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Suppose a certain type of fertilizer has an expected yield per acre of \(\mu_{1}\) with variance \(\sigma^{2}\), whereas the expected yield for a second type of fertilizer is \(\mu_{2}\) with the same variance \(\sigma^{2}\). Let \(S_{1}^{2}\) and \(S_{2}^{2}\) denote the sample variances of yields based on sample sizes \(n_{1}\) and \(n_{2}\), respectively, of the two fertilizers. Show that the pooled (combined) estimator $$ \hat{\sigma}^{2}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2}}{n_{1}+n_{2}-2} $$ is an unbiased estimator of \(\sigma^{2}\).

Short Answer

Expert verified
The pooled estimator \( \hat{\sigma}^2 \) is unbiased because its expected value equals \( \sigma^2 \).

Step by step solution

01

Understanding Unbiased Estimator

An estimator \( \hat{\theta} \) is considered unbiased for a parameter \( \theta \) if the expected value of the estimator is exactly the parameter itself. That means \( E(\hat{\theta}) = \theta \). For this problem, we need to show that \( E(\hat{\sigma}^2) = \sigma^2 \).
02

Writing the Expression for Pooled Estimator

We have the pooled estimator for variance given by \( \hat{\sigma}^{2} = \frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2} \). This expression is a weighted average of the sample variances from the two samples.
03

Using Properties of Expectation

By properties of expectation and the fact that sample variance \( S_i^2 \) is an unbiased estimator of population variance \( \sigma^2 \), we have \( E(S_1^2) = \sigma^2 \) and \( E(S_2^2) = \sigma^2 \). We will apply the expectation to the pooled estimator.
04

Applying Expectation to the Pooled Estimator

The expectation of the pooled estimator is: \[ E(\hat{\sigma}^{2}) = E\left(\frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}\right) \] The linearity of expectation allows us to separate this into: \[ \frac{(n_1 - 1)E(S_1^2) + (n_2 - 1)E(S_2^2)}{n_1 + n_2 - 2} \]
05

Calculating Expectation

Substitute \( E(S_1^2) = \sigma^2 \) and \( E(S_2^2) = \sigma^2 \) into the expression: \[ E(\hat{\sigma}^{2}) = \frac{(n_1 - 1)\sigma^2 + (n_2 - 1)\sigma^2}{n_1 + n_2 - 2} \] Factor out \( \sigma^2 \): \[ E(\hat{\sigma}^{2}) = \sigma^2 \cdot \frac{(n_1 - 1) + (n_2 - 1)}{n_1 + n_2 - 2} \]
06

Simplifying the Expression

The numerator \( (n_1 - 1) + (n_2 - 1) \) simplifies to \( (n_1 + n_2 - 2) \), so: \[ E(\hat{\sigma}^{2}) = \sigma^2 \] Thus, \( \hat{\sigma}^2 \) is an unbiased estimator of \( \sigma^2 \).

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Key Concepts

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

Pooled Variance Estimator
In statistics, the pooled variance estimator is a method used to estimate the common variance of two or more different populations when we assume that these populations have the same variance. This method is particularly useful when comparing groups to determine if they come from populations with similar variance.Think of the pooled variance as a way to combine the variances from two different samples into one estimate. Suppose we have two types of fertilizers, and each type yields some variance when applied across several acres. The pooled variance takes these separate variances and fuses them into one measure, providing a collective idea of the variability across both fertilizers.The formula for the pooled variance estimator is given by: \[ \hat{\sigma}^{2} = \frac{(n_1 - 1) S_1^2 + (n_2 - 1) S_2^2}{n_1 + n_2 - 2} \] Here, \(n_1\) and \(n_2\) represent the sample sizes for each fertilizer, while \(S_1^2\) and \(S_2^2\) denote their respective sample variances.
  • Pooled variance provides a more stable estimate compared to using individual sample variances.
  • It is commonly used in the analysis of variance tests where variance equality is assumed.
Sample Variance
Sample variance is a measure that describes the spread of data points in a sample. It's essentially a way of quantifying how much individual data points differ from the sample mean. This concept is crucial in statistics as it gives insight into the variability within a dataset.When you collect data, you're often interested in understanding not just the average outcome (like the average yield of a fertilizer) but how much deviation or variability exists around that average. Calculating the sample variance involves the following steps:1. Compute the mean of the sample.2. Subtract the mean from each data point to find the deviation of each.3. Square each deviation.4. Sum these squared deviations.5. Divide by the sample size minus one (\(n - 1\)).The formula for sample variance is:\[ S^2 = \frac{\sum{(x_i - \bar{x})^2}}{n - 1} \] where \(x_i\) represents each data point, and \(\bar{x}\) is the mean of the sample.
  • Sample variance helps in assessing the reliability of the sample mean.
  • It is an unbiased estimator of the population variance, assuming that the data is drawn independently and identically distributed.
Mathematical Expectation
Mathematical expectation, or expected value, is a fundamental concept in probability and statistics. It represents the average or mean value of a random variable if an experiment or process were repeated infinitely many times.The expected value is calculated by weighting each possible outcome by its probability and summing these values. For a random variable \(X\) that can take the values \(x_1, x_2, \ldots, x_n\) with probabilities \(p_1, p_2, \ldots, p_n\), the expectation is given by:\[ E(X) = \sum_{i=1}^{n} x_i \, p_i \]In the context of our problem, the expectation of the sample variances \(S_1^2\) and \(S_2^2\) is crucial because it tells us that these sample variances are unbiased estimators of the population variance \(\sigma^2\). This means:
  • The expected value of \(S_i^2\) is \(\sigma^2\), i.e., \(E(S_i^2) = \sigma^2\).
  • It ensures that on average, these estimators hit the true population value.
  • This property is essential when establishing that the pooled variance estimator is itself unbiased.
Parameter Estimation
Parameter estimation refers to the process of using sample data to estimate the parameters of a probability distribution or a model. In simpler terms, it's about using data to make educated guesses about key characteristics of a whole population.In our fertilizer yield example, we are interested in estimating a common population variance \(\sigma^2\) from sample variances \(S_1^2\) and \(S_2^2\). The success of parameter estimation relies on properties like unbiasedness, which indicates whether the estimator gives, on average, the true parameter value.Key aspects of parameter estimation include:
  • Unbiasedness: An estimator is unbiased if its expected value equals the parameter being estimated. This applies to both sample variances \(S_i^2\) and the pooled variance estimator \(\hat{\sigma}^2\).
  • Efficiency: Among unbiased estimators, an efficient one has the smallest variance.
  • Consistency: As the sample size increases, a consistent estimator approaches the true value of the parameter.
These aspects ensure that the pooled variance, when used as an estimator, yields accurate and reliable results when interpreting the data from agricultural experiments or other fields of research.

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Most popular questions from this chapter

A sample of \(n\) captured Pandemonium jet fighters results in serial numbers \(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\). The CIA knows that the aircraft were numbered consecutively at the factory starting with \(\alpha\) and ending with \(\beta\), so that the total number of planes manufactured is \(\beta-\alpha+1\) (e.g., if \(\alpha=17\) and \(\beta=29\), then \(29-17+1=13\) planes having serial numbers 17 , \(18,19, \ldots, 28,29\) were manufactured). However, the CIA does not know the values of \(\alpha\) or \(\beta\). A CIA statistician suggests using the estimator \(\max \left(X_{j}\right)-\min \left(X_{j}\right)+1\) to estimate the total number of planes manufactured. a. If \(n=5, x_{1}=237, x_{2}=375, x_{3}=202, x_{4}=525\), and \(x_{5}=418\). what is the corresponding estimate? b. Under what conditions on the sample will the value of the estimate be exactly equal to the true total number of planes? Will the estimate ever be larger than the true total? Do you think the estimator is unbiased for estimating \(\beta-\alpha+1\) ? Explain in one or two sentences.

The accompanying data on flexural strength (MPa) for concrete beams of a certain type was introduced in Example 1.2. \(\begin{array}{rrrrrrr}5.9 & 7.2 & 7.3 & 6.3 & 8.1 & 6.8 & 7.0 \\ 7.6 & 6.8 & 6.5 & 7.0 & 6.3 & 7.9 & 9.0 \\ 8.2 & 8.7 & 7.8 & 9.7 & 7.4 & 7.7 & 9.7 \\\ 7.8 & 7.7 & 11.6 & 11.3 & 11.8 & 10.7 & \end{array}\) a. Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion, and state which estimator you used. b. Calculate a point estimate of the strength value that separates the weakest \(50 \%\) of all such beams from the strongest \(50 \%\), and state which estimator you used. c. Calculate and interpret a point estimate of the population standard deviation \(\sigma\). Which estimator did you use? [Hint: \(\left.\sum x_{i}^{2}=1860.94 .\right]\) d. Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds \(10 \mathrm{MPa}\). [Hint: Think of an observation as a "success" if it exceeds 10.] e. Calculate a point estimate of the population coefficient of variation \(\sigma / \mu\), and state which estimator you used.

In a random sample of 80 components of a certain type, 12 are found to be defective. a. Give a point estimate of the proportion of all such components that are not defective. b. A system is to be constructed by randomly selecting two of these components and connecting them in series, as shown here. The series connection implies that the system will function if and only if neither component is defective (i.e., both components work properly). Estimate the proportion of all such systems that work properly. [Hint: If \(p\) denotes the probability that a component works properly, how can \(P\) (system works) be expressed in terms of \(p\) ?]

Consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from the shifted exponential pdf $$ f(x ; \lambda, \theta)=\left\\{\begin{array}{cl} \lambda e^{-\lambda(x-\theta)} & x \geq \theta \\ 0 & \text { otherwise } \end{array}\right. $$ Taking \(\theta=0\) gives the pdf of the exponential distribution considered previously (with positive density to the right of zero). An example of the shifted exponential distribution appeared in Example \(4.5\), in which the variable of interest was time headway in traffic flow and \(\theta=.5\) was the minimum possible time headway. a. Obtain the maximum likelihood estimators of \(\theta\) and \(\lambda\). b. If \(n=10\) time headway observations are made, resulting in the values \(3.11, .64,2.55,2.20,5.44,3.42,10.39,8.93\), \(17.82\), and \(1.30\), calculate the estimates of \(\theta\) and \(\lambda\).

Each of 150 newly manufactured items is examined and the number of scratches per item is recorded (the items are supposed to be free of scratches), yielding the following data: Number of scratches \begin{tabular}{lllllllll} per item & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline Observed frequency & 18 & 37 & 42 & 30 & 13 & 7 & 2 & 1 \end{tabular} Let \(X=\) the number of scratches on a randomly chosen item, and assume that \(X\) has a Poisson distribution with parameter \(\lambda\). a. Find an unbiased estimator of \(\lambda\) and compute the estimate for the data. [Hint: \(E(X)=\lambda\) for \(X\) Poisson, so \(E(\bar{X})=\) ?] b. What is the standard deviation (standard error) of your estimator? Compute the estimated standard error. [Hint: \(\sigma_{X}^{2}=\lambda\) for \(X\) Poisson.]

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