Answer :
Answer:
Let's go through each part step by step for the given population and the sampling process.
**Population:**
\[ \{ 2, 3, 6, 8, 11 \} \]
**Sample size (n):** 2 (samples drawn with replacement)
**Number of elements in population (N):** 5
### (i) Population Mean
The population mean (μ) is calculated as:
\[ \mu = \frac{\sum_{i=1}^{N} x_i}{N} \]
where \( x_i \) are the elements of the population.
Calculate the population mean:
\[ \mu = \frac{2 + 3 + 6 + 8 + 11}{5} = \frac{30}{5} = 6 \]
So, the population mean is \( \mu = 6 \).
### (ii) Population Standard Deviation
The population standard deviation (σ) is calculated using:
\[ \sigma = \sqrt{\frac{\sum_{i=1}^{N} (x_i - \mu)^2}{N}} \]
Calculate deviations from the mean:
\[ (2-6)^2 = 16, \; (3-6)^2 = 9, \; (6-6)^2 = 0, \; (8-6)^2 = 4, \; (11-6)^2 = 25 \]
Sum of squared deviations:
\[ \sum (x_i - \mu)^2 = 16 + 9 + 0 + 4 + 25 = 54 \]
Population standard deviation:
\[ \sigma = \sqrt{\frac{54}{5}} \approx \sqrt{10.8} \approx 3.29 \]
So, the population standard deviation is \( \sigma \approx 3.29 \).
### (iii) Mean of Sampling Distribution of Means
The mean of the sampling distribution of means (μ_{\bar{x}}) is equal to the population mean:
\[ \mu_{\bar{x}} = \mu = 6 \]
### (iv) Standard Deviation of Sampling Distribution of Means
The standard deviation of the sampling distribution of means (standard error, SE) is calculated as:
\[ SE = \frac{\sigma}{\sqrt{n}} \]
Where \( \sigma \) is the population standard deviation and \( n \) is the sample size.
For \( n = 2 \):
\[ SE = \frac{3.29}{\sqrt{2}} \approx \frac{3.29}{1.414} \approx 2.33 \]
So, the standard deviation of the sampling distribution of means (SE) is approximately \( 2.33 \).
### (v) Correction Factor
The correction factor (also known as the finite population correction factor, FPC) is given by:
\[ \sqrt{\frac{N-n}{N-1}} \]
Where \( N \) is the population size and \( n \) is the sample size.
For \( N = 5 \) and \( n = 2 \):
\[ FPC = \sqrt{\frac{5-2}{5-1}} = \sqrt{\frac{3}{4}} = \sqrt{0.75} \approx 0.866 \]
### (vi) Standard Error and Verification
The standard error (SE) is \( SE = \frac{\sigma}{\sqrt{n}} \). We calculated \( \sigma \approx 3.29 \) and for \( n = 2 \), \( SE \approx 2.33 \).
Verification:
\[ SE = \frac{3.29}{\sqrt{2}} \approx 2.33 \]
This matches our previous calculation, confirming the standard error.
### Summary
- Population mean (μ): \( 6 \)
- Population standard deviation (σ): \( \approx 3.29 \)
- Mean of sampling distribution of means (μ_{\bar{x}}): \( 6 \)
- Standard deviation of sampling distribution of means (SE): \( \approx 2.33 \)
- Correction factor (FPC): \( \approx 0.866 \)
- Standard error (SE): \( \approx 2.33 \)
These values represent the statistical characteristics of the population and the sampling distribution based on samples of size 2 drawn with replacement from the given population.