Question

A past survey of students taking a standardized test revealed that ​% of the students were planning on studying engineering in college. In a recent survey of students taking the​ SAT, ​% of the students were planning to study engineering. Construct a ​% confidence interval for the difference between proportions by using the following inequality. Assume the samples are random and independent.
The confidence interval is __

Answer 1

Complete Question

The  complete question is shown on the first uploaded image  

Answer:

The  95% confidence interval is  $-0.00870$

Step-by-step explanation:

From the question we are told that

     The first sample  size  is  $n_1 = 1068000$

     The first proportion  $\r p_1 = 0.084$

     The second  sample size is  $n_2 = 1476000$

     The  second  proportion is  $\r p_2 = 0.092$

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

      $\alpha = (100 - 95)\%$

     $\alpha = 0.05$

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

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

Now using the formula from the question to construct the 95% confidence interval we have  

  $(\r p_1 - \r p_2 )- z_c \sqrt{ \frac{\r p_1 \r q_1 }{n_1} + \frac{\r p_2 \r q_2 }{n_2} }$

Here $\r q_1 = 1 - \r p_1$

  =>   $\r q_1 = 1 - 0.084$

 =>    $\r q = 0.916$

and  

   $\r q_2 = 1 - \r p_2$

 =>   $\r q_2 = 1 - 0.092$

=>   $\r q_2 = 0.908$

So  

 $(0.084 - 0.092 )- (1.96)* \sqrt{ \frac{0.092* 0.916 }{1068000} + \frac{0.084* 0.908 }{1476000} }$

  $-0.00870$