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Chapter 7: Problem 4

Two populations are surveyed with simple random samples. A sample of size \(n_{1}\) is used for population I, which has a population standard deviation \(\sigma_{1} ;\) a sample of size \(n_{2}=2 n_{1}\) is used for population II, which has a population standard deviation \(\sigma_{2}=2 \sigma_{1} .\) Ignoring finite population corrections, in which of the two samples would you expect the estimate of the population mean to be more accurate?

Short Answer

Expert verified
The estimate from Population I is more accurate.

Step by step solution

01

Define Accuracy of Estimate

The accuracy of the sample mean estimate is typically gauged by the standard error of the mean, which represents how much the sample mean is expected to vary from the true population mean.
02

Determine Standard Error for Population I

The standard error for the sample from Population I is calculated using the formula for the standard error: \[ SE_1 = \frac{\sigma_1}{\sqrt{n_1}} \]
03

Determine Standard Error for Population II

For the sample from Population II, using the given relationships, the standard error is: \[ SE_2 = \frac{\sigma_2}{\sqrt{n_2}} \] Substitute \( \sigma_2 = 2\sigma_1 \) and \( n_2 = 2n_1 \) into the formula: \[ SE_2 = \frac{2\sigma_1}{\sqrt{2n_1}} \]Simplifying further, \[ SE_2 = \frac{2\sigma_1}{\sqrt{2}\sqrt{n_1}} = \frac{2\sigma_1}{\sqrt{2} \times \sqrt{n_1}} = \frac{2\sigma_1}{\sqrt{2}\sqrt{n_1}} = \sqrt{2} \times \frac{\sigma_1}{\sqrt{n_1}} = \sqrt{2} \times SE_1 \]
04

Compare Standard Errors

Population I has a standard error of \( SE_1 \), and Population II has a standard error of \( \sqrt{2} \times SE_1 \). Since \( \sqrt{2} > 1 \), \( SE_2 \) is larger than \( SE_1 \). This implies the sample mean for Population I is expected to vary less compared to Population II, indicating it is more accurate.

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Key Concepts

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

Standard Error
The standard error of the mean helps determine the accuracy of your sample mean as an estimate of the population mean. It quantifies the variability of the sample mean from one sample to another. In essence, standard error tells us how much the sample mean can fluctuate from the actual population mean.

To calculate the standard error (\( SE \) itself) for any population, we use the formula:
  • \( SE = \frac{\sigma}{\sqrt{n}} \)
Where \( \sigma \) represents the population standard deviation, and \( n \) is the size of the sample.

In the case of two different populations, as seen in the exercise, we calculate their standard errors to gauge which population's sample mean estimate would be more precise. A smaller standard error indicates a more reliable estimate of the population mean because it suggests that repeated samples would give similar results.
Simple Random Sampling
Simple random sampling is a foundational concept in statistics. It ensures that every individual or element in a population has an equal chance of being selected for the sample. This method is the cornerstone of producing unbiased and representative data in research.

The processes involved ensure randomness and are vital for the validity of inferential statistics. For example:
  • Each member of the population is assigned a unique number.
  • Random numbers are generated to choose the sample members.
  • This method minimizes biases and helps maintain external validity.
Simple random sampling is crucial when estimating the population mean because it gives an accurate reflection of the population's characteristics, which in turn affects the quality of the standard error and overall accuracy of your estimates.
Population Standard Deviation
Understanding population standard deviation is key to interpreting data variability within a population. It is a measure of how spread out numbers are around the average (mean) of a population.

In mathematical terms, the population standard deviation (\( \sigma \) itself)) is the square root of the variance. It is calculated using:
  • Summing up all squared deviations from the mean
  • Dividing by the population size (N)
  • Finding the square root of the result
For instance, if we consider one sample having a population standard deviation of \( \sigma_1 \) and another with \( \sigma_2 = 2\sigma_1 \), the larger the standard deviation, the greater the spread in data, leading to different standard errors as we saw in the exercise.

High population standard deviation signifies more data variability, making it harder to pinpoint an accurate population mean from a sample. Consequently, this affects the precision of the estimates derived from your samples.

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Most popular questions from this chapter

Suppose that a simple random sample is used to estimate the proportion of families in a certain area that are living below the poverty level. If this proportion is roughly \(.15,\) what sample size is necessary so that the standard error of the estimate is.02?

Suppose that of a population of \(N\) items, \(k\) are defective in some way. For example, the items might be documents, a small proportion of which are fraudulent. How large should a sample be so that with a specified probability it will contain at least one of the defective items? For example, if \(N=10,000, k=50,\) and \(p=.95,\) what should the sample size be? Such calculations are useful in planning sample sizes for acceptance sampling.

In order to halve the width of a \(95 \%\) confidence interval for a mean, by what factor should the sample size be increased? Ignore the finite population correction.

Two surveys were independently conducted to estimate a population mean, \(\mu\) Denote the estimates and their standard errors by \(\bar{X}_{1}\) and \(\bar{X}_{2}\) and \(\sigma_{\bar{X}_{1}}\) and \(\sigma_{\bar{X}_{2}}\) Assume that \(\bar{X}_{1}\) and \(\bar{X}_{2}\) are unbiased. For some \(\alpha\) and \(\beta,\) the two estimates can be combined to give a better estimator: $$X=\alpha \bar{X}_{1}+\beta \bar{X}_{2}$$ a. Find the conditions on \(\alpha\) and \(\beta\) that make the combined estimate unbiased. b. What choice of \(\alpha\) and \(\beta\) minimizes the variances, subject to the condition of unbiasedness?

In a simple random sample of 1,500 voters, \(55 \%\) said they planned to vote for a particular proposition, and \(45 \%\) said they planned to vote against it. The estimated margin of victory for the proposition is thus \(10 \% .\) What is the standard error of this estimated margin? What is an approximate \(95 \%\) confidence interval for the margin?
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