Question

According to the U.S. Census Bureau, 42% of men who worked at home were college graduates. In a sample of $14 women who worked at home, 158 were college graduates. 1. Find a point estimate for the proportion of college graduates among women who work at home. Round the answer to three decimal places. The point estimate for the proportion of college graduates among women who work at home Is .2. Construct an 80% confidence interval for the proportion of women who work at home who are college graduates. Round the answer to three decimal places. An 80% confidence interval for the proportion of women who work at home.

Answer 1

Answer:

Step-by-step explanation:

1) The point estimate, p for the proportion of college graduates among women who work at home is

p = x/n

Where

x = number of success = 158

n = number of samples = 514

p = 158/514 = 0.307

2) Confidence interval is written as

Sample proportion ± margin of error

Margin of error = z × √pq/n

Where

z represents the z score corresponding to the confidence level

p = sample proportion. It also means probability of success

q = probability of failure

q = 1 - p

Probability of success, p = 42/100 = 0.42

Probability of failure = 1 - p = 1 - 0.42 = 0.58

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

α = 1 - 0.8 = 0.2

α/2 = 0.2/2 = 0.1

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

1 - 0.1 = 0.9

The z score corresponding to the area on the z table is 1.282. Thus, the z score for a confidence level of 80% is 1.282

Margin of error = 1.282 × √(0.42 × 0.58)/514

Margin of error = 0.028

Therefore, the 80% confidence interval is

0.307 ± 0.028

The lower limit of the confidence interval is

0.307 - 0.028 = 0.279

The upper limit of the confidence interval is

0.307 + 0.028 = 0.335

Therefore, an 80% confidence interval for the proportion of women who work at home is between 0.279 and 0.335