Question

Suppose a random sample of size 53 is selected from a population with σ = 9. Find the value of the standard error of the mean in each of the following cases (use the finite population correction factor if appropriate). a. The population size is infinite (to 2 decimals). b. The population size is N = 50,000 (to 2 decimals). c. The population size is N = 5000 (to 2 decimals). d. The population size is N = 500 (to 2 decimals).

Answer 1

a. The population size is infinite (to 2 decimals).

σ = 9

n = 53

The standard deviation of the sample mean x̄ is the population standard deviation divided by the square root of the sample size

σx̄ = σ / sqrt(n)

= 9 / sqrt(53)

=1.2362

=1.24

b. The population size is N = 50,000 (to 2 decimals).

The standard deviation of the sample mean x̄ is the population standard deviation divided by the square root of the sample size, multiplying by the infinite population correction factor:

σx̄ = Sqrt (N - n / N-1) * σ / sqrt (n)

=SQRT(50000 - 53)/SQRT(50000 - 1) * 9/SQRT(53)

=1.235603

=1.24

c. The population size is N = 5000 (to 2 decimals).

The standard deviation of the sample mean x̄ is the population standard deviation divided by the square root of the sample size, multiplying by the infinite population correction factor:

σx̄ = Sqrt (N - n / N-1) * σ / sqrt (n)

=SQRT(5000 - 53)/SQRT(5000 - 1) * 9/SQRT(53)

=1.22979

=1.23

d. The population size is N = 500 (to 2 decimals).

The standard deviation of the sample mean x̄ is the population standard deviation divided by the square root of the sample size, multiplying by the infinite population correction factor:

σx̄ = Sqrt (N - n / N-1) * σ / sqrt (n)

=SQRT(500 - 53)/SQRT(500 - 1) * 9/SQRT(53)

=1.1701

=1.17