Question

A simple random sample of 100 8th graders at a large suburban middle school indicated that 81% of them are involved with some type of after school activity. Find the 98% confidence interval that estimates the proportion of them that are involved in an after school activity.

Answer 1

Answer:

The  interval is  $0.7187 < p < 2.421$

Step-by-step explanation:

From the question we are told that

      The  sample size is  $n = 100$

       The  population  proportion is $p = 0.81$

       The  confidence level is  C =  98%

The level of significance is mathematically evaluated as

     $\alpha = 100 -98$

    $\alpha = 2%$%

    $\alpha = 0.02$

Here this level of significance represented the left and the right tail

The degree of  freedom is evaluated as

     $df = n-1$

substituting value  

    $df = 100 - 1$

     $df = 99$

Since we require the critical value of one tail in order to evaluate the  98% confidence interval that estimates the proportion of them that are involved in an after school activity. we will divide the level of significance by 2

The  critical value of  $\frac{\alpha}{2}$ and the evaluated degree of freedom is  

      $t_{df , \alpha } = t_{99 , \frac{0.02}{2} } = 2.33$

this is obtained from the critical value table  

The standard error is mathematically evaluated as

             $SE = \sqrt{\frac{p(1-p )}{n} }$  

substituting value  

           $SE = \sqrt{\frac{0.81(1-0.81 )}{100} }$  

           $SE = 0.0392$  

The 98%  confidence interval is evaluated as

      $p - t_{df , \frac{\alpha }{2} } * SE < p < p + t_{df , \frac{\alpha }{2} }$

substituting value  

     $0.81 - 2.33 * 0.0392 < p < 0.81 +2.33 * 0.0392$

      $0.7187 < p < 2.421$