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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 \(=90 \%\)

Short Answer

Expert verified
The sample size n for part a is approximately 233, for part b it's approximately 16, and for part c it's approximately 18.

Step by step solution

01

Solution for part a

Given \(E = .17\), \( \sigma = .90\), and the confidence level is \(99\% \). The critical Z value for \(99\% \) confidence level is \(2.58\). Now use the formula \( n = (Z_{\alpha /2} \cdot \sigma / E) ^ 2 \) to find the sample size. So, \( n = (2.58 \cdot .90 / .17) ^ 2 \). Calculate to get the value of n.
02

Solution for part b

Given \(E = 1.45\), \( \sigma = 5.82\), and the confidence level is \(95\% \). The critical Z value for \(95\% \) confidence level is \(1.96\). Now use the formula \( n = (Z_{\alpha /2} \cdot \sigma / E) ^ 2 \) to find the sample size. So, \( n = (1.96 \cdot 5.82 / 1.45) ^ 2 \). Calculate to get the value of n.
03

Solution for part c

Given \(E = 5.65\), \( \sigma = 18.20\), and the confidence level is \(90\% \). The critical Z value for \(90\% \) confidence level is \(1.645\). Now use the formula \( n = (Z_{\alpha /2} \cdot \sigma / E) ^ 2 \) to find the sample size. So, \( n = (1.645 \cdot 18.20 / 5.65) ^ 2 \). Calculate to get the value of n.

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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 an essential concept in statistics that provides a range, within which we expect the population parameter to lie. It reflects how confident we are in the estimate. Confidence intervals are typically expressed as percentages, such as 90%, 95%, or 99% confidence levels. These percentages indicate how often the true parameter would fall within the interval if we were to take numerous samples. The higher the confidence level, the broader the interval.
  • It is calculated using the formula: \( \text{Confidence Interval} = \bar{x} \pm Z_{\alpha /2} \cdot \left( \frac{\sigma}{\sqrt{n}} \right) \)
  • Here, \(\bar{x}\) is the sample mean, \(Z_{\alpha /2}\) is the critical Z value, \(\sigma\) is the standard deviation, and \(n\) is the sample size.
Choosing the right confidence level is critical: it involves a trade-off between precision and reliability. A higher confidence level means we are more certain about our interval, but it also makes the interval wider, providing less precision.
Critical Z Value
The critical Z value is a pivotal part of calculating confidence intervals in statistics. It represents the number of standard deviations a point must be from the mean to ensure the chosen confidence level. Different confidence levels have corresponding critical Z values.
  • For a 90% confidence level, the critical Z value is approximately 1.645.
  • For a 95% confidence level, the critical Z value is approximately 1.96.
  • For a 99% confidence level, the critical Z value is approximately 2.58.
These values emanate from the standard normal (Z) distribution. It's essential to understand that the critical Z value reflects how far data points must extend above and below the sample mean to encompass the designated percentage of the total area under the curve. This is foundational for deciding how reliable your interval estimate is.
Margin of Error
The margin of error is an indispensable concept when discussing estimates in statistics. It quantifies the uncertainty around the sample statistic, allowing you to gauge how precise your sample estimate is in relation to the true population parameter. Essentially, it's an expression of how much you might expect your estimate to vary merely by sampling variability.The calculation of margin of error (\(E\)) relies on three core components:
  • The critical Z value (\(Z_{\alpha/2}\))
  • The standard deviation (\(\sigma\))
  • The sample size (\(n\))
The formula used is: \( E = Z_{\alpha /2} \cdot \left( \frac{\sigma}{\sqrt{n}} \right) \).This measure helps researchers and analysts understand the range within which the actual population parameter is expected to lie, considering sample variability. A smaller margin of error indicates more precision, which is generally preferred.
Standard Deviation
Standard deviation is a fundamental statistic that measures the amount of variation or dispersion within a set of values. In simpler terms, it tells us how spread out the data points are around the mean (average) of the dataset.
  • A small standard deviation means that the data points tend to be close to the mean, implying less variability.
  • A large standard deviation indicates that the data points are spread out over a wider range, showing more variability.
In the context of confidence intervals and sample size determination, standard deviation helps us understand the inherent variability of the data. It directly influences the width of the confidence interval: higher standard deviation leads to wider intervals, reflecting more uncertainty in our estimation of the population mean.

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Most popular questions from this chapter

Determine the most conservative sample size for the estimation of the population proportion for the following. a. \(E=.025\), confidence level \(=95 \%\) b. \(E=.05, \quad\) confidence level \(=90 \%\) c. \(E=.015\), confidence level \(=99 \%\)

Determine the sample size for the estimate of \(\mu\) for the following. a. \(E=2.3, \quad \sigma=15.40\), confidence level \(=99 \%\) b. \(E=4.1, \quad \sigma=23.45\), confidence level \(=95 \%\) c. \(E=25.9, \quad \sigma=122.25, \quad\) confidence level \(=90 \%\)

A jumbo mortgage is a mortgage with a loan amount above the industry-standard definition of conyentional conforming loan limits. As of January 2009, approximately \(2.57 \%\) of people who took out a jumbo mortgage during the previous 12 months were at least 60 days late on their payments. Suppose that this percentage is based on a random sample of 1430 people who took out a jumbo mortgage during the previous 12 months. a. Construct a \(95 \%\) confidence interval for the proportion of all people who took out a jumbo mortgage during the previous 12 months and were at least 60 days late on their payments. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

The mean time taken to design a house plan by 40 architects was found to be 23 hours with a standard deviation of \(3.75\) hours. a. Construct a \(98 \%\) confidence interval for the population mean \(\mu\). b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Describe all possible alternatives. Which alternative is the best and why?

Suppose, for a sample selected from a normally distributed population, \(\bar{x}=68.50\) and \(s=8.9\). a. Construct a \(95 \%\) confidence interval for \(\mu\) assuming \(n=16\). b. Construct a \(90 \%\) confidence interval for \(\mu\) assuming \(n=16 .\) Is the width of the \(90 \%\) confidence interval smaller 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 \(n=25 .\) Is the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=25\) smaller than the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=16\) calculated in part a? If so, why? Explain.

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