Question

We need to find the homeownership proportion in Southern California. The homeownership proportion is believed to be around 50%. How many householders would you survey for a 93% CI where the maximum likely sampling error is 1.5%?

Answer 1

The value is
`n = 3648`

Explanation:

From the question we are told that

The sample proportion is
`\^ p = 0.5`

The margin of error is
`E = 0.015`

From the question we are told the confidence level is 93% , hence the level of significance is

`\alpha = (100 - 93 ) \%`

=>
`\alpha = 0.07`

Generally from the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.812`

Generally the sample size is mathematically represented as

`n = [\frac{Z_{(\alpha )/(2) }}{E} ]^2 * \^ p (1 - \^ p )`

=>
`n = [( 1.812 )/( 0.015) ]^2 *0.5 (1 - 0.5 )`

=>
`n = 3648`