Question

A past survey of 1, 068,000 students taking a standardized test revealed that 8.9% of the students were planning on studying engineering in college.
In a recent survey of 1, 476,000 students taking the SAT. 9.2% of the students were planning to study engineering.
Construct a 95% confidence interval for the difference between proportions ^p1−^p2 by using the following inequality. Assume the samples are random and independent.
(^p1−^p2)−zc√^p1^q1n1+^p2^q2n2 The confidence interval is _____

Answer 1

Complete Question

The  complete question is shown on the first uploaded image  

Answer:

The interval is  $-0.0037 < p_1-p_2$

Step-by-step explanation:

From the question we are told that

   The first  sample size is  $n _1 = 1068000$

    The  first sample proportion is $\r p_1 = 0.089$

    The  second sample  size is  $n_2 = 1476000$

    The  second sample proportion is  $\r p_2 = 0.092$

   

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

             $\alpha = (100 - 95 )\%$

              $\alpha = 0.05$

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

The value is  

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

Generally the 95% confidence interval is mathematically represented as

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

Here  $\r q_1$ is mathematically evaluated as $\r q_1 = (1 - \r p_1)= 1-0.089 =0.911$

and  $\r q_2$  is mathematically evaluated as  $\r q_2 = (1 - \r p_2) = 1- 0.092 = 0.908$

So

       $(0.089 - 0.092 ) -1.96 \sqrt{ \frac{0.089* 0.911 }{1068000} + \frac{0.092* 0.908 }{1476000}} < (p_1 - p_2 ) < (0.089 - 0.092 ) +1.96 \sqrt{ \frac{0.089* 0.911 }{1068000} + \frac{0.092* 0.908 }{1476000}}$

$-0.0037 < p_1-p_2$