Question

A sample of students that had stuffy or runny noses were blindly given a single Skittles candy to put in their mouth. They were told the five possible flavors and then were asked which flavor they had. Of the 118 students tested, 44 gave the correct answer. Find a 95% confidence interval for the proportion of all students with stuffy or runny noses that could correctly identify a Skittles flavor blindly.

Answer 1

Answer:

The confidence interval is  $0.2857< p < 0.4601$

Step-by-step explanation:

From the question we are told that

  The sample size is  n  =  118  

   The  number that gave the correct answer is  k =  44

Generally the sample proportion is mathematically represented as

      $\^ p = \frac{44}{118}$

=>     $\^ p =0.3729$

Generally given that the confidence level is  95% the level of significance is mathematically represented as

       $\alpha = (100 -95) \%$

=>     $\alpha = 0.05$

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

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as  

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

=>   $E = 1.96 * \sqrt{\frac{0.3729(1 - 0.3729)}{118} }$

=>    $E = 0.0872$

Generally 95% confidence interval is mathematically represented as  

             $\r p -E < p < \r p +E$

=>    $0.3729 -0.0872< p < 0.3729 +0.0872$

=>  $0.2857< p < 0.4601$