Question

May 10, 12:38:01 AMA survey was given to a random sample of 195 residents of a town todetermine whether they support a new plan to raise taxes in order toincrease education spending. Of those surveyed, 39 respondents saidthey were in favor of the plan. At the 95% confidence level, what is themargin of error for this survey expressed as a proportion to thenearest thousandth? (Do not write +).Submit AnswerAnswer:

Answer 1

It is given as,

x= 39.

n= 195.

Estimate for sample proportion= 0.75

Z critical value(using Z table)=1.96

Confidence interval formula is ,

`p\pm Z*\frac{\sqrt[]{p*(1-p)}}{\sqrt[]{n}}`
`0.75\pm1.96*\frac{\sqrt[]{0.75*(1-0.75)}}{\sqrt[]{195}}`
`0.75\pm1.96*(0.433)/(1.396)`
`1.358\text{ , 0.14206}`

Lower limit for confidence interval=0.14206

Upper limit for confidence interval= 1.38.

The margin error is determined as,

`1.38-0.14206=\text{ 1.237.}`