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The designer of a sample survey stratifies a population into two strata, H and L. H contains 100,000 people, and L contains \(500,000 .\) He decides to allocate 100 samples to stratum \(\mathrm{H}\) and 200 to stratum \(\mathrm{L},\) taking a simple random sample in each stratum. a. How should the designer estimate the population mean? b. Suppose that the population standard deviation in stratum \(\mathrm{H}\) is 20 and the standard deviation in stratum L is 10. What will be the standard error of his estimate? c. Would it be better to allocate 200 samples to stratum \(\mathrm{H}\) and 100 to stratum \(\mathrm{L} ?\) d. Would it be better to use proportional allocation?

Short Answer

Expert verified
Weighted mean estimates the population mean; standard error depends on sample allocation where the optimal allocation minimizes the error. Compare results for each method to determine the best allocation.

Step by step solution

01

Understanding Population Composition

The total population is stratified into two strata: H and L. Stratum H has 100,000 people and stratum L has 500,000 people. The sample size is 300, split between the strata with 100 samples for H and 200 for L.
02

Estimating the Population Mean

To estimate the population mean using stratified random sampling, use: \[ \bar{X} = \frac{N_H}{N} \bar{X}_H + \frac{N_L}{N} \bar{X}_L \] where \( \bar{X}_H \) and \( \bar{X}_L \) are the mean estimates of strata H and L respectively, and \( N_H \) and \( N_L \) are the number of individuals in strata H and L.
03

Calculating Standard Error with Allocated Samples

The standard error for a stratified sample mean is given by: \[ SE = \sqrt{\left(\frac{N_H^2}{N^2}\right)\frac{S_H^2}{n_H} + \left(\frac{N_L^2}{N^2}\right)\frac{S_L^2}{n_L}} \] with \( S_H = 20 \), \( S_L = 10 \), \( n_H = 100 \), and \( n_L = 200 \). Calculate \( SE = \sqrt{\left(\frac{100000^2}{600000^2}\right)\frac{20^2}{100} + \left(\frac{500000^2}{600000^2}\right)\frac{10^2}{200}} \).
04

Evaluating Alternative Samples Allocation

If we allocate 200 samples to H and 100 to L and calculate the standard error similarly, we can compare this to the standard error from Step 3 to check if this reduces the standard error.
05

Considering Proportional Allocation

Proportional allocation involves setting the sample sizes proportional to the strata sizes, meaning each stratum gets allocated samples in proportion to its size. This means \( n_H = \frac{100000}{600000} \cdot 300 = 50 \) and \( n_L = \frac{500000}{600000} \cdot 300 = 250 \). Calculate the standard error with these values and compare it to previous configurations.

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Key Concepts

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

Population Mean Estimation
Population mean estimation is essential when using stratified sampling in surveys. In this context, the process involves breaking down the entire population into separate strata, collecting samples from those strata, and estimating the overall mean based on these. For the given problem, we know there are two strata: H and L. Each stratum contributes to the overall population mean weighted by its size.

To calculate the population mean, the formula is:
  • \[ \bar{X} = \frac{N_H}{N} \bar{X}_H + \frac{N_L}{N} \bar{X}_L \]
This means multiplying the mean of each stratum by the proportion its population size contributes to the total, and summing these values. Here, \( N_H \) and \( N_L \) are the sizes of strata H and L, respectively, while \( \bar{X}_H \) and \( \bar{X}_L \) are their respective mean estimates. Using this approach ensures that the population mean is accurately estimated, reflecting the distribution and contribution of each stratum.

Understanding this method is vital for anyone conducting surveys, as it ensures that each subgroup is properly represented in statistical models and outcomes.
Standard Error Calculation
Calculating the standard error of the population mean in stratified sampling is crucial for understanding the estimate's precision. The standard error provides a measure of how much sampling variability is expected in the estimation procedure.

The formula used is:
  • \[ SE = \sqrt{\left(\frac{N_H^2}{N^2}\right)\frac{S_H^2}{n_H} + \left(\frac{N_L^2}{N^2}\right)\frac{S_L^2}{n_L}} \]
Here, \( S_H \) and \( S_L \) represent the standard deviations for strata H and L, and \( n_H \) and \( n_L \) are the sample sizes allocated to these strata. This expression combines the variances of each stratum, weighted by their respective sizes, scaled by the inverse of their sample sizes.

When you apply this formula, you get the standard error, which helps you understand the reliability of the population mean estimate. A smaller standard error indicates a more precise estimate, crucial for making informed decisions based on sample data.
Proportional Allocation
Proportional allocation is a technique that assigns samples in a survey in proportion to the strata size, ensuring that each subgroup is equally represented. This method helps to minimize bias and potentially reduce sampling error.

In our scenario, the proportional allocation would involve allocating samples based on the size of strata H and L out of the total population. This is done using:
  • \( n_H = \frac{100000}{600000} \times 300 = 50 \)
  • \( n_L = \frac{500000}{600000} \times 300 = 250 \)
By assigning 50 samples to stratum H and 250 to stratum L, each stratum's relative size is acknowledged, and thus properly represented in the sample. This method is often preferred as it ensures that the sample reflects the diversity within the population, which can improve the accuracy and reliability of the population mean estimate.

Implementing proportional allocation leads to more representative survey data and plays a crucial role in the analysis of stratified samples.
Sample Allocation in Surveys
Sample allocation in surveys is the process of deciding how to distribute the total number of samples across different strata in a population. This is a critical decision that impacts the survey's accuracy and standard error.

In the context of this exercise, two potential allocations were tested:
  • Allocating 100 samples to stratum H and 200 to stratum L.
  • Allocating 200 samples to stratum H and 100 to stratum L.
Each allocation impacts the standard error calculation differently because reallocating samples changes the weights and variances used in the formula.

The goal of sample allocation is to reduce the standard error, providing a precise estimate while acknowledging practical constraints like resources. The decision about sample allocation often balances proportions, variances, and logistical considerations, aiming to provide the most reliable estimate given all available information.

In practice, evaluating different allocations helps to find a configuration that provides the best trade-off between cost and accuracy, ensuring the survey delivers meaningful results.

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Most popular questions from this chapter

(Computer Exercise) Construct a population consisting of the integers from 1 to 100\. Simulate the sampling distribution of the sample mean of a sample of size 12 by drawing 100 samples of size 12 and making a histogram of the results.

The data set families contains information about 43,886 families living i the city of Cyberville. The city has four regions: the Northern region has 10,14 families, the Eastern region has 10,390 families, the Southern region has 13,45 families, and the Western region has \(9,890 .\) For each family, the following infor mation is recorded: 1\. Family type 1: Husband-wife family 2: Male-head family 3: Female-head family 2\. Number of persons in family 3\. Number of children in family 4\. Family income 5\. Region 1: North 2: East 3: South 4: West 6\. Education level of head of household 31: Less than 1 st grade 32: 1st, 2nd, 3rd, or 4th grade 33: 5th or 6th grade 34: 7 th or 8 th grade 35: 9th grade 36: 10 th grade 37: 11th grade 38: 12 th grade, no diploma 39: High school graduate, high school diploma, or equivalent 40: Some college but no degree 41: Associate degree in college (occupation/vocation program) 42: Associate degree in college (academic program) 43: Bachelor's degree (e.g., B.S., B.A., A.B.) 44: Master's degree (e.g., M.S., M.A., M.B.A.) 45: Professional school degree (e.g., M.D., D.D.S., D.V.M., LL.B., J.D.) 46: Doctoral degree (e.g., Ph.D., Ed.D.) In these exercises, you will try to learn about the families of Cyberville by using sampling. a. Take a simple random sample of 500 families. Estimate the following population parameters, calculate the estimated standard errors of these estimates, and form \(95 \%\) confidence intervals: i. The proportion of female-headed families ii. The average number of children per family iii. The proportion of heads of households who did not receive a high school diploma iv. The average family income Repeat the preceding parameters for five different simple random samples of size 500 and compare the results. b. Take 100 samples of size 400 i. For each sample, find the average family income. ii. Find the average and standard deviation of these 100 estimates and make a histogram of the estimates. iii. Superimpose a plot of a normal density with that mean and standard deviation of the histogram and comment on how well it appears to fit. iv. Plot the empirical cumulative distribution function (see Section 10.2 ). On this plot, superimpose the normal cumulative distribution function with mean and standard deviation as earlier. Comment on the fit. v. Another method for examining a normal approximation is via a normal probability plot (Section 9.9). Make such a plot and comment on what it shows about the approximation. vi. For each of the 100 samples, find a \(95 \%\) confidence interval for the population average income. How many of those intervals actually contain the population target? vii. Take 100 samples of size \(100 .\) Compare the averages, standard deviations, and histograms to those obtained for a sample of size 400 and explain how the theory of simple random sampling relates to the comparisons. c. For a simple random sample of \(500,\) compare the incomes of the three family types by comparing histograms and boxplots (see Chapter 10.6 ). d. Take simple random samples of size 400 from each of the four regions. i. Compare the incomes by region by making parallel boxplots. ii. Does it appear that some regions have larger families than others? iii. Are there differences in education level among the four regions? e. Formulate a question of your choice and attempt to answer it with a simple random sample of size 400 . f. Does stratification help in estimating the average family income? From a simple random sample of size 400 , estimate the average income and also the standard error of your estimate. Form a \(95 \%\) confidence interval. Next, allocate the 400 observations proportionally to the four regions and estimate the average income from the stratified sample. Estimate the standard error and form a \(95 \%\) confidence interval. Compare your results to the results of the simple random sample.

This problem presents an algorithm for drawing a simple random sample from a population in a sequential manner. The members of the population are considered for inclusion in the sample one at a time in some prespecified order (for example, the order in which they are listed). The \(i\) th member of the population is included in the sample with probability \(\frac{n-n_{i}}{N-i+1}\) "where \(n_{i}\) is the number of population members already in the sample before the ith member is examined. Show that the sample selected in this way is in fact a simple random sample; that is, show that every possible sample occurs with probability $$\frac{1}{\left(\begin{array}{l}N \\\n\end{array}\right)}$$

In a wildlife survey, an area of desert land was divided into 1000 squares, or "quadrats," a simple random sample of 50 of which were surveyed. In each surveyed quadrat, the number of birds, \(Y,\) and the area covered by vegetation, \(X\) were determined. It was found that $$\begin{aligned}\sum X_{i} &=3000 \\\\\sum & Y_{i}=150 \\\\\sum X_{i}^{2} &=225,000 \\ \sum Y_{i}^{2} &=650 \\\\\sum X_{i} Y_{i} &=11,000\end{aligned}$$ a. Estimate the ratio of the average number of birds per quadrat to the average vegetation cover per quadrat. b. Estimate the standard error of your estimate and find an approximate \(90 \%\) confidence interval for the population average. c. Estimate the total number of birds and find an approximate \(95 \%\) confidence interval for the population total. d. Suppose that from an aerial survey, the total area covered by vegetation could easily be determined. How could this information be used to provide another estimate of the number of birds? Would you expect this estimate to be better than or worse than that found in part (c)?

Show that the population correlation coefficient is less than or equal to 1 in absolute value.

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