Question

confidence interavls for a population proportion. suppose that a random sample of 1000 mortgage loans that were defaulted within the first year reveals 410 of these loans were approved on hte basis of falsified applications. what is point estiamte of and a 95% confidence interval for p, the proportion of all first year defaults that are approved on the basis of flsified application

Answer 1

Answer:

The 95% confidence interval is  $0.3795 < p < 0.4405$

Step-by-step explanation:

From the question we are told that

     The sample size is  $n = 1000$

      The  number of approved loan is  k =  410

       

Generally the sample proportion is mathematically represented as

       $\r p = \frac{k}{n}$

substituting values

      $\r p = \frac{410}{1000}$

       $\r p = 0.41$

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

       $\alpha = 100 - 95$

        $\alpha = 5\%$

       $\alpha = 0.05$

Next we obtain the critical value of  $\frac{\alpha }{2}$ from the normal distribution table,the value is  

         $Z_{\frac{\alpha }{2} } =Z_{\frac{0.05 }{2} }= 1.96$

Generally the margin of error is mathematically represented as

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

substituting values

        $E = 1.96 * \sqrt{\frac{ 0.41(1- 0.41)}{1000} }$

        $E = 0.03048$

The 95% confidence interval for p is mathematically represented  as

     $\r p - E < p < \r p + E$

substituting values

     $0.41 - 0.03048 < p < 0.41 + 0.03048$

    $0.3795 < p < 0.4405$