Question

A simple random sample of college students was performed recently in which the students were asked to self identify as to whether they are binge drinkers or not. In a 1995 study it was found that about 44% of students were self identified as such and the researcher believed that this number has increased. In this study 7851 of the 17592 students interviewed self identified as binge drinkers. A 90% confidence interval of the estimated binge drinking rate is

Answer 1

Answer:

The 90% confidence interval is  $0.445< p < 0.456$

Step-by-step explanation:

From the  question we are told that

  The sample  size is  $n = 17592$

  The number of  binge  drinkers is  $k = 7851$

 

Given that the  confidence level is  90%  then the level of significance is mathematically represented as

          $\alpha = (100 - 90)\%$

          $\alpha = 0.10$

The critical value  of  $\frac{\alpha }{2}$  from the normal distribution table  is  $Z_{\frac{\alpha }{2} } = 1.645$

The  sample  proportion is mathematically represented as

      $\r p = \frac{ 7851}{ 17592}$

      $\r p = 0.45$

Generally the margin of error is mathematically represented as  

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

       $E =1.645 * \sqrt{\frac{ 0.45 (1 - 0.45 )}{17592} }$

      $E =0.00617$

The  90%  confidence interval is mathematically represented as  

      $\r p -E < p < \r p +E$

      $0.45 -0.00617 < p < 0.45 + 0.00617$

      $0.445< p < 0.456$