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Estimate the minimum sample size needed to form a confidence interval for the proportion of a population that has a particular characteristic, meeting the criteria given. a. \(p=0.81,95 \%\) confidence, \(E=0.02\) b. \(\quad p=0.81,99 \%\) confidence, \(E=0.02\) c. \(\quad p=0.81,95 \%\) confidence, \(E=0.01\)

Short Answer

Expert verified
a. 1397, b. 2312, c. 5588.

Step by step solution

01

Determine the Z-score

For each confidence level, you will need to find the corresponding Z-score which represents the number of standard deviations from the mean to the start of the interval. - For 95% confidence level, the Z-score is 1.96. - For 99% confidence level, the Z-score is 2.576.
02

Understand the Formula

The formula to estimate the minimum sample size for a proportion is given by: \[ n = \left( \frac{Z^2 \cdot p \cdot (1-p)}{E^2} \right) \]where- \(Z\) is the Z-score,- \(p\) is the population proportion,- \(E\) is the margin of error.
03

Calculate Sample Size for Case a

Substitute the values for case (a), which are \(p = 0.81\), \(Z = 1.96\), \(E = 0.02\):\[n = \left( \frac{ (1.96)^2 \times 0.81 \times (1-0.81)}{0.02^2} \right)\]Calculate \(n\):\(n \approx 1396.86\).Rounded up, the sample size \(n\) is 1397.
04

Calculate Sample Size for Case b

For (b), use \(p = 0.81\), \(Z = 2.576\), \(E = 0.02\):\[n = \left( \frac{ (2.576)^2 \times 0.81 \times (1-0.81)}{0.02^2} \right)\]Calculate \(n\):\(n \approx 2311.69\).Rounded up, the sample size \(n\) is 2312.
05

Calculate Sample Size for Case c

For (c), use \(p = 0.81\), \(Z = 1.96\), \(E = 0.01\):\[n = \left( \frac{ (1.96)^2 \times 0.81 \times (1-0.81)}{0.01^2} \right)\]Calculate \(n\):\(n \approx 5587.46\).Rounded up, the sample size \(n\) is 5588.

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Key Concepts

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

Sample Size Calculation
When you want to estimate a population proportion with confidence, you need to determine the correct sample size. This ensures your confidence interval is both accurate and precise. The aim is to collect enough data to reflect true population characteristics, without surveying the whole group. To figure out the sample size, you use a specific formula. This formula takes into account the desired confidence level, the expected proportion of the population (\(p\)), and the margin of error (\(E\)). The formula is:\[n = \left( \frac{Z^2 \cdot p \cdot (1-p)}{E^2} \right)\]Some key points on sample size calculation:- **Confidence Level**: Reflects how sure you can be about your results.- **Margin of Error (E)**: Indicates the expected range of error in your estimation.- **Proportion (p)**: Represents the trait you're measuring in the population.In practice:- Increase the confidence level, and you'll need a larger sample.- Decrease the margin of error, and the sample size must rise.Thus, choosing the right sample size is crucial, allowing you to balance accuracy and practicality in data collection.
Proportion Estimation
Estimating a proportion involves determining how large a segment of the population exhibits a certain characteristic. This is important for surveys where you want to know how common a particular feature or opinion is among people.Proportion estimation relies heavily on sample surveys. Once you have a sample collected, you determine the proportion (\( \hat{p} \)) that exhibits the trait of interest by dividing the number of occurrences by the total sample size. Common considerations:- **Sample Representativeness**: Ensure your sample reflects the broader population to avoid bias.- **Sample Variability**: Variability affects confidence in the reliability of the estimate.For reliable estimates across different scenarios, you calculate a confidence interval (CI) around the estimated proportion. The CI gives a range believed to encompass the true population proportion.An accurate proportion estimate helps guide decisions, offering insights into population behavior or opinions, and aiding in planning and policy-making endeavors.
Z-score
The Z-score is a statistic that indicates the number of standard deviations an element is from the mean of a data set. For confidence intervals, the Z-score helps you understand how data points relate to the average, providing a metric for standard deviation in normally distributed data. In confidence interval calculations, the Z-score corresponds with the confidence levels (e.g., 95% or 99%). Commonly used Z-scores are: - **95% Confidence Level**: Z-score of 1.96 - **99% Confidence Level**: Z-score of 2.576 These scores are derived from the standard normal distribution. Using the Z-score:
  • Helps determine the width of the confidence interval.
  • Is crucial in calculating the necessary sample size for accurate estimations.
  • Assists in risk assessment and decision-making processes.
To increase the precision of your statistical analysis, comprehend these Z-score values, as they are pivotal to understanding confidence in proportion estimates.

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Most popular questions from this chapter

In order to estimate the speaking vocabulary of three-year-old children in a particular socioeconomic class, a sociologist studies the speech of four children. The mean and standard deviation of the sample are \(\bar{x}-1120\) and \(s=215\) words. Assuming that speaking vocabularies are normally distributed, construct an \(80 \%\) confidence interval for the mean speaking vocabulary of all three-year-old children in this socioeconomic group.

Large Data Sets 4 and 4 Alist the results of 500 tosses of a die. Let \(p\) denote the proportion of all tosses of this die that would result in a four. Use the sample data to construct a \(90 \%\) confidence interval for \(p\).

A software engineer wishes to estimate, to within 5 seconds, the mean time that a new application takes to start up, with \(95 \%\) confidence. Estimate the minimum size sample required if the standard deviation of start up times for similar software is 12 seconds.

Wild life researchers trapped and measured six adult male collared lemmings. The data (in millimeters) are: \(104,99,112,115,96,109 .\) Assume the lengths of all lemmings are normally distributed. a. Construct a \(90 \%\) confidence interval for the mean length of all adult male collared lemmings using these data. b. Convert the six lengths in millimeters to lengths in inches using the conversion \(1 \mathrm{~mm}=0.039\) in (so the first datum changes to \((104)(0.039)=4.06\) ) . Use the converted data to construct a \(90 \%\) confidence interval for the mean length of all adult male collared lemmings expressed in inches. c. Convert your answer in part (a) into inches directly and compare it to your answer in (b). This illustrates that if you construct a confidence interval in one system of units you can convert it directly into another system of units without having to convert all the data to the new units.

For all settings a packing machine delivers a precise amount of liquid; the amount dispensed always has standard deviation 0.07 ounce. To calibrate the machine its setting is fixed and it is operated 50 times. The mean amount delivered is 6.02 ounces with sample standard deviation 0.04 ounce. Construct a \(99.5 \%\) confidence interval for the mean amount delivered at this setting. Hint: Not all the information provided is needed.

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