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The amount of lateral expansion (mils) was determined for a sample of \(n=9\) pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was \(s=2.81\) mils. Assuming normality, derive a \(95 \% \mathrm{CI}\) for \(\sigma^{2}\) and for \(\sigma\).

Short Answer

Expert verified
The 95% CI for \( \sigma^2 \) is \( (3.60, 28.95) \) and for \( \sigma \) is \( (1.90, 5.38) \).

Step by step solution

01

Identify the Given Information

We are given a sample with \( n = 9 \) and a sample standard deviation \( s = 2.81 \text{ mils} \). We need to find a 95% confidence interval (CI) for the population variance \( \sigma^2 \) and the population standard deviation \( \sigma \).
02

Determine the Chi-Square Distribution Parameters

Since the population is normally distributed, the sample variance \( s^2 \) follows a chi-square distribution. The degrees of freedom (df) are calculated as \( df = n - 1 = 9 - 1 = 8 \).
03

Find Chi-Square Critical Values

For a 95% confidence interval, we find the chi-square critical values \( \chi_{\alpha/2}^2 \) and \( \chi_{1-\alpha/2}^2 \) using a chi-square distribution table. Here, \( \alpha = 0.05 \), so \( \chi_{0.025, 8}^2 \approx 2.18 \) and \( \chi_{0.975, 8}^2 \approx 17.53 \).
04

Calculate the Confidence Interval for \( \sigma^2 \)

The 95% CI for \( \sigma^2 \) is given by:\[ \left( \frac{(n-1)s^2}{\chi_{0.975}^2}, \frac{(n-1)s^2}{\chi_{0.025}^2} \right) \]Plugging in the values, \( \frac{8 \cdot 2.81^2}{17.53} \approx 3.60 \) and \( \frac{8 \cdot 2.81^2}{2.18} \approx 28.95 \). Thus, the CI for \( \sigma^2 \) is \( (3.60, 28.95) \).
05

Calculate the Confidence Interval for \( \sigma \)

The CI for \( \sigma \) is obtained by taking the square root of the endpoints of the CI for \( \sigma^2 \):\[ (\sqrt{3.60}, \sqrt{28.95}) \]Calculating gives \( (1.90, 5.38) \). Thus, the CI for \( \sigma \) is \( (1.90, 5.38) \).

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Key Concepts

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

Chi-Square Distribution
When dealing with statistics, the chi-square distribution is a reliable tool for analyzing variation within sample data. Imagine you have a standard deviation or variance calculated from a sample, like in our exercise. If the original data follows a normal distribution, the sample variance will tend to follow a chi-square distribution. This connection allows us to form confidence intervals for things like population variance.

The chi-square distribution depends heavily on degrees of freedom, which is usually calculated as the sample size minus one, expressed as \( df = n - 1 \). For example, with a sample size of 9, you have 8 degrees of freedom.
  • This distribution is skewed, meaning it is not symmetrical like a normal distribution.
  • The behavior changes with different degrees of freedom.
  • Lower degrees of freedom lead to a more skewed distribution, while higher ones are more symmetrical.
Understanding critical points in this distribution helps us to calculate boundaries for confidence intervals. It serves as a foundational element in estimating parameters that aid in predictions within statistics.
Sample Standard Deviation
The sample standard deviation, denoted as \(s\), is a valuable measure of how spread out numbers in your sample data are. It gives the average distance of each data point from the mean of a sample. For our exercise, we dealt with a sample standard deviation of 2.81 mils, indicating the variability among the weld expansions.

Calculating the sample standard deviation involves these steps:
  • Find the mean (average) of your sample data.
  • Subtract the mean from each number to find the individual deviations.
  • Square these deviations to get positive values.
  • Sum all the squared deviations.
  • Divide this by the sample size minus one (\( n - 1 \)) - this gives you the variance.
  • Take the square root of the variance to get the standard deviation.
This measure helps in understanding the reliability of the data and is critical when formulating confidence intervals, especially when comparing sample statistics to population parameters. It reflects the expected range of values lesser affected by outliers in the context of statistical assessments.
Population Variance
Population variance, represented as \( \sigma^2 \), is an essential statistical measure. It tells you about the spread or dispersion of a set of data points in a population from the population mean. In our exercise, we focus on finding a confidence interval for this variance using sample data.

Why is variance important?
  • Variance gives insights into how much data points can deviate from the average.
  • It is crucial for probability calculations and hypothesis testing.
  • Helps in understanding data consistency and reliability.
In practical scenarios, calculating the exact population variance is often not feasible due to large data sizes. Therefore, we rely on confidence intervals derived from sample data as a range in which the true population variance is expected to lie. The chi-square distribution plays a pivotal role in creating these intervals since it accounts for variability in sample estimates.
Normal Distribution
The normal distribution, also known as the bell curve, is fundamental in statistics due to its properties of symmetry and predictable behavior. Most of the phenomena measured, like test scores or heights, follow this pattern, which makes understanding and utilizing this concept crucial.

Characteristics of a normal distribution include:
  • It's symmetric about the mean, meaning it has an equal spread on both sides.
  • Approximately 68% of the data falls within one standard deviation of the mean, about 95% within two, and about 99.7% within three.
  • The mean, median, and mode all coincide at the center of the distribution.
In our exercise, the assumption of normality is crucial to using the chi-square distribution for variance estimations. The underlying presumption is that with a normal distribution, the calculations for confidence intervals for population variance and standard deviation are credible and reflect the true population parameters accurately. Understanding normal distribution thus provides a groundwork for exploring more complex statistical models.

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Most popular questions from this chapter

The alternating current \((\mathrm{AC})\) 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 \((\mathrm{kV})\) 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 \(95 \%\) CI to have a width of \(2 \mathrm{kV}\) (so that \(\mu\) is estimated to within \(1 \mathrm{kV}\) with \(95 \%\) confidence)?

The technology underlying hip replacements has changed as these operations have become more popular (over 250,000 in the United States in 2008). Starting in 2003 , highly durable ceramic hips were marketed. Unfortunately, for too many patients the increased durability has been counterbalanced by an increased incidence of squeaking. The May 11, 2008, issue of the New York Times reported that in one study of 143 individuals who received ceramic hips between 2003 and 2005,10 of the hips developed squeaking. a. Calculate a lower confidence bound at the \(95 \%\) confidence level for the true proportion of such hips that develop squeaking. b. Interpret the \(95 \%\) confidence level used in (a).

On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with \(\sigma=100\). The composition of bars has been slightly modified, but the modification is not believed to have affected either the normality or the value of \(\sigma\). a. Assuming this to be the case, if a sample of 25 modified bars resulted in a sample average yield point of \(8439 \mathrm{lb}\), compute a \(90 \%\) CI for the true average yield point of the modified bar. b. How would you modify the interval in part (a) to obtain a confidence level of \(92 \%\) ?

A more extensive tabulation of \(t\) critical values than what appears in this book shows that for the \(t\) distribution with 20 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. \((\bar{x}-1.064 s / \sqrt{21}, \bar{x}+1.064 s / \sqrt{21})\)

Ultra high performance concrete (UHPC) is a relatively new construction material that is characterized by strong adhesive properties with other materials. The article "Adhesive Power of Ultra High Performance Concrete from a Thermodynamic Point of View" \((J\). of Materials in Civil Engr., 2012: 1050-1058) described an investigation of the intermolecular forces for UHPC connected to various substrates. The following work of adhesion measurements (in \(\mathrm{mJ} / \mathrm{m}^{2}\) ) for UHPC specimens adhered to steel appeared in the article: \(\begin{array}{lllll}107.1 & 109.5 & 107.4 & 106.8 & 108.1\end{array}\) a. Is it plausible that the given sample observations were selected from a normal distribution? b. Calculate a two-sided \(95 \%\) confidence interval for the true average work of adhesion for UHPC adhered to steel. Does the interval suggest that 107 is a plausible value for the true average work of adhesion for UHPC adhered to steel? What about 110 ? c. Predict the resulting work of adhesion value resulting from a single future replication of the experiment by calculating a \(95 \%\) prediction interval, and compare the width of this interval to the width of the CI from (b). d. Calculate an interval for which you can have a high degree of confidence that at least \(95 \%\) of all UHPC specimens adhered to steel will have work of adhesion values between the limits of the interval.

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