Question

The 2010 General Social Survey reported a sample where about 48% of US residents thought marijuana should be made legal. If we wanted to limit the margin of error of a 95% confidence interval to 4%, about how many Americans would we need to survey

Answer 1

The sample size is
`n = 600`

Explanation:

From the question we are told that

The sample proportion is
`\r p = 0.48`

The margin of error is
`MOE = 0.04`

Given that the confidence level is 95% the level of significance is mathematically represented as

`\alpha = 100 - 95`

`\alpha = 5 \%`

`\alpha = 0.05`

Next we obtain the critical value of
`(\alpha )/(2)` from the normal distribution table , the values is

`Z_{(\alpha )/(2) } = 1.96`

The reason we are obtaining critical value of
`(\alpha )/(2)` instead of
`\alpha` is because

`\alpha` represents the area under the normal curve where the confidence level interval (
`1-\alpha`) did not cover which include both the left and right tail while

`(\alpha )/(2)` is just the area of one tail which what we required to calculate the margin of error

Generally the margin of error is mathematically represented as

`MOE = Z_{(\alpha )/(2) } * \sqrt{ (\r p(1- \r p ))/(n) }`

substituting values

`0.04= 1.96* \sqrt{ (0.48(1- 0.48 ))/(n) }`

`0.02041 = \sqrt{ (0.48(52 ))/(n) }`

`0.02041 = \sqrt{ ( 0.2496)/(n) }`

`0.02041^2 = ( 0.2496)/(n)`

`0.0004166 = ( 0.2496)/(n)`

=>
`n = 600`