off we will get
n=384.
Hence the sample size needed to obtain a margin of error of 0.05 is 384.
Learn more about margin of error at https://brainly.com/question/10218601
#SPJ4
