eu (european union) countries report that 46% of their labor force is female. The United Nations wnats to determine if the percentage of femailes in the U. S. labor force is the same. Based on sample, representatives from the United States department of labor find that the 95% confidence interval for the proportion of females in the U. S. labor force is .357 to .443. if the department of labores wishes to tighten it's interval they should

Step-by-step explanation:

The question is incomplete. The complete question is:

EU (European Union) countries report that 46% of their labor force is female. The United Nations wants to determine if the percentage of females in the U.S. labor force is the same. Based on a sample of 500 employment records, representatives from the United States Department of Labor found that the 95% confidence interval for the proportion of females in the U.S. labor force is 0.357 to 0.443. If the Department of Labor wishes to tighten its interval, they should:

A. increase the confidence level

B. decrease the sample size

C. increase the sample size

D. Both A and B

E. Both A and C

Solution:

Confidence interval for population proportion is written as

Sample proportion ± margin of error

Where sample proportion is the point estimate for the population proportion.

Margin of error = z × √pq/n

The z score for 95% confidence level is 1.96

p = 46/100 = 0.46

q = 1 - p = 1 - 0.46

q = 0.54

n = 500

Margin of error = 1.96√0.46 × 0.54/500 = 0.044

To tighten it's interval, the margin of error needs to be reduced.

If we increase the confidence level, say to 99%, z = 2.58

Then

Margin of error = 2.58√0.46 × 0.54/500 = 0.058

It increased

Also, If we increase the sample size, say to 700, then

Margin of error = 1.96√0.46 × 0.54/700 = 0.037

It has reduced

Therefore, the correct options is

C. increase the sample size

the answer is 9.8975

explanation:

when you multiply the "39.59" with the 25% it reduces it and shows its final discounted price. hope this !

step-by-step explanation: