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Find the sample size required for a margin of error of 3 percentage points, and then find one for a margin of error of \(1.5\) percentage points; for both, use a \(95 \%\) confidence level. Find the ratio of the larger sample size to the smaller sample size. To reduce the margin of error to half, by what do you need to multiply the sample size?

Short Answer

Expert verified
The ratio of the larger sample size to the smaller sample size is 2:1. Therefore, to reduce the margin of error to half, you would multiply the original sample size by 2.

Step by step solution

01

Understanding and Defining Parameters

Margin of error formula is given by \(E = z\frac{\sigma}{\sqrt{n}}\) and the confidence level translates to a z-score. For this problem, we have two margins of errors, \(E1 = 0.03\) and \(E2 = 0.015\), and a confidence level of \(95 \%\), which corresponds to a z score of \(z = 1.96\). The population standard deviation, \(\sigma\), is unknown but we can resolve the margin error formula for \(n\) at \(n = \left(\frac{z\sigma}{E}\right)^2\). Given \(\sigma\) is an unknown constant, we can compare the sample sizes using the ratios of the error margins.
02

Computing the Sample Sizes

Let’s denote \(n_1\) and \(n_2\) as the sample sizes for error margins \(E1\) and \(E2\) respectively. Using the rearranged error margin formula, we get \(n_1 = \left(\frac{z\sigma}{E1}\right)^2\) and \(n_2 = \left(\frac{z\sigma}{E2}\right)^2\). We do not need the exact values as we're asked for the ratio between these sample sizes.
03

Calculating the Ratio

To find the ratio of the larger sample size to the smaller sample size, divide \(n_1\) by \(n_2\). \(Ratio = \frac{n_1}{n_2}\) = \(\frac{z\sigma/E1}{z\sigma/E2}\) = \(\frac{E2}{E1}\) = \( \frac{1.5}{3}\), simplifying to \(0.5\). This means the larger sample size is twice the smaller sample size.
04

Multiplication Factor to Reduce Margin of Error

Clearly, when the margin of error was halved the sample size doubled. Therefore, to reduce the margin of error to half (given all other things remain constant), you would need to multiply the sample size by 2.

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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 in statistics, serving as a measure of the uncertainty around a sample estimate. It represents the range within which we expect the true population parameter to lie. In simpler terms, it's like saying, "We think the true value is this, but it could be a little bit more or less."

The formula for computing the margin of error is given by:

\[E = z\frac{\sigma}{\sqrt{n}}\]

where:
  • \(E\) is the margin of error,
  • \(z\) is the Z-score corresponding to your chosen confidence level,
  • \(\sigma\) is the population standard deviation, and
  • \(n\) is the sample size.
Given a fixed Z-score and population standard deviation, the margin of error decreases as the sample size increases. This is because a larger sample size tends to give more precise estimates, hence reducing uncertainty. Understanding the margin of error is pivotal in ensuring that sample size calculations are appropriately made for reliable results.
Confidence Level
A confidence level indicates how confident we are in our statistical results, specifically in relation to the interval estimate derived from a sample. Common confidence levels include 90%, 95%, and 99%, with a higher confidence level indicating greater assurance in the statistical findings.

For instance, a 95% confidence level implies that if we were to take 100 different samples and compute a confidence interval for each one, we expect that 95 of those intervals will contain the true population parameter. It is important to select an appropriate confidence level based on the context of the study, the desired precision, and field standards.

When calculating sample sizes, the confidence level directly influences the Z-score used in the margin of error formula. A higher confidence level results in a larger Z-value, which in turn increases the margin of error if the sample size remains constant.
Z-score
The Z-score forms a bridge between statistical samples and their corresponding confidence intervals. It is essentially a measure of how many standard deviations an element is from the mean of the population.

In the context of sample size calculation and error margins, the Z-score translates a given confidence level into a numerical value. For example, a 95% confidence level correlates with a Z-score of approximately 1.96. This number indicates how far we need to reach out from the mean to capture 95% of potential outcomes.

Here's a brief understanding of some common confidence levels and their associated Z-scores:
  • 90% Confidence Level - Z-score: 1.645
  • 95% Confidence Level - Z-score: 1.96
  • 99% Confidence Level - Z-score: 2.576
Knowing the correct Z-score is key to determining the accurate margin of error and thereby, the correct sample size.
Population Standard Deviation
Population standard deviation, denoted as \(\sigma\), is a measure that quantifies the amount of variation or dispersion present in a set of data values. It provides insights into how spread out data is from the mean of the population, and it's pivotal in understanding the precision of a sample estimate.

In sample size calculations, the population standard deviation impacts the margin of error directly. A larger standard deviation indicates more variability in the data, which often necessitates a larger sample size to achieve a desired margin of error. This is because more variability means the sample mean might not closely match the true population mean, unless controlled by a sufficiently large sample.

If the true value of the population standard deviation is unknown, it might be estimated from a preliminary study or a similar population. When determining sample sizes, precise knowledge or a good estimate of \(\sigma\) is crucial to ensuring results are both accurate and reliable.

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Most popular questions from this chapter

Suppose it is known that \(60 \%\) of employees at a company use a Flexible Spending Account (FSA) benefit. a. If a random sample of 200 employees is selected, do we expect that exactly \(60 \%\) of the sample uses an FSA? Why or why not? b. Find the standard error for samples of size 200 drawn from this population. What adjustments could be made to the sampling method to produce a sample proportion that is more precise?

Bob Ross hosted a weekly television show, The Joy of Painting, on PBS in which he taught viewers how to paint. During each episode, he produced a complete painting while teaching viewers how they could produce a similar painting. Ross completed 30,000 paintings in his lifetime. Although it was an art instruction show, PBS estimated that only \(10 \%\) of viewers painted along with Ross during his show based on surveys of viewers. For each of the following, also identify the population and explain your choice. a. Is the number 30,000 a parameter or a statistic? b. Is the number \(10 \%\) a parameter or a statistic?

The Gallup poll reported that \(45 \%\) of Americans have tried marijuana. This was based on a survey of 1021 Americans and had a margin of error of plus or minus 5 percentage points with a \(95 \%\) level of confidence. a. State the survey results in confidence interval form and interpret the interval. b. If the Gallup Poll was to conduct 100 such surveys of 1021 Americans, how many of them would result in confidence intervals that did not include the true population proportion? c. Suppose a student wrote this interpretation of the interval: "We are \(95 \%\) confident that the percentage of Americans who have tried marijuana is between \(40 \%\) and \(50 \% .\) " What, if anything, is incorrect in this interpretation?

According to a 2017 Pew Research survey, \(60 \%\) of young Americans aged 18 to 29 say the primary way they watch television is through streaming services on the Internet. Suppose a random sample of 200 Americans from this age group is selected. a. What percentage of the sample would we expect to watch television primarily through streaming services? b. Verify that the conditions for the Central Limit Theorem are met. c. Would it be surprising to find that 125 people in the sample watched television primarily through streaming services? Why or why not? d. Would it be surprising to find that more than \(74 \%\) of the sample watched television primarily through streaming services? Why or why not?

Assume your class has 30 students and you want a random sample of 10 of them. Describe how to randomly select 10 people from your class using the random number table.

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