Question

The labor force participation rate is the number of people in the labor force divided by the number of people in the country who are of working age and not institutionalized. The BLS reported in February 2012 that the labor force participation rate in the United States was 63.7% (Calculatedrisk). A marketing company asks 120 working-age people if they either have a job or are looking for a job, or, in other words, whether they are in the labor force.What are the expected value and the standard error for a labor participation rate in the company’s sample? Multiple Choice : a) 0.637, b) 0.0019, c) 0.637, d) 0.04397, e) 6.44, f) 0.0019

Answer 1

Answer: a) 0.637

d) 0.04397

Explanation:

The formula to calculate the expected value and standard error for proportion (p) is given by :-

`E(\overline{P})=p`

`SD(\overline{P})=\sqrt{(p(1-p))/(n)}` , where n is the sample size.

Given : The BLS reported in February 2012 that the labor force participation rate in the United States was 63.7% .

i.e. the proportion of labor force participation rate in the United States : p=0.637

A marketing company asks 120 working-age people if they either have a job or are looking for a job, or, in other words, whether they are in the labor force.

i.e. n= 120

Then, the expected value and the standard error for a labor participation rate in the company’s sample will be :-

`E(\overline{P})=0.637`

`SD(\overline{P})=\sqrt{(0.637(1-0.637))/(120)}\\\\=√(0.001926925)=0.0438967538663\approx0.043897`

Hence, the correct answers are : a) 0.637 (For expected value)

d) 0.04397 (For standard error .)

Answer 2

The expected participation rate is 0.637.

The standard error is 0.04397

Explanation:

For each working age people asked, there are only two possible outcomes. Either they are in the labor force, or they are not. This means that we can solve this problem using binomial distribution probability concepts.

Binomial probability:

Expected value for the participation rate: The expected value is the probability of a success. In this problem, a success is a working age people being in the labor force. 63.7% of them are. So
`\pi = 0.637`. This means that the expected participation rate is 0.637.

Standard error for the participation rate:

The standard error is given by the following formula:

`\sigma = \sqrt{(\pi(1-\pi))/(n)}`.

In this problem, 120 people are asked, so
`\pi = 120`

`s = √(n\pi(1-\pi)) = \sqrt{(0.637*(1-0.637))/(120)} = 0.04397`.

So the standard error is 0.04397