Answer:
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
