Question

An investigator computes a 95% confidence interval for a population mean on the basis of a sample of size 65. If she wishes to compute a 95% confidence interval that is half as wide, how large a sample does she need

Answer 1

The value is
`n_2 = 260`

Explanation:

From the question we are told the confidence level is 95% , hence the level of significance is

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

=>
`\alpha = 0.05`

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

`Z_{(\alpha )/(2) } =  1.96`

Generally the width of a confidence interval is dependent on the margin of error which is mathematically represented as

`E = Z_{(\alpha )/(2) } * ( \sigma)/(√(n) )`

Here
`\sigma` is the standard deviation

Let assume that
`\sigma` and
`Z_{(\alpha )/(2) }` are constant for the width of the confidence interval when the sample size is n =65 and now that it has been divided to two

=>
`E = ( K )/(√(n) )`

Here K is a constant

=>
`E * √(n) = K`

=>
`E_1 * √(n)_1 = E_2 * √(n)_2`

Now let
`E_1 \ and \ n_1` be the margin of error and sample size before the reduction

So
`n_1 = 65`

and let
`E_2 \ and \ n_2` be the margin of error and sample size after the reduction

So
`E_2 = (1)/(2) E_1`

=>
`E_1 * √(65) = (E_1)/(2) * √(n)_2`

=>
`√(65) = (1)/(2) * √(n)_2`

=>
`n_2 = 260`