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Chapter 7: Problem 5

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation \(.75\). a. Compute a \(95 \%\) CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was \(4.85\). b. Compute a \(98 \%\) CI for true average porosity of another seam based on 16 specimens with a sample average porosity of \(4.56 .\) c. How large a sample size is necessary if the width of the \(95 \%\) interval is to be \(.40\) ? d. What sample size is necessary to estimate true average porosity to within \(.2\) with \(99 \%\) confidence?

Short Answer

Expert verified
a) CI: (4.522, 5.178); b) CI: (4.123, 4.997); c) 55 samples; d) 150 samples.

Step by step solution

01

Understanding Confidence Interval (CI)

A Confidence Interval (CI) gives a range within which we expect the true parameter (here, the average porosity) to lie. The formula for a confidence interval for the mean when the population standard deviation is known is \[ \bar{x} \pm z_\alpha \left( \frac{\sigma}{\sqrt{n}} \right) \] where \( \bar{x} \) is the sample mean, \( \sigma \) is the population standard deviation, and \( n \) is the sample size. The value \( z_\alpha \) is the z-score corresponding to the desired confidence level.
02

Compute a 95% CI for Part (a)

For a 95% confidence interval, we use \( z_{0.025} \approx 1.96 \). Let's plug in the values: \( \bar{x} = 4.85 \), \( \sigma = 0.75 \), and \( n = 20 \).\[ 4.85 \pm 1.96 \times \left( \frac{0.75}{\sqrt{20}} \right) \]Calculate:\[ 4.85 \pm 1.96 \times 0.167 \approx 4.85 \pm 0.328 \]The 95% CI is approximately \((4.522, 5.178)\).
03

Compute a 98% CI for Part (b)

For a 98% confidence interval, use \( z_{0.01} \approx 2.33 \). Given: \( \bar{x} = 4.56 \), \( \sigma = 0.75 \), and \( n = 16 \).\[ 4.56 \pm 2.33 \times \left( \frac{0.75}{\sqrt{16}} \right) \]Calculate:\[ 4.56 \pm 2.33 \times 0.1875 \approx 4.56 \pm 0.437 \]The 98% CI is approximately \((4.123, 4.997)\).
04

Determine Sample Size for 95% CI Width of 0.40 (Part c)

To find the sample size \( n \), we need the formula for the width of a CI: \[ 2 \times z_{0.025} \left( \frac{\sigma}{\sqrt{n}} \right) = 0.40 \]Rearranging for \( n \):\[ n = \left( \frac{2 \times 1.96 \times 0.75}{0.40} \right)^2 \]Calculate:\[ n \approx \left( \frac{2.94}{0.40} \right)^2 = 54.04 \]Therefore, at least 55 samples are needed.
05

Determine Sample Size for 99% CI and Margin of 0.20 (Part d)

For a width of 0.40, the margin of error is \( 0.20 \). For a 99% confidence level, use \( z_{0.005} \approx 2.576 \).Using the formula for required sample size:\[ n = \left( \frac{2 \times 2.576 \times 0.75}{0.40} \right)^2 \]Calculate:\[ n = \left( \frac{3.864}{0.40} \right)^2 \approx 149.62 \]Thus, at least 150 samples are needed.

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Key Concepts

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

Normal Distribution
Normal distribution is a continuous probability distribution that is symmetric around its mean. Most values cluster around the central peak, and probabilities for values further from the mean taper off equally in both directions. This makes the well-known bell-shaped curve.
  • The normal distribution is defined by two parameters: the mean (\( \mu \)) and standard deviation (\( \sigma \)).
  • In our exercise, the normal distribution helps assess the porosity of coal samples, which are assumed to vary around their average in a predictable way.
  • Understanding normal distribution is foundational for calculating confidence intervals, as many statistical tests and estimations assume a normal distribution for underlying data.
Normal distribution is critical in this context because it allows us to confidently infer characteristics of a population (all coal samples from a seam) based on a sample (20 or 16 specimens).
Sample Size Determination
Sample size determination is about deciding how many observations of a population are needed to meet certain statistical goals. Knowing the correct sample size is vital for making valid and reliable inferences about a population's parameters.
  • For this exercise, sample size is determined for different confidence levels and width of the confidence interval (CI).
  • The more samples you take, the more precise your estimate, but this requires more resources and effort.
  • The formula for determining sample size when you know the population standard deviation (\( \sigma \)) and want a certain CI width is used in the solution.
  • It involves rearranging the CI formula to solve for \( n \), where \( n \) needs to be rounded up to ensure sufficiency.
Accurate sample size determination ensures the CI is neither too wide (imprecise) nor too narrow (too resource-intensive to obtain).
Z-scores
Z-scores measure how many standard deviations an element is from the mean. In the context of normal distribution, z-scores are used to calculate the probability of a sample score occurring and are crucial for determining confidence intervals.
  • In our exercise, z-scores correspond to desired confidence levels, such as 95% or 98%.
  • Common z-scores are \( z_{0.025} = 1.96 \) for 95% CI and \( z_{0.01} = 2.33 \) for 98% CI.
  • These numbers come from standard normal distribution tables or software capable of estimating them.
Understanding and using z-scores not only help in constructing confidence intervals but also play a role in other statistical methodologies like hypothesis testing.
Confidence Level
A confidence level represents the frequency with which the confidence interval, if you were to repeat the same study over and over, would contain the true parameter. It is expressed as a percentage, such as 95% or 99%.
  • A higher confidence level means that there is a greater chance the interval contains the true mean, but it also results in a wider interval.
  • It's important to balance confidence level with precision; a too wide interval may offer little practical insight.
  • In our exercise, different confidence levels (\( 95\% \) and \( 98\% \)) are explored to show how it affects the interval width and sample size calculation.
  • The choice of confidence level often depends on how critical the decision based on the interval is.
Understanding confidence levels is crucial for interpreting the reliability of statistical estimates and decision-making based on them.

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Most popular questions from this chapter

High concentration of the toxic element arsenic is all too common in groundwater. The article "Evaluation of Treatment Systems for the Removal of Arsenic from Groundwater" (Practice Periodical of Hazardous, Toxic, and Radioactive Waste Mgmt., 2005: 152-157) reported that for a sample of \(n=5\) water specimens selected for treatment by coagulation, the sample mean arsenic concentration was \(24.3 \mu \mathrm{g} / \mathrm{L}\), and the sample standard deviation was 4.1. The authors of the cited article used \(t\)-based methods to analyze their data, so hopefully had reason to believe that the distribution of arsenic concentration was normal. a. Calculate and interpret a \(95 \%\) CI for true average arsenic concentration in all such water specimens. b. Calculate a \(90 \%\) upper confidence bound for the standard deviation of the arsenic concentration distribution. c. Predict the arsenic concentration for a single water specimen in a way that conveys information about precision and reliability.

Consider the next \(100095 \%\) CIs for \(\mu\) that a statistical consultant will obtain for various clients. Suppose the data sets on which the intervals are based are selected independently of one another. How many of these 1000 intervals do you expect to capture the corresponding value of \(\mu\) ? What is the probability that between 940 and 960 of these intervals contain the corresponding value of \(\mu\) ? [Hint: Let \(Y=\) the number among the 1000 intervals that contain \(\mu\). What kind of random variable is \(Y\) ?]

A triathlon consisting of swimming, cycling, and running is one of the more strenuous amateur sporting events. The article "Cardiovascular and Thermal Response of Triathlon Performance" (Medicine and Science in Sports and Exercise, 1988: \(385-389)\) reports on a research study involving nine male triathletes. Maximum heart rate (beats/min) was recorded during performance of each of the three events. For swimming, the sample mean and sample standard deviation were \(188.0\) and \(7.2\), respectively. Assuming that the heart rate distribution is (approximately) normal, construct a \(98 \%\) \(\mathrm{CI}\) for true mean heart rate of triathletes while swimming.

Determine the confidence level for each of the following large-sample one- sided confidence bounds: a. Upper bound: \(\bar{x}+.84 s / \sqrt{n}\) b. Lower bound: \(\bar{x}-2.05 \mathrm{~s} / \sqrt{n}\) c. Upper bound: \(\bar{x}+.67 \mathrm{~s} / \sqrt{n}\)

It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work in temperatures of \(200-500^{\circ} \mathrm{F}\). In a test of one type of mask, 11 of 55 masks had lenses pop out at \(250^{\circ}\). Construct a \(90 \%\) CI for the true proportion of masks of this type whose lenses would pop out at \(250^{\circ}\).
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