Understanding Population Variance vs. Sample Variance

The question asks about the key differences in how population variance and sample variance are calculated or represented. Population variance describes the spread of data points in an entire population, while sample variance estimates the spread based on a subset (sample) of that population.

Formulas for Variance

Let's look at the standard formulas for both:

• Population Variance (\(\sigma^2\)): \( \sigma^2 = \frac{\sum_{i=1}^{N} (x_i - \mu)^2}{N} \)
• Sample Variance (\(s^2\)): \( s^2 = \frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1} \)

Where:

• \(x_i\) is the i-th data point.
• \(\mu\) is the population mean.
• \(\bar{x}\) is the sample mean.
• \(N\) is the total number of observations in the population.
• \(n\) is the total number of observations in the sample.

Analyzing the Given Statements

Now let's examine each statement provided in the options:

• A. \(\mu \pm 3\sigma\) is replaced by \(\bar{x} + 9\sigma^2\): This statement is not accurate. \(\mu \pm 3\sigma\) relates to ranges around the mean, often used in the context of the empirical rule for normal distributions. Replacing it with \(\bar{x} + 9\sigma^2\) doesn't describe a standard difference between population and sample variance formulas. The term \(9\sigma^2\) also seems out of context here.
• B. \(\mu\) is replaced by \(\bar{x}\): This statement is correct. In the numerator of the variance formula, the population mean (\(\mu\)) is used when calculating population variance, while the sample mean (\(\bar{x}\)) is used when calculating sample variance.
• C. \(\mu^2\) is replaced by \(\frac{\mu^2}{(n-1) \alpha}\): This statement is incorrect. This replacement doesn't reflect how variance formulas differ. It introduces \(\alpha\), which is usually related to significance levels in hypothesis testing, not the fundamental variance calculation difference.
• D. N is replaced by n - 1: This statement is correct. In the denominator, the population size (\(N\)) is used for population variance. For sample variance, the sample size minus one (\(n-1\)) is used. This is known as Bessel's correction and is used to provide an unbiased estimate of the population variance from a sample.
• E. N is replaced by n - 1 - \(\alpha\): This statement is incorrect. The denominator for sample variance is \(n-1\), not \(n-1-\alpha\). Again, \(\alpha\) is irrelevant to the basic calculation of sample variance.

Identifying the Correct Differences

Based on the analysis of the formulas, the key differences mentioned in the statements are:

1. The population mean (\(\mu\)) is used for population variance, while the sample mean (\(\bar{x}\)) is used for sample variance (Statement B).
2. The population size (\(N\)) is the denominator for population variance, while the sample size minus one (\(n-1\)) is the denominator for sample variance (Statement D).

Therefore, statements B and D accurately describe how population variance differs from sample variance in terms of their calculation components.

Conclusion

The correct differences described by the statements are that the population mean \(\mu\) is replaced by the sample mean \(\bar{x}\), and the population size \(N\) is replaced by the sample size minus one \(n-1\).

Revision Table: Population Variance vs. Sample Variance

Additional Information: Bessel's Correction (n-1)

The use of \(n-1\) in the denominator for sample variance is called Bessel's correction. If we used \(n\) instead of \(n-1\), the sample variance would, on average, underestimate the true population variance. Dividing by \(n-1\) corrects this bias, making the sample variance (\(s^2\)) an unbiased estimator of the population variance (\(\sigma^2\)). This is particularly important when working with small samples.