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Chapter 7: Problem 16

\( \bar{X}_{1}\) and \(S_{1}^{2}\) are the sample mean and sample variance from a population with mean \(\mu_{1}\) and variance \(\sigma_{1}^{2}\). Similarly, \(\bar{X}_{2}\) and \(S_{2}^{2}\) are the sample mean and sample variance from a second independent population with mean \(\mu_{2}\) and variance \(\sigma_{2}^{2}\). The sample sizes are \(n_{1}\) and \(n_{2},\) respectively. (a) Show that \(\bar{X}_{1}-\bar{X}_{2}\) is an unbiased estimator of \(\mu_{1}-\mu_{2}\). (b) Find the standard error of \(\bar{X}_{1}-\bar{X}_{2}\). How could you estimate the standard error? (c) Suppose that both populations have the same variance; that is, \(\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma^{2}\). Show that $$ S_{p}^{2}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2}}{n_{1}+n_{2}-2} $$ is an unbiased estimator of \(\sigma^{2}\).

Short Answer

Expert verified
(a) Yes, it's unbiased. (b) Standard error: \( \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \). (c) Yes, \( S_p^2 \) is unbiased for \( \sigma^2 \).

Step by step solution

01

Define Sample Mean Properties

The sample mean \( \bar{X}_1 \) is calculated from the first population with true mean \( \mu_1 \) and sample size \( n_1 \). Therefore, \( E(\bar{X}_1) = \mu_1 \). Similarly, for the second population, \( E(\bar{X}_2) = \mu_2 \). This follows because the expected value of a sample mean is equal to the population mean by definition.
02

Prove Unbiased Estimator Property for Part (a)

To show that \( \bar{X}_1 - \bar{X}_2 \) is an unbiased estimator for \( \mu_1 - \mu_2 \), we find:\[ E(\bar{X}_1 - \bar{X}_2) = E(\bar{X}_1) - E(\bar{X}_2) = \mu_1 - \mu_2. \]Since the expected value of the difference is equal to the difference of the population means, the estimator is unbiased.
03

Find Standard Error for Part (b)

The standard error of the difference in sample means is given by:\[ SE(\bar{X}_1 - \bar{X}_2) = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}. \]
04

Estimate Standard Error

The standard error can be estimated using the sample variances as:\[ SE_{\text{estimated}} = \sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}. \]Substituting sample variances in place of population variances provides the estimate of standard error.
05

Show Pooled Variance Unbiased for Part (c)

Given the assumption that both populations have the same variance \( \sigma_1^2 = \sigma_2^2 = \sigma^2 \), the pooled variance estimator \( S_p^2 \) is given by:\[ S_p^2 = \frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}. \]We need to show \( E(S_p^2) = \sigma^2 \). Since \( S_1^2 \) and \( S_2^2 \) are unbiased estimators of \( \sigma_1^2 \) and \( \sigma_2^2 \) respectively, \( E(S_1^2) = \sigma^2 \) and \( E(S_2^2) = \sigma^2 \). Thus,\[ E(S_p^2) = \frac{(n_1 - 1)\sigma^2 + (n_2 - 1)\sigma^2}{n_1 + n_2 - 2} = \sigma^2. \]Therefore, \( S_p^2 \) 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.

Unbiased Estimator
In statistics, an unbiased estimator is a statistical tool or formula that provides estimates that, on average, are equal to the actual parameter of a population. When an estimator is unbiased, it means that the expected value of the estimator is equal to the true value of the parameter being estimated.

For example, in the original exercise, the expression \( \bar{X}_1 - \bar{X}_2 \) is presented as an unbiased estimator for \( \mu_1 - \mu_2 \). This is concluded by calculating the expectation, \( E(\bar{X}_1 - \bar{X}_2) \), which simplifies to \( \mu_1 - \mu_2 \), matching the true value of the difference of the population means. This shows no systemic error, as the mean of the estimator equals the parameter it aims to measure.

Unbiasedness is essential in statistical analysis because it ensures that we are not consistently overestimating or underestimating the parameter of interest. This quality makes the estimator reliable over many samples, giving confidence in its accuracy for inference.
Sample Mean
The sample mean is a key concept in statistics, utilized to estimate the central tendency of a sample taken from a population. It is calculated by summing all the observed values in the sample and then dividing by the number of observations. The formula for the sample mean \( \bar{X} \) is: \[ \bar{X} = \frac{X_1 + X_2 + \ldots + X_n}{n} \] where \( X_1, X_2, \ldots, X_n \) are the sample observations and \( n \) is the sample size.

The sample mean is an unbiased estimator of the population mean \( \mu \). This means that the expected value of the sample mean, \( E(\bar{X}) \), is equal to the population mean. As such, the sample mean provides a good approximation of \( \mu \), especially when the sample size is large.
  • It is simple to compute and interpret.
  • It is sensitive to extreme values or outliers, which can affect the mean greatly.
  • It's incredibly useful when applied to infer and generalize about the population mean.
Through the sample mean, one can gain initial insight into the distribution and characteristics of a larger population.
Sample Variance
Sample variance quantifies the variability or spread of the sample data around the sample mean. It is calculated as the average of the squared differences from the mean, providing a measure of how spread out the data points are. The formula for the sample variance \( S^2 \) is: \[ S^2 = \frac{\sum_{i=1}^n (X_i - \bar{X})^2}{n-1} \] where \( X_i \) represents each data point, \( \bar{X} \) is the sample mean, and \( n \) denotes the sample size.

This variance is considered an unbiased estimator of the population variance when dividing by \( n-1 \) instead of \( n \). This correction is known as Bessel's correction, and it helps adjust the bias in the estimation of the population variance caused by the fact we are using the sample mean rather than the real population mean.
  • Sample variance is vital in assessing the reliability and precision of the sample mean.
  • It plays essential roles in further statistical calculations, like the standard deviation and variance of combined samples.
  • It forms the basis for further statistical analyses, such as hypothesis testing and confidence intervals.
Understanding sample variance gives a deeper insight into the structure and behavior of the data being analyzed, facilitating better data-driven decisions.

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Most popular questions from this chapter

Consider the probability density function: $$ f(x)=c(1+\theta x), \quad-1 \leq x \leq 1 $$ (a) Find the value of the constant \(c\). (b) What is the moment estimator for \(\theta\) ? (c) Show that \(\hat{\theta}=3 \bar{X}\) is an unbiased estimator for \(\theta .\) (d) Find the maximum likelihood estimator for \(\theta\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample of size \(n\) from a population with mean \(\mu\) and variance \(\sigma^{2}\) (a) Show that \(\bar{X}^{2}\) is a biased estimator for \(\mu^{2}\) (b) Find the amount of bias in this estimator. (c) What happens to the bias as the sample size \(n\) increases?

PVC pipe is manufactured with a mean diameter of 1.01 inch and a standard deviation of 0.003 inch. Find the probability that a random sample of \(n=9\) sections of pipe will have a sample mean diameter greater than 1.009 inch and less than 1.012 inch.

Researchers in the Hopkins Forest (see Exercise 7.16 ) also count the number of maple trees (genus acer) in plots throughout the forest. The following is a histogram of the number of live maples in 1002 plots sampled over the past 20 years. The average number of maples per plot was 19.86 trees with a standard deviation of 23.65 trees. (a) If we took the mean of a sample of eight plots, what would be the standard error of the mean? (b) Using the central limit theorem, what is the probability that the mean of the eight would be within 1 standard error of the mean? (c) Why might you think that the probability that you calculated in (b) might not be very accurate?

A lot consists of \(N\) transistors. and of these \(M(M \leq N)\) are defective. We randomly select two transistors without replacement from this lot and determine whether they are defective or non defective. The random variable $$ X_{i}=\left\\{\begin{array}{ll} 1, & \text { if the } i \text { th transistor } \\ & \text { is nondefective } \\ 0, & \text { if the ith transistor } \\\ & \text { is defective } \end{array} \quad i=1,2\right. $$ Determine the joint probability function for \(X_{1}\) and \(X_{2}\). What are the marginal probability functions for \(X_{1}\) and \(X_{2}\) ? Are \(X_{1}\) and \(X\), independent random variables?
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