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

Answer:

The  margin of error is  $E = 0.021$

Step-by-step 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

        $\r p = \frac{ 1214}{2161}$

=>     $\r p = 0.56$

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 $\frac{\alpha }{2}$ from the normal distribution table is  

      $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

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

=>      $E = 1.96 * \sqrt{\frac{ 0.56(1 - 0.56 )}{2161} }$

=>      $E = 0.021$