Consider a population that is finite, and sampling is with replacement. If the variance of the population is 2176.8 with a sample size of 16, then the variance of the sampling distribution of means is:
136.05
Calculating the Variance of the Sampling Distribution of the Mean
The question asks us to determine the variance of the sampling distribution of the mean ($\sigma_{\bar{x}}^2$) given the population variance ($\sigma^2$) and the sample size (n), specifically when sampling is done with replacement from a finite population.
Understanding the Concepts
- Population Variance ($\sigma^2$): This measures the spread or dispersion of the data points in the entire population. It is given as 2176.8.
- Sample Size (n): This is the number of observations included in the sample. It is given as 16.
- Sampling Distribution of the Mean: This is the probability distribution of sample means from all possible samples of a given size taken from a population.
- Variance of the Sampling Distribution of the Mean ($\sigma_{\bar{x}}^2$): This measures the spread of the sample means around the true population mean. It tells us how much the sample means are expected to vary from one sample to another.
- Sampling With Replacement: This means that once an item is selected for the sample, it is put back into the population and can be selected again. This process ensures that the probability of selecting any item remains constant for each draw.
Formula for Variance of Sampling Distribution of Mean (With Replacement)
When sampling is done with replacement (or from an infinite population), the formula for the variance of the sampling distribution of the mean is:
$$ \sigma_{\bar{x}}^2 = \frac{\sigma^2}{n} $$
Where:
- $\sigma_{\bar{x}}^2$ is the variance of the sampling distribution of the mean.
- $\sigma^2$ is the population variance.
- $n$ is the sample size.
The fact that the population is finite is noted, but for sampling with replacement, the size of the finite population does not influence the variance of the sampling distribution of the mean directly in the formula, unlike sampling without replacement where the finite population correction factor is used.
Step-by-Step Calculation
Given the values:
- Population Variance ($\sigma^2$) = 2176.8
- Sample Size (n) = 16
Using the formula $\sigma_{\bar{x}}^2 = \frac{\sigma^2}{n}$, we substitute the given values:
$$ \sigma_{\bar{x}}^2 = \frac{2176.8}{16} $$
Now, we perform the division:
$$ \sigma_{\bar{x}}^2 = 136.05 $$
Result and Conclusion
The calculated variance of the sampling distribution of the means is 136.05.
Comparing this result with the provided options:
- Option 1: 137.04
- Option 2: 135.21
- Option 3: 136.05
- Option 4: 134.12
The calculated value matches Option 3.
| Parameter | Value |
|---|---|
| Population Variance ($\sigma^2$) | 2176.8 |
| Sample Size (n) | 16 |
| Formula (Sampling With Replacement) | $\sigma_{\bar{x}}^2 = \frac{\sigma^2}{n}$ |
| Calculated Variance ($\sigma_{\bar{x}}^2$) | 136.05 |
Revision Table: Sampling Distribution Variance
| Concept | Formula | Conditions |
|---|---|---|
| Variance of Sample Mean ($\sigma_{\bar{x}}^2$) | $\frac{\sigma^2}{n}$ | Sampling With Replacement OR Infinite Population |
| Variance of Sample Mean ($\sigma_{\bar{x}}^2$) | $\frac{\sigma^2}{n} \times \left(\frac{N-n}{N-1}\right)$ | Sampling Without Replacement from a Finite Population (size N) |
| Standard Error of Mean ($\sigma_{\bar{x}}$) | $\sqrt{\sigma_{\bar{x}}^2}$ | Square root of the variance of the sample mean |
Additional Information: Sampling and Statistics
Understanding sampling distributions is fundamental in inferential statistics. The variance of the sampling distribution of the mean is a key component in calculating the standard error of the mean, which is used in constructing confidence intervals and performing hypothesis tests about the population mean.
- Standard Error of the Mean ($\sigma_{\bar{x}}$): This is the standard deviation of the sampling distribution of the mean. It indicates the typical distance between a sample mean and the true population mean. It is simply the square root of the variance of the sampling distribution of the mean: $\sigma_{\bar{x}} = \sqrt{\sigma_{\bar{x}}^2}$. In this case, the standard error would be $\sqrt{136.05} \approx 11.66$.
- Finite Population Correction Factor (FPC): When sampling without replacement from a finite population of size N, the samples are not independent. The FPC, $\left(\frac{N-n}{N-1}\right)$, is multiplied by $\frac{\sigma^2}{n}$ to adjust for this lack of independence. If the population size N is much larger than the sample size n (a common rule of thumb is n/N < 0.05), the FPC is close to 1, and the formula for sampling with replacement provides a good approximation even for sampling without replacement. However, the question explicitly states sampling *with* replacement, making the FPC unnecessary.
- Central Limit Theorem: This crucial theorem states that, regardless of the shape of the population distribution, the sampling distribution of the mean will be approximately normally distributed if the sample size is sufficiently large (usually n > 30). This allows us to use normal distribution properties for inference even if the original population is not normal, provided the sample size is large enough. If the population itself is normally distributed, the sampling distribution of the mean will be normal for any sample size n.
Important Questions from Variance and Standard Deviation
Mean of 100 observations is 50 and standard deviation is 10. If 5 is added to each observation, then what will be the new mean and new standard deviation respectively?
When sampling is done without replacement then standard error of mean is:
If \(\rm \displaystyle \sum_{i = 1}^{10}x_i = 110\) and \(\rm \displaystyle \sum_{i = 1}^{10}x_i^2 = 1540\) then what is the variance?
The sum of deviations of n numbers from 10 and 20 are p and q respectively. If (p - q)2 = 10000, then what is the value of n?
The fourth central moment of a mesokurtic distribution is 243. Its standard deviation is:
