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<-0.0023`

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
`(\alpha )/(2)` from the normal distribution table

The value is

`Z_{(\alpha )/(2) } =z_c= 1.96`

Generally the 95% confidence interval is mathematically represented as

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

`-0.0037 < p_1-p_2<-0.0023`