Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 200 adult residents sampled, 86 had kids. Based on this, construct a 99% confidence interval for the proportion π π of adult residents who are parents in this county. Give your answers as decimals, to three places. < π π

Answer 1

We wish to estimate what percent of adult residents in a certain county are parents.

Out of 200 adult residents sampled, 86 had kids.

Thus n = 200      ( sample size )

          x = 86         ( number of success i.e dult residents having kids )

So   = x/n = 86/200

       = 0.43         ( probability of success )

Now 100(1-)% confidence interval CI for population proportion is given by

CI = { - * , + * }

Calculation :-

For 99% confidence critical z-value is = 2.576             ......(i)

     = 0.43                                                                           ......(ii)        

Thus   = =

      = 0.03500714                                    ......(iii)

Hence

99% confidence interval CI is given by

CI = { - * , + * }

= { 0.43   - 2.576 * 0.03500714   , 0.43   + 2.576 * 0.03500714 }

CI = { 0.3398216 , 0.5201784 }

CI = { 0.340 , 0.520 }                    [ rounding to three decimals ]

Thus 99% confidence interval for the proportion π π of adult residents who are parents in this country is { 0.340 , 0.520 }