Suppose a random sample of size 50 is selected from a population with σ = 10. 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).
Some notes on finite population correction factor:
If the population is small and the sample is large (more than 5% of the small population), use the finite population correction factor.
For standard error of the mean, use:
sd/√n
If you need to adjust for the finite population correction factor, use:
sd/√n * √(N-n)/N-1)
N = number in population
n = number in sample
Therefore:
a. Use the formula sd/√n (population is infinite)
b. Is the sample size of 50 more than 5% of the population? No, it isn't; your answer will be the same as part a).
c. Is the sample size of 50 more than 5% of the population? No, it isn't; your answer will be the same as part a) and b).
d. Is the sample size of 50 more than 5% of the population? Yes, it is, so use the finite population correction factor.
I hope this will help get you started.
To find the value of the standard error of the mean in each case, we need to use the formula:
Standard Error of the Mean (SE) = σ /√n
Where:
- σ is the population standard deviation
- n is the sample size
a. The population size is infinite:
In this case, we assume the population is infinite. The formula remains the same and the standard error of the mean is given by:
SE = σ /√n
= 10 /√50
≈ 1.41 (to 2 decimals)
b. The population size is N = 50,000:
When the population size is finite, we need to use the finite population correction factor. The formula for standard error of the mean with finite population correction factor is:
SE = σ /√n * √(N-n) / √(N-1)
Substituting the values into the formula:
SE = 10 /√50 * √(50000-50) / √(50000-1)
≈ 1.41 * 99.98 / 223.61
≈ 0.63 (to 2 decimals)
c. The population size is N = 5000:
Using the same formula with the given values:
SE = 10 /√50 * √(5000-50) / √(5000-1)
≈ 1.41 * 99.75 / 70.68
≈ 1.99 (to 2 decimals)
d. The population size is N = 500:
Using the formula with the given values:
SE = 10 /√50 * √(500-50) / √(500-1)
≈ 1.41 * 99.5 / 22.36
≈ 6.28 (to 2 decimals)
Therefore, the value of the standard error of the mean in each case is as follows:
a. Approximately 1.41
b. Approximately 0.63
c. Approximately 1.99
d. Approximately 6.28
To find the value of the standard error of the mean in each case, we can use the formula:
Standard Error of the Mean (SE) = σ / √n
Where σ is the population standard deviation and n is the sample size.
a. When the population size is infinite:
In this case, we can assume that the sample is representative of the entire population, and the finite population correction factor does not need to be used.
SE = σ / √n
= 10 / √50
≈ 1.41
b. When the population size is N = 50,000:
Since the population size is large, we can assume that the sample size is small relative to the population size, and the finite population correction factor does not need to be used.
SE = σ / √n
= 10 / √50
≈ 1.41
c. When the population size is N = 5,000:
In this case, the sample size is relatively large compared to the population size. We need to use the finite population correction factor:
SE = σ / √(n * (N - n) / N)
= 10 / √(50 * (5,000 - 50) / 5,000)
= 10 / √(50 * (4,950) / 5,000)
≈ 1.40
d. When the population size is N = 500:
Similar to the previous case, the sample size is relatively large compared to the population size, so we need to use the finite population correction factor:
SE = σ / √(n * (N - n) / N)
= 10 / √(50 * (500 - 50) / 500)
= 10 / √(50 * (450) / 500)
= 10 / √(22.5)
≈ 1.91
So the value of the standard error of the mean in each case is approximately:
a. 1.41
b. 1.41
c. 1.40
d. 1.91