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Suppose a random sample of size is selected from a population with . 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 (to 2 decimals). c. The population size is (to 2 decimals). d. The population size is (to 2 decimals).

Respuesta :

Answer:

A) σ_x' = 1.4142

B) σ_x' = 1.4135

C) σ_x' = 1.4073

D) σ_x' = 1.343

Step-by-step explanation:

We are given;

σ = 10

n = 50

A) when size is infinite, the standard deviation of the sample mean is given by the formula;

σ_x' = σ/√n

Thus,

σ_x' = 10/√50

σ_x' = 1.4142

B) size is given, thus, the standard deviation of the sample mean is given by the formula;

σ_x' = (σ/√n)√((N - n)/(N - 1))

Thus, with size of N = 50,000, we have;

σ_x' = 1.4142 x √((50000 - 50)/(50000 - 1))

σ_x' = 1.4142 x 0.9995

σ_x' = 1.4135

C) at N = 5000;

σ_x' = 1.4142 x √((5000 - 50)/(5000 - 1))

σ_x' = 1.4073

D) at N = 500;

σ_x' = 1.4142 x √((500 - 50)/(500 - 1))

σ_x' = 1.343

The value of the standard error in each of the following are:

A) σ_x' = 1.4142

B) σ_x' = 1.4135

C) σ_x' = 1.4073

D) σ_x' = 1.343

Standard error calculation:

Given:

σ = 10

n = 50

A) when size is infinite, the standard deviation of the sample mean is given by the formula;

σ_x' = σ/√n

Thus,

σ_x' = 10/√50

σ_x' = 1.4142

B) size is given, thus, the standard deviation of the sample mean is given by the formula;

σ_x' = (σ/√n)√((N - n)/(N - 1))

Thus, with size of N = 50,000, we have;

σ_x' = 1.4142 x √((50000 - 50)/(50000 - 1))

σ_x' = 1.4142 x 0.9995

σ_x' = 1.4135

C) at N = 5000;

σ_x' = 1.4142 x √((5000 - 50)/(5000 - 1))

σ_x' = 1.4073

D) at N = 500;

σ_x' = 1.4142 x √((500 - 50)/(500 - 1))

σ_x' = 1.343

Find more information about Standard errors here:

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