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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?

Short Answer

Expert verified
A sample size of 319 is needed.

Step by step solution

01

Understand the Problem

We are given a proportion of families below the poverty level, \( p = 0.15 \), and want to find a sample size that results in a standard error of 0.02 for this proportion.
02

Recall the Formula for Standard Error

The standard error \( SE \) of a proportion \( p \) is given by the formula: \[ SE = \sqrt{\frac{p(1 - p)}{n}} \] where \( n \) is the sample size.
03

Set Up the Equation

We need the standard error to be 0.02, so set \( SE = 0.02 \) and substitute the given proportion \( p = 0.15 \) into the formula: \[ 0.02 = \sqrt{\frac{0.15(1 - 0.15)}{n}} \]
04

Solve for the Sample Size \( n \)

To solve for \( n \), first square both sides of the equation to remove the square root: \[ (0.02)^2 = \frac{0.15 \times 0.85}{n} \] which simplifies to \[ 0.0004 = \frac{0.1275}{n} \] Next, solve for \( n \) by multiplying both sides by \( n \) and then dividing both sides by 0.0004: \[ n = \frac{0.1275}{0.0004} \]
05

Calculate \( n \)

Execute the division: \[ n = 318.75 \] Since sample size cannot be a fraction, round up to the nearest whole number.
06

Round to the Nearest Whole Number

The sample size required is \( n = 319 \) people.

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Key Concepts

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

Understanding the Standard Error
Standard error is a measure of how much the sampling proportion is expected to vary from the actual proportion in the population. When conducting a survey or a study, it is normal to estimate the proportion of a certain characteristic, such as families living below the poverty level in a specific area.

The standard error helps us understand the reliability of this estimate. It is calculated from the sample data and gives insight into how much the sample proportion might differ from the true proportion in the entire population. The formula for the standard error of a proportion, given as \( SE = \sqrt{\frac{p(1 - p)}{n}} \), considers both the estimated proportion \( p \) and the sample size \( n \).

  • A smaller standard error indicates a more precise estimate of the population proportion.
  • To achieve a smaller standard error, either a larger sample size \( n \) is needed, or the variability \( p(1-p) \) must be minimized (though \( p \) is often fixed in practical situations).
Proportion Estimation
Proportion estimation is a method used in statistics to determine what portion of a population exhibits a particular characteristic or attribute. In the context of sample surveys, this involves selecting a sample and using its data to estimate a population parameter.

For example, if researchers want to estimate the percentage of families in a particular area living below the poverty level, they might draw a simple random sample of households and determine the proportion among those sampled. This estimate is expressed with respect to the population to understand the extent of poverty more broadly.

  • The estimated proportion \( p \) is key in calculating the standard error, influencing the reliability of the estimation.
  • Accurate proportion estimation relies heavily on correctly calculating the sample size \( n \), which ensures an appropriate level of precision needed in the sampling results.
Simple Random Sample
A simple random sample (SRS) is a fundamental concept in statistics for selecting a subset of a population where each member has an equal chance of being chosen. This method is essential because it underpins the validity of statistical inference, helping ensure that the sample represents the broader population effectively.

In the context of determining proportions, like those living below the poverty level, using a simple random sample means each family in the area has the same probability of being selected. This helps eliminate biases that could affect the accuracy of estimating the population proportion.

  • Simple random sampling helps support the calculation of an accurate standard error, as it assumes each observation is equally independent and identically distributed.
  • The approach of using simple random samples helps avoid systematic errors and provides a foundation for estimating the proportion with a known level of precision.

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Most popular questions from this chapter

In a survey of a very large population, the incidences of two health problems are to be estimated from the same sample. It is expected that the first problem will affect about \(3 \%\) of the population and the second about \(40 \% .\) Ignore the finite population correction in answering the following questions. a. How large should the sample be in order for the standard errors of both estimates to be less than .01? What are the actual standard errors for this sample size? b. Suppose that instead of imposing the same limit on both standard errors, the investigator wants the standard error to be less than \(10 \%\) of the true value in each case. What should the sample size be?

Which of the following is a random variable? a. The population mean b. The population size, \(N\) c. The sample size, \(n\) d. The sample mean e. The variance of the sample mean f. The largest value in the sample g. The population variance h. The estimated variance of the sample mean

A simple random sample of a population of size 2000 yields the following 25 values: \(\begin{array}{rrrrr}104 & 109 & 111 & 109 & 87 \\ 86 & 80 & 119 & 88 & 122 \\\ 91 & 103 & 99 & 108 & 96 \\ 104 & 98 & 98 & 83 & 107 \\ 79 & 87 & 94 & 92 & 97\end{array}\) a. Calculate an unbiased estimate of the population mean. b. Calculate unbiased estimates of the population variance and \(\operatorname{Var}(\bar{X})\) c. Give approximate \(95 \%\) confidence intervals for the population mean and total.

The value of a population mean increases linearly through time: \(\mu(t)=\alpha+\beta t\) while the variance remains constant. Independent simple random samples of size \(n\) are taken at times \(t=1,2,\) and 3 a. Find conditions on \(w_{1}, w_{2},\) and \(w_{3}\) such that $$\hat{\beta}=w_{1} \bar{X}_{1}+w_{2} \bar{X}_{2}+w_{3} \bar{X}_{3}$$ is an unbiased estimate of the rate of change, \(\beta .\) Here \(\bar{X}_{i}\) denotes the sample mean at time \(t_{i}\) b. What values of the \(w_{i}\) minimize the variance subject to the constraint that the estimate is unbiased?

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.

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