/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 117 Refer to Exercise \(8.116\). Ass... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 117

Refer to Exercise \(8.116\). Assume that a preliminary sample has shown that \(63 \%\) of the adults in this city favor legalized casino gambling. How large should the sample size be so that the \(95 \%\) confidence interval for the population proportion has a margin of error of \(.05 ?\)

Short Answer

Expert verified
To get a \(95\%\) confidence interval with a margin of error of \(0.05\) based on the preliminary sample, the sample size would need to be \(n\). Remember to round up your answer to a whole human count.

Step by step solution

01

Understand the Given Information

The preliminary sample showed that \(63\%\) or \(0.63\) of adults in the city favor legalized casino gambling. The desired margin of error (E) for the \(95\%\) confidence interval is \(0.05\). Also note that the confidence interval of \(95\%\) corresponds to a \(Z\) score of \(1.96\), obtained from the standard normal distribution table.
02

Identify the Formula

The formula for determining the sample size in proportion estimation is given by: \( n = \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \). This formula is derived from the concept of the confidence interval of a sample proportion.
03

Substitute the Values into the Formula

Substitute \( Z = 1.96, p = 0.63, E = 0.05 \) into the formula to find the sample size: \( n = \frac{(1.96)^2 \cdot 0.63 \cdot (1 - 0.63)}{(0.05)^2} \)
04

Calculate the Sample Size

Do the arithmetic in the formula to get the value of the required sample size. Remember, since we can't survey partial individuals, round up to the next whole number if the answer is in decimal form.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values, derived from the sample data, that is believed to contain the true value of an unknown population parameter. In this exercise, the confidence interval is used to estimate the population proportion of adults who favor legalized casino gambling. The key idea is that if we were to take a large number of random samples and compute the confidence interval for each sample, a certain percentage (in this case, 95%) of those intervals would contain the actual population proportion. Confidence intervals are essential when dealing with statistical data as they indicate the reliability of an estimate, giving us a clear range within which the true value is expected to lie.
Population Proportion
The population proportion is an important concept in statistics that refers to the fraction of the entire population that has a particular characteristic. In this example, the population proportion is the percentage of adults in the city who support legalized casino gambling, which the preliminary sample suggests is 63% or 0.63. Understanding the population proportion helps researchers and statisticians understand the makeup of the population segment being studied. With a reliable sample, assumptions about the whole population can be made with more confidence, which is crucial when planning resources or making decisions based on that segment.
Margin of Error
The margin of error is a crucial aspect of any statistical estimate, providing a boundary within which the true value is expected to fall. In this context, a margin of error of 0.05 signifies that the estimation of the population proportion can vary by plus or minus 5%. Margin of error is significant because it quantifies the uncertainty of the estimate. A smaller margin indicates a more precise estimate, while a larger margin reflects more uncertainty. For this problem, the goal is to determine the sample size that will keep the margin of error within 5%, thus ensuring the estimate is both accurate and reliable.
Z Score
The Z score plays a pivotal role in calculating the required sample size for statistical estimates. It is a measure indicating how many standard deviations an element is from the mean and, in statistics, is used to determine the confidence level of an estimate. For a 95% confidence interval, the corresponding Z score is 1.96. This value is derived from the standard normal distribution and can be found in Z tables or with statistical software. In sample size determination for the population proportion, the Z score ensures that the probability of the estimate falling within the confidence interval is as desired. This mathematical standardization helps us to confidently predict the reliability of our estimates.

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Most popular questions from this chapter

For each of the following, find the area in the appropriate tail of the \(t\) distribution. a. \(t=-1.302\) and \(d f=42\) b. \(t=2.797\) and \(n=25\) c. \(t=1.397\) and \(n=9\) d. \(t=-2.383\) and \(d f=67\)

Tony's Pizza guarantees all pizza deliveries within 30 minutes of the placement of orders. An agency wants to estimate the proportion of all pizzas delivered within 30 minutes by Tony's. What is the most conservative estimate of the sample size that would limit the margin of error to within \(.02\) of the population proportion for a \(99 \%\) confidence interval?

Suppose, for a sample selected from a population, \(\bar{x}=25.5\) and \(s=4.9\). a. Construct a \(95 \%\) confidence interval for \(\mu\) assuming \(n=47\). b. Construct a \(99 \%\) confidence interval for \(\mu\) assuming \(n=47\). Is the width of the \(99 \%\) confidence interval larger than the width of the \(95 \%\) confidence interval calculated in part a? If yes, explain why. c. Find a \(95 \%\) confidence interval for \(\mu\) assuming \(n=32\). Is the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=32\) larger than the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=47\) calculated in part a? If so, why? Explain.

Determine the sample size for the estimate of \(\mu\) for the following. a. \(E=.17, \quad \sigma=.90\), confidence level \(=99 \%\) b. \(E=1.45, \quad \sigma=5.82\), confidence level \(=95 \%\) c. \(E=5.65, \quad \sigma=18.20\), confidence level \(=90 \%\)

A company that produces 8 -ounce low-fat yogurt cups wanted to estimate the mean number of calories for such cups. A random sample of 10 such cups produced the following numbers of calories. \(\begin{array}{lllllllll}147 & 159 & 153 & 146 & 144 & 148 & 163 & 153 & 143 & 158\end{array}\) Construct a \(99 \%\) confidence interval for the population mean. Assume that the numbers of calories for such cups of yogurt produced by this company have an approximately normal distribution.
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