Question

It has been found that 26% of men 20 years and older suffer from hypertension (high blood pressure) and 31.5% of women are hypertensive. A random sample 150 of each gender was selected from recent hospital records, and the following results were obtained. Construct 95% confidence interval for the difference of the two proportion. Round your answer to nearest ten-thousandth. Interpret the result.

Answer 1

Complete Question

It has been found that 26% of men 20 years and older suffer from hypertension (high blood pressure) and 31.5% of women are hypertensive. A random sample 150 of each gender was selected from recent hospital records, and the following results were obtained.

Men. 43 patients had high blood pressure

Woman. 52 patients had high blood pressure.

Answer:

The  95% confidence interval is  

      $- 0.1651 < p_m - p_f$

This mean that there is a 95 % confidence that the difference between the true proportions of male and  female that are hypertensive  is within this interval and given that the interval contains zero then there is no statistically significant difference between the genders that are hypertensive        

Step-by-step explanation:

From the question we are told that

    The  sample size for male is  $n_1 = 150$

    The  number of male that are hypertensive is  $m = 42$

    The  sample size of female is  $n_2 = 150$

     The  number of female that are hypertensive is $q = 52$

The proportion of male that are hypertensive is mathematically represented as

         $\r p_m = \frac{43}{150}$

         $\r p_m = 0.287$          

The proportion of female that are hypertensive is mathematically represented as

       $p_f = \frac{52}{150}$

      $p_f = 0.347$

From the question we are told that confidence level is 95%, hence the level of significance is mathematically represented as

       $\alpha = 100 -95$

      $\alpha =5\%$

     $\alpha =0.05$

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

           $Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

          $E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{ \r p_m (1- \r p_m )}{n_1} + \frac{ \r p_f (1- \r p_f )}{n_2} }$

substituting value

         $E = 1.96 * \sqrt{\frac{ 0.287 (1- 0.287 )}{150} + \frac{ 0.347 (1- 0.347 )}{150} }$

         $E = 0.1051$

The  95% confidence interval is mathematically represented as  

           $(\r p_m - \r p_f ) - E < p_m - p_f < (\r p_m - \r p_f ) + E$

substituting values

          $( 0.287 - 0.347 ) - 0.1051 < p_m - p_f$    

           $- 0.1651 < p_m - p_f$

This mean that there is a 95 % confidence that the difference between the true proportion is within this interval and given that the interval contains zero then there is no statistically significant difference between the genders that are hypertensive.