Suppose a random sample of size 40 is selected from a population with = 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).
39) Suppose a random sample of size 40 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).
σ/√n
9/√40 = 1.42
b. The population size is N = 50,000 (to 2 decimals). No
n/N > 0.05
40/50000
0.0008 < 0.05
σ/√n * √(N-n)/N-1)
c. The population size is N = 5000 (to 2 decimals). No
n/N > 0.05
40/5000
0.008 < 0.05
σ/√n * √(N-n)/N-1)
d. The population size is N = 500 (to 2 decimals). Yes
n/N > 0.05
40/500
0.08 > 0.05
σ/√n * √(N-n)/N-1)
1.42 * √(500-40)/(500-1)
1.42 * √460/499)
= 1.36
bnv
D= 1.34
a. The value of the standard error of the mean when the population size is infinite is approximately 1.42 (to 2 decimals). Because, let's face it, when the population is infinite, who needs to worry about standard errors anyway? It's like worrying about the number of hairs on a clown's wig. It's just not worth losing sleep over.
b. When the population size is N = 50,000, the value of the standard error of the mean is approximately 0.4 (to 2 decimals), assuming you choose to use the finite population correction factor. Just remember, in a population of 50,000, everyone is a star, or at least a tiny speck in an enormous galaxy of statistical possibilities.
c. Alright, let's get serious for a moment. When the population size is N = 5000, the value of the standard error of the mean is approximately 0.45 (to 2 decimals), taking into account the finite population correction factor. It's like walking a tightrope, trying to estimate the mean with a smaller population size. But don't worry, I've seen clowns do scarier things.
d. When the population size is N = 500, the value of the standard error of the mean is approximately 0.71 (to 2 decimals), accounting for the finite population correction factor. It's like juggling 500 balls in the air and hoping none of them fall. But hey, as a clown, I'm used to dropping things, so you'll be just fine.
To find the value of the standard error of the mean in each of the given cases, we need to consider the population size and whether or not to apply the finite population correction factor.
The standard error of the mean (SE) is given by the formula:
SE = σ / √n
where σ represents the population standard deviation and n is the sample size.
a. The population size is infinite (to 2 decimals):
When the population size is infinite, we assume an approximate infinite population correction factor. In this case, we can use the formula without any correction factor:
SE = σ / √n
b. The population size is N = 50,000 (to 2 decimals):
For a finite population, if the sample size is no more than 5% of the population size (n ≤ 0.05N), we use the finite population correction factor. In this case, the sample size (n) is 40, which is less than 5% of 50,000. Therefore, we should use the finite population correction factor (sqrt((N - n) / (N - 1))):
SE = σ / √n * sqrt((N - n) / (N - 1))
Substituting the given values: σ = 9, n = 40, and N = 50,000, we can calculate the standard error of the mean using the formula above.
c. The population size is N = 5000 (to 2 decimals):
Similar to case b, we verify whether the sample size is less than or equal to 5% of the population size. Since n = 40, which is less than 5% of 5000, we should apply the finite population correction factor:
SE = σ / √n * sqrt((N - n) / (N - 1))
Substituting the given values: σ = 9, n = 40, and N = 5000, we can calculate the standard error of the mean using the formula above.
d. The population size is N = 500 (to 2 decimals):
Again, we check if the sample size is less than or equal to 5% of the population size. Since n = 40, which is more than 5% of 500, we do not need to apply the finite population correction factor:
SE = σ / √n
Substituting the given values: σ = 9 and n = 40, we can calculate the standard error of the mean using the formula above.
Please note that to get the final values, you need to substitute σ (the population standard deviation) into the formulas and perform the necessary calculations.