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Chapter 9: Problem 17

In an opinion poll it is assumed that an unknown proportion \(p\) of the people are in favor of a proposed new law and a proportion \(1-p\) are against it. A sample of \(n\) people is taken to obtain their opinion. The proportion \(\bar{p}\) in favor in the sample is taken as an estimate of \(p\). Using the Central Limit Theorem, determine how large a sample will ensure that the estimate will, with probability \(.95,\) be correct to within .01 .

Short Answer

Expert verified
A sample size of 9604 is needed.

Step by step solution

01

Understanding the Problem

We have an unknown population proportion \(p\) and want to estimate it using a sample of size \(n\). We require the sample proportion \(\bar{p}\) to be within \(0.01\) of \(p\) with \(95\%\) confidence. This involves determining an appropriate \(n\) such that the margin of error is no more than \(0.01\).
02

Margin of Error Formula

The margin of error (E) for an estimate of a population proportion at a given confidence level is given by \(E = Z_{\alpha/2} \times \sqrt{\frac{\bar{p} (1-\bar{p})}{n}}\), where \(Z_{\alpha/2}\) is the z-score for the desired confidence level.
03

Finding the Z-score

Since we need a \(95\%\) confidence level, we find the z-score \(Z_{\alpha/2}\). For \(\alpha = 0.05\), the z-score is approximately 1.96 (from standard normal distribution tables or calculators).
04

Setting up the Equation for Sample Size

Set the margin of error formula to \(0.01\): \(0.01 = 1.96 \times \sqrt{\frac{\bar{p}(1-\bar{p})}{n}}\). We want to solve for the sample size \(n\).
05

Finding the Worst Case Scenario for Sample Proportion

Since the true \(\bar{p}\) is unknown and to ensure a conservative sample estimation, we assume \(\bar{p} = 0.5\). This maximizes the product \(\bar{p}(1-\bar{p})\), making our sample size calculation as large as possible. Using \(\bar{p} = 0.5\), \(\bar{p}(1-\bar{p}) = 0.25\).
06

Substituting and Solving for n

Plug \(\bar{p}(1-\bar{p}) = 0.25\) into the equation: \[0.01 = 1.96 \times \sqrt{\frac{0.25}{n}}\]. Solve for \(n\): \[(0.01)^2 = (1.96)^2 \times \frac{0.25}{n}\] \[0.0001 = 3.8416 \times \frac{0.25}{n}\] \[n = \frac{3.8416 \times 0.25}{0.0001}\] \[n = \frac{0.9604}{0.0001} = 9604\].
07

Conclusion on Sample Size

To ensure that the estimate \(\bar{p}\) is within \(0.01\) of the true proportion \(p\) with \(95\%\) confidence, a sample size of at least \(9604\) is needed.

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Key Concepts

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

Sample Size Calculation
Determining the appropriate sample size is crucial when estimating a population parameter, like the proportion of people favoring a new law. We aim for our sample estimate to closely reflect the true population proportion with a given level of confidence. In this scenario, to ensure our estimate is within 0.01 of the true proportion with 95% certainty, we must carefully calculate the sample size.
Utilizing the Central Limit Theorem, which allows us to assume that the sampling distribution of the sample proportion is approximately normal for large samples, we can derive the necessary sample size.
  • The central formula involved is \( E = Z_{\alpha/2} \times \sqrt{\frac{\bar{p} (1-\bar{p})}{n}} \), where \( E \) is the margin of error and \( Z_{\alpha/2} \) is the z-score corresponding to the desired confidence level.
  • In our case, we want \( E \) to be no more than 0.01 to satisfy our accuracy requirement.
  • Using a worst-case scenario, with \( \bar{p} = 0.5 \), maximizes the product \( \bar{p}(1-\bar{p}) \) to make a conservative estimate for the sample size.
  • This calculation leads us to find that at least 9604 participants are needed in our sample to achieve the desired accuracy and confidence.
Confidence Interval
A confidence interval provides a range of values within which we expect the true population parameter鈥攊n this case, the proportion of people in favor of a law change鈥攖o lie. This range is determined based on the sample data and the desired level of confidence.
  • For a 95% confidence interval, we want to be 95% sure that our interval contains the true proportion \( p \).
  • The interval is centered around the sample proportion \( \bar{p} \), and its width is determined by the margin of error \( E \).
  • The general formula for the confidence interval for a proportion is \( \bar{p} \pm E \).
  • In practice, determining the confidence interval helps in understanding the reliability and precision of an estimate derived from sample data.
Z-score
Understanding the z-score is essential in the context of confidence intervals and sample size calculations. The z-score represents the number of standard deviations a data point is from the mean of a set of values. In statistics, it is utilized to understand standard normal data distributions.
  • The z-score for a specified confidence level reflects how extreme the values in a normal distribution are, relative to the mean.
  • For a 95% confidence level, the z-score used is approximately 1.96, found from a standard normal distribution table.
  • This value underpins the calculation of the confidence interval. It directly influences the width of the interval and, in turn, the sample size necessary to achieve the required accuracy.
  • Calculating the z-score is crucial for determining the appropriate margin of error and ensuring the statistical validity of the interval estimate.

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Most popular questions from this chapter

Let \(S_{200}\) be the number of heads that turn up in 200 tosses of a fair coin. Estimate (a) \(P\left(S_{200}=100\right)\). (b) \(P\left(S_{200}=90\right)\). (c) \(P\left(S_{200}=80\right)\).

A piece of rope is made up of 100 strands. Assume that the breaking strength of the rope is the sum of the breaking strengths of the individual strands. Assume further that this sum may be considered to be the sum of an independent trials process with 100 experiments each having expected value of 10 pounds and standard deviation \(1 .\) Find the approximate probability that the rope will support a weight (a) of 1000 pounds. (b) of 970 pounds.

Recall that if \(X\) is a random variable, the cumulative distribution function of \(X\) is the function \(F(x)\) defined by $$ F(x)=P(X \leq x) $$ (a) Let \(S_{n}\) be the number of successes in \(n\) Bernoulli trials with probability \(p\) for success. Write a program to plot the cumulative distribution for \(S_{n}\). (b) Modify your program in (a) to plot the cumulative distribution \(F_{n}^{*}(x)\) of the standardized random variable $$ S_{n}^{*}=\frac{S_{n}-n p}{\sqrt{n p q}} $$ (c) Define the normal distribution \(N(x)\) to be the area under the normal curve up to the value \(x\). Modify your program in (b) to plot the normal distribution as well, and compare it with the cumulative distribution of \(S_{n}^{*} .\) Do this for \(n=10,50,\) and 100 .

Write a computer program to simulate 10,000 Bernoulli trials with probability .3 for success on each trial. Have the program compute the 95 percent confidence interval for the probability of success based on the proportion of successes. Repeat the experiment 100 times and see how many times the true value of .3 is included within the confidence limits.

A rookie is brought to a baseball club on the assumption that he will have a .300 batting average. (Batting average is the ratio of the number of hits to the number of times at bat.) In the first year, he comes to bat 300 times and his batting average is .267 . Assume that his at bats can be considered Bernoulli trials with probability .3 for success. Could such a low average be considered just bad luck or should he be sent back to the minor leagues? Comment on the assumption of Bernoulli trials in this situation.
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