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 US 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 US labor force.

(a) The representatives from the Department of Labor want to estimate a percentage of females in the US labor force to within +/- 5%, with 90% confidence. How many employment records should they sample?

(b) They actually select a random sample of 525 employment records and find that 229 of the people are females. Create a confidence interval.

(c) Interpret the confidence interval in this context.

(d) Explain what 90% confidence means in this context.

(e) 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

a) At 90% confidence level, the critical value is z_0.05 = 1.645

Margin of error = 0.05

or, z_0.05 * sqrt(P(1 - P)/n) = 0.05

or, 1.645 * sqrt(0.46(1 - 0.46)/n) = 0.05

or, n = (1.645 * sqrt(0.46 * 0.54)/0.05)^2

or, n = 269

b) = 229/525 = 0.4362

The 90% confidence interval is

+/- z_0.05 * sqrt((1 - )/n)

= 0.4362 +/- 1.645 * sqrt(0.4362(1 - 0.4362)/525)

= 0.4362 +/- 0.0356

= 0.4006, 0.4718

c) We are 90% confidence that the true proportion of females lies in the above confidence interval.

d) The 90% confidence means, the 90% of the intervals would contain the true population proportion.

e) No, the representatives from the Department of Labor cannot conclude that the percentage of females in their labor force is lower than Europe's rate of 46%. Because 0.46 lies in the confidence interval.