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Chapter 6: Problem 50

A survey is planned to estimate the proportion of voters who support a proposed gun control law. The estimate should be within a margin of error of \(\pm 2 \%\) with \(95 \%\) confidence, and we do not have any prior knowledge about the proportion who might support the law. How many people need to be included in the sample?

Short Answer

Expert verified
The minimum number of people that need to be surveyed to achieve a margin of error of \(\pm 2\%\) with \(95\%\) confidence level is calculated as approximately 2401. This figure should be rounded up to 2401 as we cannot survey a fraction of a person.

Step by step solution

01

Understanding the problem

We need to determine the minimum number of people to survey in order to confidently guess the percentage of voters that support a gun control law, with a margin of error of \(2\%\) and a \(95\%\) confidence level. Since we don't have any initial estimates, the most conservative approach is to assume that the proportion (p) is \(50\%\), which maximizes variability.
02

Use formula for sample size

Since the unknown proportion that maximizes variance is \(50\%\), it will require the largest sample size for a given confidence level and margin of error. This is the formula for sample size: n = \(\frac{Z^2_{\alpha/2} *p*(1-p)}{E^2}\) where n is the sample size, \(Z_{\alpha/2}\) is the z-value from the standard normal distribution for the desired confidence level, p is the proportion (estimated as \(0.5\) here), and E is the margin of error as a decimal.
03

Given conditions

Since the confidence interval is \(95\%\), the z-value (\(Z_{\alpha/2}\)) from the standard normal distribution is approximately \(1.96\). The margin of error E given is \(2\%\). This must be converted into a decimal, i.e., \(0.02\). Now, we can substitute these values into the formula.
04

Substitute values into formula

Substitute the values into the formula: n = \(\frac{(1.96)^2 * 0.5 * (1-0.5)}{(0.02)^2}\)
05

Calculate the sample size

Calculate the value for 'n'. This will result in an exact number. However, since we can't survey fractions of individuals, always round up to the next whole number. This provides the minimum number of survey participants to meet the requirements.

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Most popular questions from this chapter

Is a Normal Distribution Appropriate? In Exercises 6.13 and \(6.14,\) indicate whether the Central Limit Theorem applies so that the sample proportions follow a normal distribution. In each case below, does the Central Limit Theorem apply? (a) \(n=80\) and \(p=0.1\) (b) \(n=25\) and \(p=0.8\) (c) \(n=50\) and \(p=0.4\) (d) \(n=200\) and \(p=0.7\)

Standard Error from a Formula and Simulation In Exercises 6.15 to \(6.18,\) find the mean and standard error of the sample proportions two ways: (a) Use StatKey or other technology to simulate at least 1000 sample proportions. Give the mean and standard error and comment on whether the distribution appears to be normal. (b) Use the formulas in the Central Limit Theorem to compute the mean and standard error. Are the results similar to those found in part (a)? Sample proportions of sample size \(n=100\) from a population with \(p=0.4\)

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 100 from a population with proportion 0.41

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Use StatKey or other technology to generate a bootstrap distribution of sample means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviation as an estimate of the population standard deviation. Mean body temperature, in \({ }^{\circ} \mathrm{F},\) using the data in BodyTemp50 with \(n=50, \bar{x}=98.26,\) and \(s=0.765\)
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