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Chapter 9: Problem 42

In a Gallup Poll, \(44 \%\) of the people polled answered "more strict" to the following question: "Do you feel that the laws covering the sale of firearms should be made more strict, less strict, or kept as they are now?" Suppose the margin of error in the poll was \(3.5 \%\) and the estimate was made with \(95 \%\) confidence. At least how many people were surveyed?

Short Answer

Expert verified
At least 774 people were surveyed.

Step by step solution

01

Identify the given information

The percentage of people who answered 'more strict' is given as \(44\%\). The margin of error is \(3.5\%\), and the confidence level is \(95\%\).
02

Understand the margin of error formula

The margin of error (E) in a proportion can be found with the formula: \(E = Z \cdot \sqrt{\dfrac{p(1-p)}{n}}\) where \(Z\) is the Z-value corresponding to the confidence level, \(p\) is the sample proportion, and \(n\) is the sample size.
03

Find the Z-value for a 95\% confidence level

For a 95\% confidence level, the Z-value is typically \(1.96\). This value comes from standard normal distribution tables.
04

Set up the margin of error formula

Substitute the given values into the margin of error formula: \(3.5\% = 1.96 \cdot \sqrt{\dfrac{0.44(0.56)}{n}}\)
05

Solve for the sample size

Rearrange the formula to solve for \(n\): \(0.035 = 1.96 \cdot \sqrt{\dfrac{0.44 \cdot 0.56}{n}}\) Square both sides of the equation to remove the square root: \(0.035^2 = (1.96)^2 \cdot \dfrac{0.44 \cdot 0.56}{n}\) \(0.001225 = 3.8416 \cdot \dfrac{0.2464}{n}\) Multiply both sides by \(n\) to isolate \(n\) on one side: \(0.001225n = 0.9466\) Divide both sides by \(0.001225\) to solve for \(n\): \(n = \dfrac{0.9466}{0.001225}\) \(n \approx 773.22\)
06

Round up to the nearest whole number

Since the number of people surveyed must be a whole number, round up \(773.22\) to the nearest whole number: \(n \approx 774\)

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Key Concepts

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

Margin of Error
The margin of error (often abbreviated as MoE) tells you how much you can expect the results of a survey to reflect the true views of a population. It's essentially a buffer zone around your sample statistic, like the 44% in the original problem. This buffer zone accounts for potential sampling errors.

To calculate the margin of error, you use the formula: \( E = Z \cdot \sqrt{\dfrac{p(1-p)}{n}} \), where:
  • \textbf{E} is the margin of error
  • \textbf{Z} is the Z-value associated with your confidence level
  • \textbf{p} is the sample proportion
  • \textbf{n} is the sample size

The margin of error shows the range within which you can expect the true population parameter to fall. For instance, if 44% say
Z-value
Z-value comes from the Z-distribution or standard normal distribution. In a normal distribution, about 95% of data lies within approximately 1.96 standard deviations from the mean. That's why for a 95% confidence level, the Z-value is 1.96. Other common confidence levels are 90% (with a Z-value around 1.645) and 99% (with a Z-value around 2.576).

To find the Z-value, you can use statistical tables or Z-value calculators. Always match your desired confidence level with its corresponding Z-value to apply in your margin of error and sample size calculations.

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Most popular questions from this chapter

Construct a confidence interval of the population proportion at the given level of confidence. \(x=30, n=150,90 \%\) confidence

A simple random sample of size \(n\) is drawn from a population that is normally distributed. The sample mean, \(\bar{x},\) is found to be \(50,\) and the sample standard deviation, \(s,\) is found to be \(8 .\) (a) Construct a \(98 \%\) confidence interval for \(\mu\) if the sample size, \(n,\) is 20 (b) Construct a \(98 \%\) confidence interval for \(\mu\) if the sample size, \(n\), is \(15 .\) How does decreasing the sample size affect the margin of error, \(E ?\) (c) Construct a \(95 \%\) confidence interval for \(\mu\) if the sample size, \(n\), is 20. Compare the results to those obtained in part (a). How does decreasing the level of confidence affect the margin of error, \(E\) ? (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Why?

In March 2014, Harris Interactive conducted a poll of a random sample of 2234 adult Americans 18 years of age or older and asked, "Which is more annoying to you, tailgaters or slow drivers who stay in the passing lane?" Among those surveyed, 1184 were more annoyed by tailgaters. (a) Explain why the variable of interest is qualitative with two possible outcomes. What are the two outcomes? (b) Verify the requirements for constructing a \(90 \%\) confidence interval for the population proportion of all adult Americans who are more annoyed by tailgaters than slow drivers in the passing lane. (c) Construct a \(90 \%\) confidence interval for the population proportion of all adult Americans who are more annoyed by tailgaters than slow drivers in the passing lane.

An urban economist wishes to estimate the proportion of Americans who own their homes. What size sample should be obtained if he wishes the estimate to be within 0.02 with \(90 \%\) confidence if (a) he uses a 2010 estimate of 0.669 obtained from the U.S. Census Bureau? (b) he does not use any prior estimates?

Indicate whether a confidence interval for a proportion or mean should be constructed to estimate the variable of interest. Justify your response. A developmental mathematics instructor wishes to estimate the typical amount of time students dedicate to studying mathematics in a week. She asks a random sample of 50 students enrolled in developmental mathematics at her school to report the amount of time spent studying mathematics in the past week.
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