How to find mean and standard deviation of sampling distribution

Asked by admin @ 28/10/2021 in Mathematics viewed by 354 People

Calculate the mean and the standard deviation of the sampling distribution of possible sample proportions for each combination of sample size (n) and population proportion (p).

Answered by admin @ 28/10/2021

Here is the full question.

Calculate the mean and the standard deviation of the sampling distribution of possible sample proportions for each combination of sample size (n) and population proportion (p).

a.) n = 64, p = 0.8

b.) n = 256, p = 0.8

Answer:

a.) Mean = 0.8

Standard deviation = 0.05

b.) Mean = 0.8

Standard deviation = 0.025

Step-by-step explanation:

From the given information:

a.)

The population parameter (p) is the mean of the sampling distribution for the sample proportion.

Thus; The Mean = 0.8

The standard deviation of the distribution can be calculated as follows:

Let's first represent the standard deviation with Sd( $\hat p$ )

Then:

$sd(\hat p) = \sqrt{\dfrac{p*(1-p)}{n} }$

where;

p = 0.8

n = 64

Then:

$sd(\hat p) = \sqrt{\dfrac{0.8*(1-0.8)}{64} }$

$sd(\hat p) = \sqrt{\dfrac{0.8*(0.2)}{64} }$

$sd(\hat p) = \sqrt{\dfrac{0.16}{64} }$

$sd(\hat p) = \sqrt{0.0025}$

$\mathbf{sd(\hat p) =0.05}$

b.)

The population parameter (p) is the mean of the sampling distribution for the sample proportion.

Thus; The Mean = 0.8

The standard deviation of the distribution can be calculated as follows:

Let's first represent the standard deviation with Sd( $\hat p$ )

Then:

$sd(\hat p) = \sqrt{\dfrac{p*(1-p)}{n} }$

where;

p = 0.8

n = 256

Then:

$sd(\hat p) = \sqrt{\dfrac{0.8*(1-0.8)}{256} }$

$sd(\hat p) = \sqrt{\dfrac{0.8*(0.2)}{256} }$

$sd(\hat p) = \sqrt{\dfrac{0.16}{256} }$

$sd(\hat p) = \sqrt{6.25 \times 10^{-4}}$

$\mathbf{sd(\hat p) =0.025}$

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