Question

The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of females in the labor force is the same in the United States. Representatives from the United States Department of Labor plan to check a random sample of over 10,000 employment records on file to estimate a percentage of females in the United States labor force.

They select a random sample of 525 employment records, and find that 229 of the people are females. Check the required conditions to construct a confidence interval. Create the confidence interval, Interpret the confidence interval in this context, Explain what 90% confidence means in this context, and should the representatives from the Department of Labor conclude that the percentage of females in their labor force is lower than Europe’s rate of 46%? Explain.

Answer 1

The sample must be reasonably random

The sample must be less than 10% of the population

The sample must be large enough so that:
n• pˆ and n(1 - pˆ ) >= 5 for a confidence interval

pcap = 229/525 = 0.436

The z value at 90% confidence interval is,

alpha = 1 - 0.90 = 0.1
alpha/2 = 0.1/2 = 0.05
Zalpha/2 = Z0.05 = 1.645

Margin of error = E =z *sqrt(pcap*(1-pcap)/n)
= 1.64 *sqrt(0.436*(1-0.436)/525)
= 0.0355

CI = pcap +/- 0.035
= 0.436 +/- 0.0355

= (0.4005 , 0.4715)

We are 90% confidencnt taht the percentage of females in their labor force is between 0.4005 and 0.4715

No, because confidence interval contain 0.46