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The article "Evaluating Tunnel Kiln Performance" (Amer: Ceramic Soc. Bull., Aug. 1997: 59-63) gave the following summary information for fracture strengths (MPa) of \(n=169\) ceramic bars fired in a particular kiln: \(\bar{x}=89.10\), \(s=3.73\). a. Calculate a (two-sided) confidence interval for true average fracture strength using a confidence level of \(95 \%\). Does it appear that true average fracture strength has been precisely estimated? b. Suppose the investigators had believed a priori that the population standard deviation was about \(4 \mathrm{MPa}\). Based on this supposition, how large a sample would have been required to estimate \(\mu\) to within . \(5 \mathrm{MPa}\) with \(95 \%\) confidence?

Short Answer

Expert verified
The 95% CI for fracture strength is (88.54, 89.66) MPa; 246 samples are needed for 0.5 MPa precision with 95% confidence.

Step by step solution

01

Identify the Given Values

We are provided with the sample mean \( \bar{x} = 89.10 \), the sample standard deviation \( s = 3.73 \), and the sample size \( n = 169 \). We need to construct a two-sided confidence interval with a confidence level of 95%.
02

Determine the Critical Value

The degrees of freedom for the t-distribution are \( n - 1 = 168 \). For a 95% confidence level, we find the critical t-value \( t_{\alpha/2} \). Using a t-table or technology, \( t_{0.025, 168} \approx 1.96 \).
03

Calculate the Confidence Interval Margin of Error

The margin of error is calculated using the formula: \[ E = t_{\alpha/2} \times \frac{s}{\sqrt{n}} \]. Substituting the values, \[ E = 1.96 \times \frac{3.73}{\sqrt{169}} = 1.96 \times 0.286 \approx 0.56 \].
04

Construct the Confidence Interval

The confidence interval is given by \( \bar{x} \pm E \). Thus, the 95% confidence interval is: \[ 89.10 \pm 0.56 = (88.54, 89.66) \]. This suggests the true average fracture strength is quite closely estimated.
05

Calculate Required Sample Size for Updated Scenario

Given the population standard deviation \( \sigma \approx 4 \) and wanted margin of error \( E = 0.5 \), we use the formula for sample size: \[ n = \left( \frac{z_{\alpha/2} \sigma}{E} \right)^2 \]. Using \( z_{0.025} \approx 1.96 \), the calculation is: \[ n = \left( \frac{1.96 \times 4}{0.5} \right)^2 = \left( 15.68 \right)^2 \approx 246 \]. Thus, 246 samples are needed.

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Key Concepts

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

Understanding the t-distribution in Confidence Intervals
The t-distribution is a statistical distribution that is very closely related to the standard normal distribution. It's particularly useful when working with small sample sizes or when the population standard deviation is unknown. This makes it a staple in many confidence interval calculations.

The t-distribution has heavier tails than the normal distribution, which means it accounts for more variability and uncertainty. This feature of the t-distribution ensures that the estimation of the population parameter is more reliable when based on a sample. The degree to which these tails are heavy is determined by the degrees of freedom, typically calculated as the sample size minus one, \(n - 1\).
  • Used when the sample size is small, generally \(n < 30\).
  • More spread out than the standard normal curve, providing a more conservative estimate.
  • Utilizes degrees of freedom to determine the shape of the distribution.
  • The critical value \(t_{\alpha/2}\) is extracted from the t-table corresponding to the desired confidence level.
Calculating Sample Size for Precision
Calculating an adequate sample size is crucial to ensuring that the results of a study are both reliable and precise. Sample size calculations ensure the resulting estimates of population parameters are within a certain margin of error with a specified level of confidence.

For large samples or known variances, a normal distribution-based approach may be used, but for smaller samples or when the population variance is unknown, the t-distribution is preferred.
  • Helps determine how many samples are needed to achieve a specific margin of error.
  • Ensures the estimation of parameters like the mean is accurate.
  • Depends heavily on prior assumptions about population variance or standard deviation.
For instance, to maintain a margin of error within 0.5 MPa and a 95% confidence level, the calculation formula is:
\[ n = \left( \frac{z_{\alpha/2} \times \sigma}{E} \right)^2 \]
In this formula, \(\sigma\) is the approximated standard deviation, \(z_{\alpha/2}\) is from the z-distribution (since we assume very large sample sizes), and \(E\) is the desired margin of error. This equation guides researchers to determine how many samples are sufficient to precisely estimate their parameter of interest.
Exploring Fracture Strength in Materials
Fracture strength is the stress level at which a material will fail through fracturing. It’s a critical aspect when designing and evaluating materials, especially in industries like ceramics, metalworks, and construction.

Understanding fracture strength helps in ensuring safety, reliability, and longevity of the material under consideration. The measure is often reported in megapascals (MPa) and involves conducting stress tests to measure the maximum stress that sample materials can endure before breaking. This metric is crucial as it assists engineers and researchers in evaluating whether a material can withstand the pressures and forces it will encounter in its intended application.
  • Essential for the safety and design of structures and components.
  • Helps predict the lifespan and durability of a material.
  • A key consideration in quality control and material characterization.
When evaluating statistical summaries of fracture strength, these figures help determine whether the sample mean provides a good representation of the true average fracture strength. Doing so can signal if consistent results are being obtained or if the materials are experiencing too much variance in their strength levels.
Defining the Confidence Level in Estimations
The confidence level in statistics expresses how confident we are that a given parameter lies within a calculated confidence interval. For example, a 95% confidence level means we expect that in 95 out of 100 samples, the true population parameter will fall within the derived interval.

This concept is fundamental in hypothesis testing and establishing the reliability of parameter estimations. A higher confidence level will result in a wider confidence interval, indicating more certainty about capturing the true parameter but with less precision.
  • Reflects our degree of certainty in predicting parameter values.
  • Tied directly to the choice of z or t critical values in interval calculations.
  • A higher confidence level usually increases the margin of error.
When researchers choose a 95% confidence level, they accept a 5% risk (or significance level) that the interval does not contain the true population parameter. Understanding this concept allows for balanced decision-making between precision and confidence in estimations.

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Most popular questions from this chapter

Chronic exposure to asbestos fiber is a well-known health hazard. The article "The Acute Effects of Chrysotile Asbestos Exposure on Lung Function" (Erviron. Research, 1978: 360-372) reports results of a study based on a sample of construction workers who had been exposed to asbestos over a prolonged period. Among the data given in the article were the following (ordered) values of pulmonary compliance \(\left(\mathrm{cm}^{3} / \mathrm{cm} \mathrm{H}_{2} \mathrm{O}\right)\) for each of 16 subjects 8 months after the exposure period (pulmonary compliance is a measure of lung elasticity, or how effectively the lungs are able to inhale and exhale): \(\begin{array}{llllll}167.9 & 180.8 & 184.8 & 189.8 & 194.8 & 200.2 \\ 201.9 & 206.9 & 207.2 & 208.4 & 226.3 & 227.7 \\ 228.5 & 232.4 & 239.8 & 258.6 & & \end{array}\) a. Is it plausible that the population distribution is normal? b. Compute a \(95 \% \mathrm{CI}\) for the true average pulmonary compliance after such exposure. c. Calculate an interval that, with a confidence level of \(95 \%\), includes at least \(95 \%\) of the pulmonary compliance values in the population distribution.

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 \(95 \%\) CI for \(\sigma^{2}\) and for \(\sigma\).

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)?

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 \(98 \%\) CI for true mean heart rate of triathletes while swimming.

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. \((\bar{x}-1.064 s / \sqrt{21}, \bar{x}+1.064 s / \sqrt{21})\)

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