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Chapter 7: Problem 45

The following observations were made on fracture toughness of a base plate of \(18 \%\) nickel maraging steel ["Fracture Testing of Weldments," ASTM Special Publ. No. 381, 1965: 328-356 (in ksi \(\sqrt{\text { in., }}\) given in increasing order)]: \(\begin{array}{llllllll}69.5 & 71.9 & 72.6 & 73.1 & 73.3 & 73.5 & 75.5 & 75.7 \\\ 75.8 & 76.1 & 76.2 & 76.2 & 77.0 & 77.9 & 78.1 & 79.6 \\ 79.7 & 79.9 & 80.1 & 82.2 & 83.7 & 93.7 & & \end{array}\) Calculate a \(99 \%\) CI for the standard deviation of the fracturetoughness distribution. Is this interval valid whatever the nature of the distribution? Explain.

Short Answer

Expert verified
The 99% CI for the standard deviation is valid under the normality assumption.

Step by step solution

01

Understanding the Problem

We need to calculate a 99% confidence interval for the standard deviation of fracture toughness from a given data set. The data is composed of fracture toughness measurements of steel in ksi \(\sqrt{\text{in.}}\).
02

Organizing and Checking Data

List out the given fracture toughness data: 69.5, 71.9, 72.6, 73.1, 73.3, 73.5, 75.5, 75.7, 75.8, 76.1, 76.2, 76.2, 77.0, 77.9, 78.1, 79.6, 79.7, 79.9, 80.1, 82.2, 83.7, 93.7. Count the number of observations: 22.
03

Calculating the Sample Standard Deviation

Calculate the sample mean \( \bar{x} \) and the sample standard deviation \( s \). Mean \( \bar{x} = \frac{\Sigma x}{n} \).Sample standard deviation \( s = \sqrt{\frac{\Sigma (x_i - \bar{x})^2}{n-1}} \), where \(x_i\) are the observations, and \(n\) is the number of observations.
04

Using Chi-Square Distribution

For a 99% confidence level, determine the critical values \( \chi^2_{0.005} \) and \( \chi^2_{0.995} \) based on \( n-1 = 21 \) degrees of freedom, using a chi-square distribution table. These values are used to calculate the confidence interval for the population variance \( \sigma^2 \) as \( \left(\frac{(n-1)s^2}{\chi^2_{0.995}}, \frac{(n-1)s^2}{\chi^2_{0.005}}\right) \).
05

Calculating the Confidence Interval for the Standard Deviation

Compute the confidence interval for the variance using the critical chi-square values:\[ \left(\frac{21s^2}{43.77}, \frac{21s^2}{6.84}\right) \]Then take the square root of each bound to get the confidence interval for the standard deviation \( \sigma \).
06

Interpretation of Results

The calculated confidence interval for the standard deviation of the fracture toughness is valid only if the underlying distribution is normal since the chi-square distribution is based on this assumption.

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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 (CI) is a range of values that estimates a population parameter with a certain level of confidence. For instance, a 99% confidence interval implies we can be 99% confident that the interval contains the true population parameter. To compute a CI for the standard deviation of a dataset, we first calculate the sample standard deviation \(s\). This provides a measure of how spread out the individual data points are from the mean. From there, we use statistical distributions, like the chi-square distribution, to determine the interval based on our confidence level.A critical point to remember is that the confidence interval for the standard deviation depends on the underlying data distribution. If our data doesn't follow a normal distribution, the CI may not be reliable.
Standard Deviation
Standard deviation (SD) is a statistic that measures the amount of variability in a dataset. It tells us how much individual data points tend to deviate from the mean of the dataset. The formula for standard deviation in a sample is:\[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\]where \(x_i\) represents each data point, \(\bar{x}\) is the sample mean, and \(n\) is the number of observations.A higher standard deviation indicates that the data points spread out significantly around the mean, while a lower standard deviation means they are clustered closely.Understanding the standard deviation of a dataset is crucial for interpreting the confidence interval. It helps us predict the variability expected in a given dataset, which is essential for making informed decisions based on data analysis.
Normal Distribution
The normal distribution, often referred to as the bell curve, is a probability distribution that is symmetric about the mean. It is widely used in statistics because many datasets naturally follow this pattern.Characteristics of a normal distribution include:
  • The mean, median, and mode are all equal and located at the center.
  • It is defined by two parameters: the mean \(\mu\) and the standard deviation \(\sigma\).
  • Approximately 68% of the data falls within one standard deviation of the mean, 95% within two SDs, and 99.7% within three SDs.
For the chi-square distribution used in calculating confidence intervals, it is crucial the underlying data be normally distributed. This assumption ensures that the estimates of the confidence interval are reliable and valid. Without this assumption, we might not be able to accurately infer the variability and spread of the data.

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Most popular questions from this chapter

The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. The article "Testing Practices for the AC Breakdown Voltage Testing of Insulation Liquids" (IEEE Electrical Insulation Magazine, 1995: 21-26) gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions. \(\begin{array}{llllllllllllllll}62 & 50 & 53 & 57 & 41 & 53 & 55 & 61 & 59 & 64 & 50 & 53 & 64 & 62 & 50 & 68 \\ 54 & 55 & 57 & 50 & 55 & 50 & 56 & 55 & 46 & 55 & 53 & 54 & 52 & 47 & 47 & 55 \\ 57 & 48 & 63 & 57 & 57 & 55 & 53 & 59 & 53 & 52 & 50 & 55 & 60 & 50 & 56 & 58\end{array}\) a. Construct a boxplot of the data and comment on interesting features. b. Calculate and interpret a \(95 \%\) CI for true average breakdown voltage \(\mu\). Does it appear that \(\mu\) has been precisely estimated? Explain. c. Suppose the investigator believes that virtually all values of breakdown voltage are between 40 and 70 . What sample size would be appropriate for the \(95 \% \mathrm{Cl}\) to have a width of \(2 \mathrm{kV}\) (so that \(\mu\) is estimated to within \(1 \mathrm{kV}\) with \(95 \%\) confidence)?

A random sample of 110 lightning flashes in a certain region resulted in a sample average radar echo duration of \(.81 \mathrm{sec}\) and a sample standard deviation of \(34 \mathrm{sec}\) ("Lightning Strikes to an Airplane in a Thunderstorm," \(J .\) of Aircraft, 1984: 607-611). Calculate a \(99 \%\) (two- sided) confidence interval for the true average echo duration \(\mu\), and interpret the resulting interval.

A more extensive tabulation of \(t\) critical values than what appears in this book shows that for the \(t\) distribution with \(20 \mathrm{df}\), the areas to the right of the values \(.687, .860\), and \(1.064\) are \(.25, .20\), and .15, respectively. What is the confidence level for each of the following three confidence intervals for the mean \(\mu\) of a normal population distribution? Which of the three intervals would you recommend be used, and why? a. \((\bar{x}-.687 s / \sqrt{21}, \bar{x}+1.725 s / \sqrt{21})\) b. \((\bar{x}-.860 s / \sqrt{21}, \bar{x}+1.325 s / \sqrt{21})\) c. \((\bar{x}-1.064 s / \sqrt{21}, \bar{x}+1.064 s / \sqrt{21})\)

Determine the following: a. The 95 th percentile of the chi-squared distribution with \(v=10\) b. The 5 th percentile of the chi-squared distribution with \(v=10\) c. \(P\left(10.98 \leq \chi^{2} \leq 36.78\right)\), where \(\chi^{2}\) is a chi- squared rv with \(v=22\) d. \(P\left(\chi^{2}<14.611\right.\) or \(\left.\chi^{2}>37.652\right)\), where \(\chi^{2}\) is a chisquared rv with \(v=25\)

Silicone implant augmentation rhinoplasty is used to correct congenital nose deformities. The success of the procedure depends on various biomechanical properties of the human nasal periosteum and fascia. The article "Biomechanics in Augmentation Rhinoplasty" (J. of Med. Engr: and Tech., 2005: 14-17) reported that for a sample of 15 (newly deceased) adults, the mean failure strain (\%) was \(25.0\), and the standard deviation was \(3.5\). a. Assuming a normal distribution for failure strain, estimate true average strain in a way that conveys information about precision and reliability. b. Predict the strain for a single adult in a way that conveys information about precision and reliability. How does the prediction compare to the estimate calculated in part (a)?
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