Question

Find the indicated margin of error. In a clinical test with 2161 subjects, 1214 showed improvement from the treatment. Find the margin of error for the 95% confidence interval used to estimate the population proportion.

Answer 1

The margin of error is
`E = 0.021`

Explanation:

From the question we are told that

The population size is
`n = 2161`

The number that showed improvement is
`k = 1214`

Generally the sample proportion is mathematically represented as

`\r p = ( 1214)/(2161)`

=>
`\r p = 0.56`

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

`\alpha =(100-95) \%`

=>
`\alpha =0.05`

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

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

Generally the margin of error is mathematically represented as

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

=>
`E = 1.96 * \sqrt{( 0.56(1 - 0.56 ))/(2161) }`

=>
`E = 0.021`