Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 500 adult residents sampled, 175 had kids. Based on this, construct a 99% confidence interval for the proportion p of adult residents who are parents in this county. Express your answer in tri-inequality form. Give your answers as decimals, to three places.

Answer 1

Answer:

The 99% confidence interval  is  $0.3003 < I < 0.3997$

Step-by-step explanation:

From the question we are told that

     The sample size is  $n = 500$

      The the number that are parents  x =  175

The  proportion of parents is  mathematically represented as

       $\r p = \frac{x}{n}$

substituting values

      $\r p = \frac{175}{500}$

     $\r p = 0.35$

The  level of confidence is given as 99%  which implies that the level of significance is  

       $\alpha = 100 - 99$

       $\alpha =$1%

      $\alpha = 0.01$

The critical value for this level of significance is obtained from the table of critical value as

          $t_{x, \alpha } = t_{175, 0.05} = 2.33$

Generally the margin of error is mathematically evaluated as

       $M =\frac{ t_{175, 0.01 } * \sqrt{\r p (1-\r p)} }{\sqrt{n} }$

substituting values

     $M =\frac{ 2.33 * \sqrt{\r 0.35 (1-0.35)} }{\sqrt{500} }$

     $M = 0.0497$

Generally the 99% confidence interval is mathematically represented as

        $I = \r p \pm M$

    $\r p -M < I < \r p + M$

substituting values

    $0.35 -0.0497 < I < 0.35 + 0.0497$

    $0.3003 < I < 0.3997$