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 <p_1 -p_2 < -0.007297`

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
`( \alpha )/(2)` the value is

`Z_{(\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{ (\r p_1 \r q_1 )/(n_1) + (\r p_2 \r q_2 )/(n_2) } <p_1 -p_2 < (\r p_1 - \r p_2 )+ z_c \sqrt{ (\r p_1 \r q_1 )/(n_1) + (\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{ (0.092* 0.916 )/(1068000) + (0.084* 0.908 )/(1476000) } <p_1 -p_2 < (0.084 - 0.092 )+ (1.96)* \sqrt{ (0.084* 0.916 )/(1068000) + (0.092* 0.908 )/(1476000) }`

`-0.00870 <p_1 -p_2 < -0.007297`