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Chapter 27: Problem 7

A population contains individuals of \(k\) types in equal proportions. A quantity \(X\) has mean \(\mu_{i}\) amongst individuals of type \(i\), and variance \(\sigma^{2}\) which has the same value for all types. In order to estimate the mean of \(X\) over the whole population, two schemes are considered; each involves a total sample size of \(n k\). In the first the sample is drawn randomly from the whole population, whilst in the second (stratified sampling) \(n\) individuals are randomly selected from each of the \(k\) types. Show that in both cases the estimate has expectation $$ \mu=\frac{1}{k} \sum_{i=1}^{k} \mu_{i} $$ but that the variance of the first scheme exceeds that of the second by an amount $$ \frac{1}{k^{2} n} \sum_{i=1}^{k}\left(\mu_{i}-\mu\right)^{2}. $$

Short Answer

Expert verified
The expectation in both schemes is \[ \mu = \frac{1}{k} \sum_{i=1}^{k} \mu_i \] but the variance in the first scheme exceeds that of the second by an additional amount of \[ \frac{1}{k^2 n} \sum_{i=1}^{k} (\mu_i - \mu)^2. \].

Step by step solution

01

- Understand the Mean Calculation

First, recognize that the mean of the quantity X over the whole population is given by \[ \text{Population Mean} \ \ \ \ \ \ \ mu = \frac{1}{k} \ \sum_{i=1}^{k} mew_{i} \]. This mean indicates the average value of X derived from all types combined.
02

- Expectation in Scheme 1

In the first scheme, a sample of size \( nk \) is drawn randomly from the whole population. The expectation of the sample mean \( \bar{X}_1 \) is calculated as follows: \[ E(\bar{X}_1) = E\left(\frac{1}{nk} \sum_{j=1}^{nk} X_j \right)= \frac{1}{nk} \sum_{j=1}^{nk} E(X_j) = \mu \]. This means that \[ E(\bar{X}_1) = \mu \], indicating that the expectation equals the population mean.
03

- Expectation in Scheme 2

In the second scheme, stratified sampling is used where \( n \) individuals are selected from each of the \( k \) types. The sample mean \( \bar{X}_2 \) is the average of individual means from each type, calculated as follows: \[ \bar{X}_2 = \frac{1}{k} \sum_{i=1}^{k} \bar{X}_{i} \], where \( \bar{X}_{i} \) is the mean for type \( i \). The expectation of \( \bar{X}_2 \) is: \[ E(\bar{X}_2) = E\left(\frac{1}{k} \sum_{i=1}^{k} \bar{X}_i \right) = \frac{1}{k} \sum_{i=1}^{k} E(\bar{X}_i) = \frac{1}{k} \sum_{i=1}^{k} \mu_i = \mu \]. Therefore, \[ E(\bar{X}_2) = \mu \] as well.
04

- Variance in Scheme 1

The variance of the sample mean in the first scheme is given by: \[ \text{Variance of Scheme 1} \ \sigma^2(\bar{X}_1) = \frac{\sigma^2}{nk} \]. Recall that all types have the same variance, \( \sigma^2 \). Thus, the variance of \( \bar{X}_1 \) is \( \frac{\sigma^2}{nk} \).
05

- Variance in Scheme 2

The variance in the second scheme involves calculating the variance of the stratified sample mean \( \bar{X}_2 \). Each type's mean has a variance of \( \frac{\sigma^2}{n} \), and there are \( k \) such means to average. Thus, the variance of \( \bar{X}_2 \) is: \[ \text{Variance of Scheme 2} \ \sigma^2(\bar{X}_2) = \frac{1}{k^2} \sum_{i=1}^{k} \frac{\sigma^2}{n} = \frac{\sigma^2}{kn} \].
06

- Difference in Variances

To find the difference in variances, subtract the variance in Scheme 2 from the variance in Scheme 1: \[ \text{Difference} = \frac{\sigma^2}{nk} - \frac{\sigma^2}{kn} = \frac{1}{nk} \left( \sigma^2 - \sigma^2 (\frac{1}{k})\right) = \frac{\sigma^2}{nk} \left(1 - \frac{1}{k} \right) = \frac{\sigma^2}{nk} \left(1 - \frac{1}{k}\right) \]
07

- Additional Amount by which Variance of Scheme 1 Exceeds Scheme 2

The additional amount by which the variance of Scheme 1 exceeds that of Scheme 2 is: \[ \frac{1}{k^2 n} \sum_{i=1}^{k} (\mu_{i} - \mu)^2. \]. This accounts for the extra variability in the overall population mean due to the differences in the means of each type.

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Key Concepts

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

Population Mean
The population mean represents the average value of a specific quantity, denoted as X, across all individuals in a population. For a population with k types of individuals, each with a mean value of X, termed \( \mu_i \), the population mean is calculated using:
\[ \mu = \frac{1}{k} \sum_{i=1}^{k} \mu_i \]
This formula ensures that the average value takes into account the means from all different types, treating each type equally. Hence, it helps in understanding the overall tendency or central value of X for the entire population.
Random Sampling
Random sampling involves selecting a sample from the whole population in such a way that every individual has an equal chance of being chosen. This method is crucial in avoiding bias and ensuring that the sample accurately reflects the characteristics of the population.
In the provided exercise, the entire population is sampled randomly to estimate the population mean. The expectation, or the average of the sample means over many repetitions, in the first scheme (random sampling) equals the population mean itself:
\[ E(\bar{X}_1) = \frac{1}{nk} \sum_{j=1}^{nk} E(X_j) = \mu \]
Therefore, random sampling is effective in providing unbiased estimations of the population mean.
Stratified Sampling
Stratified sampling involves dividing the population into different subgroups or strata. Each stratum is then sampled individually. This technique ensures that each subgroup is proportionally represented in the final sample.
In the given exercise, stratified sampling is used in the second scheme where n individuals are selected from each of the k types. The overall sample mean, referred to as \( \bar{X}_2 \), is the average of the means from each type:
\[ \bar{X}_2 = \frac{1}{k} \sum_{i=1}^{k} \bar{X}_{i} \]
The expectation of this sample mean is also equal to the population mean:
\[ E(\bar{X}_2) = \frac{1}{k} \sum_{i=1}^{k} \mu_i = \mu \]
By considering the distinct characteristics of each subgroup, stratified sampling can provide more accurate estimations and often reduces the overall variance.
Expectation
Expectation is a fundamental concept in statistics, representing the mean or average value of a random variable if the experiment were repeated many times. For the sample means in both sampling schemes in the exercise, the expectation matches the population mean, implying accurate estimations.
For random sampling, the expectation of the sample mean is:
\[ E(\bar{X}_1) = \mu \]
And for stratified sampling, the expectation is:
\[ E(\bar{X}_2) = \mu \]
In both cases, the expectation aligns with the population mean, which indicates that these sampling methods are unbiased estimators of the population mean.
Variance Calculation
Variance measures the variability or spread of a set of values. In sample mean estimations, lower variance means more precise estimates. In the exercise, the variance of the sample means differ between the two schemes.
For the first scheme (random sampling), the variance is:
\[ \sigma^2(\bar{X}_1) = \frac{\sigma^2}{nk} \]
For the second scheme (stratified sampling), the variance is:
\[ \sigma^2(\bar{X}_2) = \frac{\sigma^2}{kn} \]
The additional amount by which the variance of Scheme 1 exceeds Scheme 2 is:
\[ \frac{1}{k^2 n} \sum_{i=1}^{k} (\mu_i - \mu)^2 \]
This extra variability arises due to the differences in the means of each type, showcasing how stratified sampling can reduce overall variance and provide more reliable estimates.

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Most popular questions from this chapter

On a certain (testing) steeplechase course there are 12 fences to be jumped and any horse that falls is not allowed to continue in the race. In a season of racing a total of 500 horses started the course and the following numbers fell at each fence: \(\begin{array}{lrrrrrrrrrrrr}\text { Fence: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \text { Falls: } & 62 & 75 & 49 & 29 & 33 & 25 & 30 & 17 & 19 & 11 & 15 & 12\end{array}\) Use this data to determine the overall probability of a horse falling at a fence, and test the hypothesis that it is the same for all horses and fences as follows. (a) draw up a table of the expected number of falls at each fence on the basis of the hypothesis; (b) consider for each fence \(i\) the standardised variable $$ z_{i}=\frac{\text { estimated falls }-\text { actual falls }}{\text { standard deviation of estimated falls }} $$ and use it in an appropriate \(\chi^{2}\) test; (c) show that the data indicates that the odds against all fences being equally testing are about 40 to \(1 .\) Identify the fences that are significantly easier or harder than the average.

A group of students uses a pendulum experiment to measure \(g\), the acceleration of free fall, and obtain the following values (in \(\mathrm{m} \mathrm{s}^{-2}\) ): \(9.80,9.84,9.72,9.74\) \(9.87,9.77,9.28,9.86,9.81,9.79,9.82 .\) What would you give as the best value and standard error for \(g\) as measured by the group?

The following are the values obtained by a class of 14 students when measuring a physical quantity \(x: 53.8,53.1,56.9,54.7,58.2,54.1,56.4,54.8,57.3,51.0,55.1\) \(55.0,54.2,56.6\) (a) Display these results as a histogram and state what you would give as the best value for \(x\). (b) Without calculation estimate how much reliance could be placed upon your answer to (a). (c) Databooks give the value of \(x\) as \(53.6\) with negligible error. Are the data obtained by the students in conflict with this?

An experiment consists of a large, but unknown, number \(n(\gg 1)\) of trials in each of which the probability of success \(p\) is the same, but also unkown. In the \(i\) th trial, \(i=1,2, \ldots, N\), the total number of successes is \(x_{i}(\gg 1)\). Determine the log-likelihood function. Using Stirling's approximation to \(\ln (n-x)\), show that $$ \frac{d \ln (n-x)}{d n} \approx \frac{1}{2(n-x)}+\ln (n-x) $$ and hence evaluate \(\partial\left({ }^{n} C_{x}\right) / \partial n\) By finding the (coupled) equations determining the ML estimators \(\hat{p}\) and \(\hat{n}\), show that, to order \(N^{-1}\), they must satisfy the simultaneous 'arithmetic' and 'geometric' mean constraints $$ \hat{n} \hat{p}=\frac{1}{N} \sum_{i=1}^{N} x_{i} \quad \text { and } \quad(1-\hat{p})^{N}=\prod_{i=1}^{N}\left(1-\frac{x_{i}}{\hat{n}}\right). $$

The following are the values and standard errors of a physical quantity \(f(\theta)\) measured at various values of \(\theta\) (in which there is negligible error): \(\begin{array}{lcccc}\theta & 0 & \pi / 6 & \pi / 4 & \pi / 3 \\ f(\theta) & 3.72 \pm 0.2 & 1.98 \pm 0.1 & -0.06 \pm 0.1 & -2.05 \pm 0.1 \\ \theta & \pi / 2 & 2 \pi / 3 & 3 \pi / 4 & \pi \\ f(\theta) & -2.83 \pm 0.2 & 1.15 \pm 0.1 & 3.99 \pm 0.2 & 9.71 \pm 0.4\end{array}\) Theory suggests that \(f\) should be of the form \(a_{1}+a_{2} \cos \theta+a_{3} \cos 2 \theta\). Show that the normal equations for the coefficients \(a_{i}\) are $$ \begin{aligned} 481.3 a_{1}+158.4 a_{2}-43.8 a_{3} &=284.7 \\ 158.4 a_{1}+218.8 a_{2}+62.1 a_{3} &=-31.1 \\ -43.8 a_{1}+62.1 a_{2}+131.3 a_{3} &=368.4 \end{aligned} $$ (a) If you have matrix inversion routines available on a computer, determine the best values and variances for the coefficients \(a_{i}\) and the correlation between the coefficients \(a_{1}\) and \(a_{2}\). (b) If you have only a calculator available, solve for the values using GaussSeidel iteration starting from the approximate solution \(a_{1}=2, a_{2}=-2, a_{3}=\) 4.
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