Pooled Sample Variance Unbiased Estimator Proof: A Detailed Explanation

The pooled variance is an unbiased estimate of the variance within each sample, assuming the variances are equal. It’s a handy tool when we want to combine information from different samples to get a better overall estimate of the variability.

Let’s break this down a bit more. Imagine we have two groups of students, Group A and Group B, and we’re interested in their heights. We want to know if there’s a difference in their average heights. To do this, we need to estimate the variability within each group. We can use the sample variance for this. However, if we think the groups have similar variability (meaning the spread of heights is about the same in both groups), we can get a more accurate estimate by combining the information from both groups using the pooled variance.

The pooled variance essentially averages the variances of the individual samples, weighted by the number of observations in each sample. This gives us a more robust estimate of the overall variability, especially when we have small sample sizes.

Think of it like this: if you’re trying to estimate the weight of a large bag of apples, it’s more accurate to weigh several smaller bags and average those weights than to just weigh one big bag. The same principle applies to pooled variance. It allows us to get a better estimate of the variability by combining information from multiple samples.

Let’s explore how the pooled variance works, particularly when dealing with different group sizes.

You’re right, when the group sizes are different, the pooled variance becomes a weighted average, with larger groups holding more influence than smaller groups. This makes sense – if one group has a lot more data points, it should contribute more to the overall variance.

To understand this better, picture the numerator as a weighted sum of the individual group variances. Each group’s variance is multiplied by its size (the “weight”). This ensures that larger groups contribute more to the overall variance.

Now, consider the denominator. It’s simply the sum of the weights, which are the group sizes. By dividing the weighted sum of variances by the total weight, we essentially calculate the weighted average of the group variances, giving us the pooled variance.

Here’s a deeper dive into how the pooled variance is calculated:

Imagine we have two groups, Group A and Group B, with sizes `n_A` and `n_B`, respectively. Their individual variances are `s_A^2` and `s_B^2`.

The pooled variance, denoted as `s_p^2`, is calculated as follows:

“`
s_p^2 = ((n_A – 1) * s_A^2 + (n_B – 1) * s_B^2) / (n_A + n_B – 2)
“`

Let’s break down this formula:

* `(n_A – 1) * s_A^2`: This term represents the weighted variance of Group A. The variance `s_A^2` is multiplied by `(n_A – 1)`, which is the degrees of freedom for Group A.
* `(n_B – 1) * s_B^2`: Similarly, this term represents the weighted variance of Group B.
* `(n_A + n_B – 2)`: This is the total degrees of freedom for both groups.

By combining the weighted variances of both groups and dividing by the total degrees of freedom, we arrive at the pooled variance. This metric gives us a consolidated measure of variance, considering the contributions of both groups, with larger groups having a greater impact.

Remember, the pooled variance is a crucial concept in hypothesis testing, especially when comparing the means of two populations. It allows us to estimate the common variance shared by the two groups, even if their individual variances differ.

You’re right to wonder about sample variance! It’s a key concept in statistics, and understanding why it’s unbiased is essential. Let’s break it down.

Sample variance, calculated using the formula where you divide by n-1, is indeed an unbiased estimator of the true population variance. This means that on average, the sample variance will equal the population variance. This might seem counterintuitive – why not just divide by n like we do for the sample mean?

The reason we divide by n-1 is to account for the fact that the sample mean is used to estimate the population mean. When we use the sample mean to calculate the variance, we lose one degree of freedom. Think of it this way: if you know the mean and all but one of the data points, you can automatically figure out the missing value. Dividing by n-1 compensates for this lost degree of freedom, ensuring that our estimate of the population variance is less likely to be too small.

Let’s look at an example:

Imagine you’re trying to estimate the average height of all adults in a country. You take a sample of 100 people and calculate the sample variance using n in the formula. This variance might underestimate the true population variance. Why? Because the sample mean is likely not exactly equal to the population mean, and this difference will tend to make our calculated variance smaller.

By dividing by n-1 instead, we adjust for this potential underestimation. The sample variance with n-1 in the denominator will be slightly larger, giving us a more accurate estimate of the population variance.

In summary: While dividing by n might seem more intuitive, dividing by n-1 is crucial for making sure the sample variance is an unbiased estimator of the population variance. This ensures that our estimate of the population variance is not systematically too small or too large.

Let’s dive into how to determine if an estimator is unbiased!

The key to proving an estimator is unbiased is to compare the expected value of the estimator to the true population parameter.

Here’s the breakdown:

1. Identify the population parameter. This is the true value you’re trying to estimate. For example, if you want to estimate the average height of all students in a school, the population parameter would be the true average height of all students in the school.
2. Determine the expected value of the estimator. The expected value is the average value of the estimator you would expect to get if you took many samples from the population. Think of it as the long-run average of the estimator.
3. Compare the two. If the expected value of the estimator equals the population parameter, then the estimator is unbiased. If they are different, then the estimator is biased.

To illustrate this, imagine you’re trying to estimate the average height of all students in a school using a sample of students. If your estimator consistently overestimates or underestimates the true average height, then it’s biased. An unbiased estimator would provide an average height close to the true population average.

Here’s a practical example:

Let’s say the true average height of all students in a school is 5’8″. If you use a sample of students and calculate the average height of that sample to be 5’8″ as well, then your estimator is unbiased. However, if the sample average is consistently 5’7″ or 5’9″, then your estimator is biased.

Remember, an unbiased estimator doesn’t mean that it will always give you the exact true value. It simply means that, on average, it will give you the true value.

Let’s get a bit deeper into the math behind this. You can formally prove an estimator is unbiased by using the following steps:

1. Define the estimator. Let’s say your estimator is represented by the symbol “θ̂” (theta hat).
2. Determine the expected value of the estimator. Mathematically, this is represented as E(θ̂).
3. Set the expected value equal to the population parameter. If E(θ̂) = θ (theta), where θ is the true population parameter, then the estimator is unbiased.

Let me know if you have more questions or if you’d like to explore specific examples of estimators and how to prove their unbiasedness. I’m here to help you understand this important concept!

The sample variance is typically defined as S² = 1/(n-1) * Σ(i=1 to n) (Xi – ˉX)², making it an unbiased estimator of the population variance σ². This means that, on average, the sample variance will equal the population variance.

Let’s break down why this is the case. Imagine you’re trying to estimate the average height of all people in a city. You could take a random sample of people and calculate the average height of that sample. However, if you used this sample average as your estimate of the average height of everyone in the city, you’d likely underestimate the true average. This is because your sample is likely to exclude some of the tallest and shortest people in the city.

To account for this potential underestimation, we use a slightly adjusted formula for the sample variance, which divides by n-1 instead of n. This adjustment, known as Bessel’s correction, ensures that the sample variance is an unbiased estimator of the population variance.

In simpler terms, dividing by n-1 instead of n helps us to “inflate” the sample variance slightly, making it a better estimate of the population variance. This is because the sample variance is calculated using the sample mean, which itself is an estimate of the population mean. By dividing by n-1, we account for the uncertainty introduced by using a sample mean instead of the true population mean.

Let’s talk about the Mean Squared Error (MSE) and its relationship to variance. You’re right to be curious! While MSE is a crucial concept in statistics, it’s not a direct measure of variance, nor is it always an unbiased estimator of it.

Here’s the breakdown:

MSE measures the average squared difference between the estimated values and the true values. Think of it as a way to quantify how close our predictions are to reality.

Variance, on the other hand, measures how spread out the data is around its mean. It reflects the variability within a dataset.

The key difference: MSE takes into account the bias of the estimator. Bias refers to the systematic difference between the estimator’s expected value and the true value. In other words, if our estimator consistently overestimates or underestimates the true value, it has a bias.

So, why isn’t MSE always an unbiased estimator of variance?

Imagine you’re trying to estimate the average height of all students in a school. You take a sample of students and calculate the average height of your sample. Now, let’s say your sample is skewed towards taller students (maybe you sampled from the basketball team!). Your sample average will be biased upwards. This bias will be reflected in the MSE.

Here’s where it gets interesting:

* If your estimator is unbiased, meaning its expected value equals the true value, then MSE is indeed a direct measure of the estimator’s variance.
* However, if your estimator has a bias, then MSE will include the squared bias along with the variance.

Let’s illustrate with an example:

Imagine you’re trying to estimate the true mean of a population. You have two estimators:

1. Estimator 1: An unbiased estimator.
2. Estimator 2: A biased estimator.

For Estimator 1, its MSE will be equal to its variance.

For Estimator 2, its MSE will be greater than its variance because it includes the squared bias component.

In conclusion:

While MSE is closely related to variance, it’s not always an unbiased estimator of it. If your estimator is unbiased, then MSE is a direct measure of its variance. But if your estimator has a bias, then MSE will also include the squared bias, making it a more comprehensive measure of the estimator’s error.

You’re likely looking to understand the unbiased pooled estimator of variance and how it’s used in the context of a pooled t-test. Let’s break it down!

The pooled estimator is a powerful tool when you have two separate samples that you believe come from populations with the same variance. We use the information from both samples to estimate the common variance. This is especially helpful when dealing with t-tests because we can improve the accuracy of our hypothesis testing.

Here’s how the pooled estimator of variance, often represented as S p 2, is calculated:

S p 2 = ( (n – 1) S X 2 + (m – 1) S Y 2) / (m + n – 2)

S X 2 and S Y 2 represent the sample variances of the two groups, X and Y.
n and m are the sample sizes of the two groups.

Why is it called “pooled”? It’s because we’re combining the information about the variance from both samples into a single estimate. The formula essentially averages the squared deviations from the means of each sample, weighted by their respective degrees of freedom (n-1 and m-1).

Why is it “unbiased”? An unbiased estimator means that on average, the estimator will equal the true value of the population variance. This is important because we want our estimate to be as close to the true value as possible. In this case, the pooled estimator is designed to give us an unbiased estimate of the common population variance.

Let’s talk about why we’d use this in a pooled t-test: Imagine you’re studying two different types of fertilizers to see if they have a significant impact on plant growth. You’d collect data on the height of plants using each fertilizer. A pooled t-test lets you compare the average plant heights between the two groups. Since you assume the variability in plant height is the same regardless of the fertilizer used, the pooled estimator of variance is the best way to estimate that common variability.

In summary, the pooled estimator of variance is a valuable tool for combining information from two samples to estimate the common variance. This is particularly useful in the context of the pooled t-test where we’re comparing two groups that we believe share the same underlying variance.

Let’s dive into the question: Is the sample variance an unbiased estimator of the population variance?

The answer is yes! We can show this mathematically. The expected value of the sample variance, denoted as E(s²), is equal to the population variance, σ².

E(s²) = (n−1)σ² / (n−1) = σ²

This means that on average, the sample variance will equal the population variance. This is a crucial property for statistical inference, as it allows us to use the sample variance to estimate the unknown population variance.

Understanding the “i.i.d.” Assumption:

The proof that the sample variance is an unbiased estimator relies on the assumption that the data points are independent and identically distributed (i.i.d.). Let’s break down why this assumption is so critical:

Independence: This means that the value of one data point doesn’t influence the value of any other data point. Imagine you’re measuring the heights of students in a classroom. Each student’s height is independent of the heights of other students.
Identical Distribution: This means that all data points come from the same distribution. In our height example, all students are drawn from the same population distribution of heights.

If these assumptions are violated, then the sample variance might not be an unbiased estimator of the population variance. For example, if you were to measure the heights of students in a single family, the heights wouldn’t be independent (siblings tend to be similar in height).

Let’s illustrate with an example:

Suppose we have a population of 100 students with a mean height of 5’6″ and a standard deviation of 3 inches. We want to estimate the population variance using a sample of 10 students.

We collect the heights of our sample and calculate the sample variance. We repeat this process many times, drawing different samples of 10 students each time.

If the data is i.i.d., then the average of all these sample variances should be close to the true population variance (9 square inches in our example). If the data is not i.i.d., then the average sample variance may be systematically higher or lower than the true population variance.

In a nutshell, the i.i.d. assumption is essential for ensuring that the sample variance is a reliable estimator of the population variance. When we can confidently assume that our data is i.i.d., we can use the sample variance to draw accurate inferences about the underlying population.

We’ve shown that the sample variance is an unbiased estimator of the population variance. Let’s improve the answers per question metric of the site by providing a variant of FiveSigma’s answer that uses the i.i.d assumption explicitly, highlighting its necessity. We’ll prove the unbiasedness of the sample-variance estimator.

To demonstrate this, let’s delve into the core concepts. An unbiased estimator is one whose expected value equals the true value of the parameter being estimated. In our case, we want to show that the expected value of the sample variance equals the population variance.

Let’s break down the sample variance formula:

s² = ∑(xᵢ – x̄)² / (n – 1)

Where:

s² is the sample variance.
xᵢ is the i-th observation.
x̄ is the sample mean.
n is the sample size.

To prove unbiasedness, we need to calculate the expected value of s². This involves taking the expectation of the formula above.

E[s²] = E[∑(xᵢ – x̄)² / (n – 1)]

Now, we can simplify the expression by taking advantage of the i.i.d assumption. This assumption implies that each observation is independent and identically distributed, meaning they have the same distribution and are not influenced by each other.

Using the i.i.d assumption, we can rewrite the expectation as:

E[s²] = (1 / (n – 1)) * E[∑(xᵢ – x̄)²]

Expanding the summation, we get:

E[s²] = (1 / (n – 1)) * E[∑(xᵢ² – 2xᵢx̄ + x̄²)]

Since the expectation of a sum is the sum of expectations, we can separate the terms:

E[s²] = (1 / (n – 1)) * [∑E[xᵢ²] – 2∑E[xᵢx̄] + ∑E[x̄²]]

Now, let’s focus on each term individually.

E[xᵢ²]: This is simply the second moment of the distribution, which is the variance plus the square of the mean.

E[xᵢx̄]: We can expand this as E[xᵢ(∑xⱼ / n)] and use linearity of expectation to get E[xᵢ² / n] + E[xᵢxⱼ / n] (where i ≠ j). Since xᵢ and xⱼ are independent, E[xᵢxⱼ] = E[xᵢ]E[xⱼ].

E[x̄²]: We can express this as E[(∑xᵢ / n)²] and again use linearity of expectation. This results in a term with E[xᵢ² / n²] and terms with E[xᵢxⱼ / n²] for i ≠ j.

After carefully substituting these expectations and simplifying the expression, we finally arrive at:

E[s²] = σ²

where σ² is the population variance.

Therefore, we’ve proven that the expected value of the sample variance equals the population variance, demonstrating that the sample variance is an unbiased estimator of the population variance.

This proof hinges on the i.i.d assumption. Without this assumption, the independence and identical distribution of observations would be violated, leading to a biased estimator.

You’re right, the pooled variance is indeed a weighted average. It’s a combination of the two individual variances, S²c and S²t, from two separate samples.

The key here is that it’s a *weighted* average. The weights used aren’t arbitrary; they’re determined by the degrees of freedom associated with each sample. This means that the sample with more degrees of freedom contributes more to the pooled variance.

Think of it this way: a sample with more data points provides a more reliable estimate of the population variance. So, we give it more “weight” in the calculation.

Let’s break down those weights.

Understanding the Weights

The weights for the pooled variance are directly tied to the degrees of freedom. Let’s say we have a sample with *n* data points. The degrees of freedom for that sample would be *n-1*.

Here’s how the weights are calculated:

– Weight for S²c: (n_c – 1) / (n_c + n_t – 2)
– Weight for S²t: (n_t – 1) / (n_c + n_t – 2)

Notice that the denominator is the same for both weights. This ensures that the weights always add up to 1, which is a property of any weighted average.

Why are the Weights Related to Degrees of Freedom?

Remember that the degrees of freedom represent the number of independent pieces of information in a sample. The more degrees of freedom a sample has, the more information it provides about the population variance. So, it makes sense to give more weight to the sample with more degrees of freedom.

The Pooled Variance as an Unbiased Estimator

The reason for using this specific weighted average is to ensure that the pooled variance is an unbiased estimator of the common population variance. An unbiased estimator means that, on average, the pooled variance will equal the true population variance. This is crucial because we often use the pooled variance to test hypotheses about the difference in means between two populations.

In a Nutshell

The pooled variance is a powerful tool for combining information from multiple samples. By using the degrees of freedom as weights, we ensure that the pooled variance is an accurate and unbiased estimate of the underlying population variance. This makes it a reliable tool for statistical analysis and inference.

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