A Pew Research Center poll asked independent random samples of working women and men how much they value job security. Of the 798 women, 702 said job security was very or extremely important, compared with 802 of the 940 men surveyed. Construct and interpret a 95% confidence interval for the difference in the proportion of all working women and men who consider job security very or extremely important

Step-by-step explanation:

Confidence interval for the difference in the two proportions is written as

Difference in sample proportions ± margin of error

Sample proportion, p= x/n

Where x = number of success

n = number of samples

For the women,

x = 702

n1 = 798

p1 = 702/798 = 0.88

For the men

x = 802

n2 = 940

p2 = 802/940 = 0.85

Margin of error = z√[p1(1 - p1)/n1 + p2(1 - p2)/n2]

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.025 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, confidence level of 95% is 1.96

Margin of error = 1.96 × √[0.88(1 - 0.88)/798 + 0.85(1 - 0.85)/940]

= 1.96 × √0.00026796913

= 0.032

Confidence interval = (0.88 - 0.85) ± 0.032

= 0.03 ± 0.032

Lower boundary = 0.03 - 0.032 = - 0.002

Upper boundary = 0.03 + 0.032 = 0.062

The proportion of all working women who consider job security very or extremely important is higher than that of men and we are 95% confident that the population difference lies between - 0.002 and

0.062

17.08

correct me if i am wrong.

the answer for the second equation is (4,0)