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Chapter 6: Problem 46

Find the sample size needed to give, with \(95 \%\) confidence, a margin of error within \(\pm 6 \%\) when estimating a proportion. Within \(\pm 4 \%\). Within \(\pm 1 \%\). (Assume no prior knowledge about the population proportion \(p\).) Comment on the relationship between the sample size and the desired margin of error.

Short Answer

Expert verified
The sample sizes required for a margin of error of 卤6%, 卤4% and 卤1% are calculated using the formula \(n = (Z^2 * p * (1 - p)) / E^2\). Here, \(p\) is assumed to be 0.5. If we decrease the margin of error, the required sample size increases, implying that more precision requires a larger sample size.

Step by step solution

01

Calculate sample size for a margin of error of 卤6%

First, we'll calculate the sample size for a margin of error of 卤6% or 0.06. Plug \(Z = 1.96\), \(p = 0.5\), and \(E = 0.06\) into the formula \(n = (Z^2 * p * (1 - p)) / E^2\) to calculate the required sample size \(n\).
02

Calculate sample size for a margin of error of 卤4%

Next, we'll calculate the sample size for a margin of error of 卤4% or 0.04. Plug \(Z = 1.96\), \(p = 0.5\), and \(E = 0.04\) into the same formula to determine the required sample size \(n\).
03

Calculate sample size for a margin of error of 卤1%

Lastly, we'll calculate the sample size for a margin of error of 卤1% or 0.01. Plugging \(Z = 1.96\), \(p = 0.5\), and \(E = 0.01\) into the formula gives us the required sample size \(n\).
04

Comment on the relationship between the sample size and the margin of error

As we decrease the margin of error, the required sample size will increase. This is because a smaller margin of error implies that our estimate needs to be more precise, and to achieve this more precision, we need a larger sample size.

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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 is a critical concept that helps us understand the precision of an estimated statistic. When we talk about survey results, the margin of error tells us how much the results could differ from the true population value. Imagine you conduct a survey and find that 60% of respondents prefer vanilla ice cream over chocolate. If your margin of error is 5%, this means the true proportion of the population that prefers vanilla could be anywhere from 55% to 65%.

Several factors can affect the margin of error:
  • Sample Size: Larger sample sizes tend to reduce the margin of error, providing a more precise estimate.
  • Confidence Level: A higher confidence level results in a larger margin of error, reflecting a broader range of possible values.
  • Variability: More variability in a population also increases the margin of error.
Understanding the margin of error helps in making informed decisions about how large a sample should be to achieve the desired precision.
Confidence Interval
A confidence interval provides a range of values likely to contain the population parameter. It is an interval estimate, supported by the notion of confidence. For example, if we say our 95% confidence interval for a survey result is 47% to 53%, we are 95% sure that the true value lies within this range.

Key points about confidence intervals include:
  • The width of the interval is determined by both the sample size and the margin of error. A larger sample size results in a narrower interval.
  • The level of confidence, such as 95%, is chosen by the researcher and indicates how sure we are that the interval captures the true parameter.
  • Using the standard deviation and the sample mean, we calculate confidence intervals to provide a sense of accuracy.
Understanding confidence intervals helps to interpret survey data, as it tells us how much we can trust in the results reported.
Population Proportion
The population proportion is the fraction of the population that exhibits a certain attribute. For example, if we want to know the proportion of adults in a city who own a car, the population proportion is the true proportion of all adults in that city who are car owners.

When sampling, we can estimate the population proportion using the sample proportion. But, especially if we have no prior knowledge about the population proportion (assuming it to be 0.5 as a conservative estimate), we have to pay close attention to sample size and the resulting statistics:
  • Sample Estimates: Using a representative sample, we can estimate the population proportion, accounting for variability with the margin of error and confidence intervals.
  • Formula Usage: The formula \( p \) = (sample with attribute / total sample) helps to find the sample proportion, after which formulas can guide us to the confidence interval and required sample size.
  • Parameter Estimation: In cases where true parameter values aren't directly observable, sample data serves as a valuable tool to estimate these values accurately.
This understanding highlights the importance of accurately calculating sample sizes to make reliable inferences about a population's characteristics.

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Most popular questions from this chapter

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 price of used Mustang cars online (in \(\$ 1000\) s) using the data in MustangPrice with \(n=25,\) \(\bar{x}=15.98,\) and \(s=11.11\)

Use a t-distribution. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the endpoints of the t-distribution with \(5 \%\) beyond them in each tail if the samples have sizes \(n_{1}=8\) and \(n_{2}=10\)

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 number of penalty minutes for NHL players using the data in Ottawa Senators with \(n=24, \bar{x}=49.58,\) and \(s=49.14\)

Situations comparing two proportions are described. In each case, determine whether the situation involves comparing proportions for two groups or comparing two proportions from the same group. State whether the methods of this section apply to the difference in proportions. (a) In a taste test, compare the proportion of tasters who prefer one brand of cola to the proportion who prefer the other brand. (b) Compare the proportion of males who voted in the last election to the proportion of females who voted in the last election. (c) Compare the graduation rate (proportion to graduate) of students on an athletic scholarship to the graduation rate of students who are not on an athletic scholarship. (d) Compare the proportion of voters who vote in favor of a school budget to the proportion who vote against the budget.

Ron flips a coin \(n_{1}\) times and Freda flips a coin \(n_{2}\) times. We can assume all coin flips are fair: The coin has an equal chance of landing heads or tails. In each of the following cases, state whether inference for a difference in proportions is appropriate using the methods of this section. If so, give the mean and standard error for the distribution of the difference in proportions \(\left(\hat{p}_{1}-\hat{p}_{2}\right)\) and state whether the normal approximation is appropriate. (a) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land heads and \(\hat{p}_{2}\) be the proportion of Freda's flips that land heads; \(n_{1}=100\) and \(n_{2}=50\). (b) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land heads and \(\hat{p}_{2}\) be the proportion of Ron's flips that land tails; \(n_{1}=100\). (c) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land heads and \(\hat{p}_{2}\) be the proportion of Freda's flips that land tails; \(n_{1}=200\) and \(n_{2}=200\). (d) Let \(\hat{p}_{1}\) be the proportion of Ron's flips that land tails and \(\hat{p}_{2}\) be the proportion of Freda's flips that land tails; \(n_{1}=5\) and \(n_{2}=10\).
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