Question

It has been reported that 20.4% of incoming freshmen indicate that they will major in business or a related field. A random sample of 400 incoming college freshmen was asked their preference, and 95 replied that they were considering business as a major. Estimate the true proportion of freshman business majors with 98% confidence. Does your interval contain 20.4%?

Answer 1

Answer:

The  98% confidence interval

                         $0.1884 < p < 0.2876$

The confidence interval contains  20.4%

Step-by-step explanation:

From the question we are told that

    The sample size is  n =  400

The number that replied that they were considering business as a major $x = 95$

  The  sample proportion is mathematically  evaluated as

          $\r p = \frac{95}{400}$

         $\r p = 0.238$

Given that the confidence level 98% then the level of significance is evaluated as

      $\alpha = 100 - 98$

     $\alpha = 2 \%$

     $\alpha = 0.02$

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

       $Z_{\frac{ \alpha }{2} } = 2.33$

  Generally the margin of error is mathematically represented as

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

       $E = 2.33 * \sqrt{ \frac{ 0.238 (1 - 0.238 )}{400} }$

        $E = 0.0496$

The  98%  confidence interval is mathematically represented

   $\r p - E < p < \r p + E$

 =>    $0.238 - 0.0496 < p$

=>      $0.1884 < p < 0.2876$