Question

Of 118 randomly selected adults, 34 were found to have high blood pressure. Construct a 95% confidence interval for the true percentage of all adults that have high blood pressure. Construct a confidence interval for the population proportion p.

Answer 1

Answer:

The  confidence interval is     $0.20644 < p <0.36984$

Step-by-step explanation:

From the question we are told that

     The  sample is n  = 118

     The  confidence level is  C   =  95 %  

     The  number of people with high blood pressure is  k  =  34

The proportion of those with high blood pressure is evaluated as

          $\r p = \frac{k}{n}$

substituting values

        $\r p = \frac{34}{118}$

       $\r p = 0.288136$

Given that the confidence level is 95%  then the level of significance is evaluated as

       $\alpha = 100 -95$

      $\alpha = 5$%

     $\alpha = 0.05$

Now the critical values of  $\frac{\alpha }{2}$ obtained from the normal distribution table is $Z_{\frac{\alpha }{2} } = 1.96$

The reason we are obtaining values for  is because  is the area under the normal distribution curve for both the left and right tail where the 95% interval did not cover while   is the area under the normal distribution curve for just one tail and we need the  value for one tail in order to calculate the confidence interval

 Now the margin of error is evaluated as

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

substituting values

      $MOE = 1.96 * \sqrt{\frac{ 0.288136 (1- 0.288136)}{118} }$

      $MOE = 0.0817$

Thus the 95% confidence interval for the true percentage of all adults that have high blood pressure is evaluated as

      $\r p - MOE < p < \r p + MOE$

substituting values

      $0.288136 - 0.0817 < p <0.288136 + 0.0817$

     $0.20644 < p <0.36984$