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What does it mean that the variance (computed by dividing by \(\mathrm{N}\) ) is a biased statistic?

Short Answer

Expert verified
The variance computed by dividing by \(N\) is biased as it underestimates the true population variance; using \(N-1\) corrects this bias.

Step by step solution

01

Understanding Variance and Biased Statistics

In statistics, variance refers to a measure of the spread or dispersion of a set of values. We use the formula \( \text{Var}(X) = \frac{1}{N} \sum_{i=1}^{N} (x_i - \bar{x})^2 \) to calculate variance for a population, where \( N \) is the number of observations, \( x_i \) are the values, and \( \bar{x} \) is the mean. A statistic is said to be biased if its expected value does not equal the true parameter value of the population. When we compute variance using \( N \) in the formula, it can lead to a biased estimate of the population variance.
02

Exploring the Impact of Sample Size on Bias

When we compute variance from a sample, using \( N \) tends to underestimate the population variance because it assumes the sample captures all variability of the population. This is due to the fact that samples usually have less variability than the entire population. Consequently, the variance calculated this way is systematically lower than the true population variance.
03

Correcting the Bias with Dividing by \( N-1 \)

To correct this bias, the variance of a sample is often calculated using \( \text{Var}(X) = \frac{1}{N-1} \sum_{i=1}^{N} (x_i - \bar{x})^2 \). This adjustment, known as Bessel's correction, increases the variance estimate slightly, making it unbiased by accounting for the fact that the sample mean \( \bar{x} \) itself is an estimate.
04

Conclusion on Bias in Variance

In conclusion, variance computed by dividing by \( N \) directly provides a biased estimate because it does not account for the additional uncertainty introduced by estimating the mean from a sample. By dividing by \( N-1 \), we instead get an unbiased estimator of the population variance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Variance
Variance is a crucial concept in statistics used to measure how much a set of values spread out from their average or mean. It's like asking how diverse your group of numbers is. For a population, we find this by calculating the average of the squared differences between each number in the set and the mean. The formula is \[ \text{Var}(X) = \frac{1}{N} \sum_{i=1}^{N} (x_i - \bar{x})^2 \] where
  • \( N \) is the total number of observations,
  • \( x_i \) represents each value,
  • and \( \bar{x} \) is the mean of all values.
Using this formula for populations makes perfect sense since it considers the entire set of data. However, when working with just a sample of the data, dividing by \( N \) may not accurately represent the variability of the entire population. This results in what's called a "biased statistic." Simply put, it might underestimate the true variance of the population.

This brings us to the next concept of why and how we need to correct this bias.
Bessel's Correction
Bessel's correction is a neat trick statisticians use to fix the bias in variance estimation when dealing with sample data. This correction involves dividing by \( N-1 \) instead of \( N \) when calculating variance for a sample. Why \( N-1 \)? Because it accounts for the fact that a sample doesn't include all the data from a population, which might mean it misses some variability.When calculating sample variance, using \[ \text{Var}(X) = \frac{1}{N-1} \sum_{i=1}^{N} (x_i - \bar{x})^2 \] you're adjusting for that dreaded bias.
  • This makes the estimated variance slightly larger.
  • It acknowledges that the mean itself is an estimate based on limited data.
In short, Bessel's correction helps ensure our variance calculation is as accurate as possible, given the constraints of working with a sample instead of an entire population.

It allows us to aim for a more accurate view of what true variability might look like when we can't measure everyone.
Unbiased Estimator
When statisticians talk about an unbiased estimator, they're looking for a statistical formula or method that hits the true population parameter without consistently missing the mark. It's about precision and reliability.A crucial point to understand is that an unbiased estimator doesn't offer a perfect estimate every time, but rather, on average, the errors in estimation tend to cancel out over repeated samples. For variance, using Bessel's correction (i.e., dividing by \( N-1 \)) gives us an unbiased estimator of population variance. Why is this important?
  • It helps ensure that our results are not systematically skewed.
  • It means we're less likely to make faulty conclusions based on incomplete data.
When we adjust for bias using this method, we make sure our findings align as closely as possible with the real-world data they aim to represent.

In the realm of statistics, aiming for unbiased estimators like this one is crucial to drawing fair and accurate conclusions from sample data.

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Most popular questions from this chapter

True/false: In order to construct a confidence interval on the difference between means, you need to assume that the populations have the same variance and are both normally distributed.

(DH) Compute a \(95 \%\) confidence interval on the proportion of people who are healthy on the AHA diet. $$ \begin{array}{|l|l|l|l|l|l|} \hline & \text { Cancers } & \text { Deaths } & \text { Nonfatal illness } & \text { Healthy } & \text { Total } \\ \hline \text { AHA } & 15 & 24 & 25 & 239 & 303 \\ \hline \text { Mediterranean } & 7 & 14 & 8 & 273 & 302 \\ \hline \text { Total } & 22 & 38 & 33 & 512 & 605 \\ \hline \end{array} $$

How does the t distribution compare with the normal distribution? How does this difference affect the size of confidence intervals constructed using \(\mathrm{z}\) relative to those constructed using t? Does sample size make a difference?

You take a sample of 22 from a population of test scores, and the mean of your sample is 60 . (a) You know the standard deviation of the population is 10 . What is the \(99 \%\) confidence interval on the population mean. (b) Now assume that you do not know the population standard deviation, but the standard deviation in your sample is \(10 .\) What is the \(99 \%\) confidence interval on the mean now?

Imagine that there are 100 different researchers each studying the sleeping habits of college freshmen. Each researcher takes a random sample of size 50 from the same population of freshmen. Each researcher is trying to estimate the mean hours of sleep that freshmen get at night, and each one constructs a \(95 \%\) confidence interval for the mean. Approximately how many of these 100 confidence intervals will NOT capture the true mean? a. None b. 1 or 2 c. 3 to 7 d. about half e. 95 to 100 f. other

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