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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: \(x=89.10\), \(s=3.73\). 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
a) CI: [88.54, 89.66]; precise estimation. b) Sample size needed: 246 (given \(\sigma = 4\)).

Step by step solution

01

Determine the confidence interval formula

For the confidence interval of a population mean with a known sample size ( extit{n}=169), sample mean ( extit{x}=89.10), and sample standard deviation ( extit{s}=3.73), we use the formula for a two-sided confidence interval: CI = \(\bar{x} \pm t_{\alpha/2} \frac{s}{\sqrt{n}}\) where \(t_{\alpha/2}\) is the t-value from the t-distribution with \(n-1\) degrees of freedom, and \(s\) is the sample standard deviation.
02

Identify the critical t-value

With \(n = 169\), the degrees of freedom is \( df = n - 1 = 168\). For a 95% confidence level, we need to find the critical value \(t_{0.025}\) from the t-distribution table. Since 168 is a large degree of freedom, we can approximate \(t_{0.025}\) as 1.96, which is a common approximation for large sample sizes in normal distributions.
03

Calculate the margin of error

To calculate the margin of error (ME) for the confidence interval, use the formula: \( ME = t_{\alpha/2} \times \frac{s}{\sqrt{n}} \) Substituting the known values: \( ME = 1.96 \times \frac{3.73}{\sqrt{169}} \approx 0.56 \)
04

Determine the confidence interval

With the margin of error calculated, the confidence interval (CI) can be determined using: \( CI = 89.10 \pm 0.56 \) Therefore, the confidence interval is: \([88.54, 89.66]\) This interval suggests the true average fracture strength is likely within this range.
05

Analyze the precision of the estimate

The confidence interval \([88.54, 89.66]\) gives a relatively small range (over a short span of values compared to the mean), indicating that the true average fracture strength has been precisely estimated.
06

Calculate required sample size for given precision

To find the sample size needed to estimate \(\mu\) within \(0.5 \text{ MPa}\) with 95% confidence, use the formula: \( n = \left(\frac{Z_{\alpha/2} \sigma}{E}\right)^2 \) Where \(Z_{\alpha/2} = 1.96\) for 95% confidence, \(\sigma = 4 \text{ MPa}\), and \(E = 0.5 \text{ MPa}\).
07

Compute the sample size

Substitute the known values into the sample size formula: \( n = \left(\frac{1.96 \times 4}{0.5}\right)^2 \approx \left(\frac{7.84}{0.5}\right)^2 \approx \left(15.68\right)^2 \approx 246 \) This means a sample size of approximately 246 would be needed to estimate the population mean to within 0.5 MPa at 95% confidence.

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Key Concepts

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

Fracture Strength
Fracture strength is a vital property in material science, especially when evaluating materials like ceramic bars used in kiln processes. It measures the ability of a material to withstand pressure without breaking. In the context of our exercise, the fracture strength data from ceramic bars helps understand how these materials might perform under practical conditions. The value is quantitative and includes measurements in megapascals (MPa).

Knowing the fracture strength can help engineers design safer and more efficient products. It ensures that materials used won't fail under predefined stress conditions, which is crucial in manufacturing processes like firing ceramic bars.
Sample Size Calculation
Determining an appropriate sample size is essential to achieve reliable statistical conclusions. It affects how well you can confidently generalize your results to a larger population. In practical terms, for our ceramic bars' fracture strength, knowing the right sample size helps ensure the findings are accurate and representative.

To calculate sample size, you follow this formula: \[ n = \left(\frac{Z_{\alpha/2} \sigma}{E}\right)^2 \] where
  • \( n \) is the sample size,
  • \( Z_{\alpha/2} \) is the z-value corresponding to your confidence level,
  • \( \sigma \) is the population standard deviation,
  • \( E \) is the desired margin of error.
Using this formula ensures that the sample size is large enough to capture the true state of the population while maintaining a desired level of precision.
T-Distribution
The t-distribution is an important concept in estimating a population mean, especially with smaller sample sizes. It is similar in shape to the normal distribution but has heavier tails. This means it takes into account the variability that could occur in smaller samples.

In our exercise, the t-distribution plays a key role in calculating the confidence interval for the mean fracture strength. We select a t-value based on the sample size's degrees of freedom (df). Here, the degrees of freedom are calculated as \( n - 1 \), which equals 168 for a sample size of 169. This is then matched against the t-distribution table to find the critical value needed for constructing the interval.

This choice of t-distribution ensuring accuracy in our confidence intervals, particularly when dealing with small or moderately-sized datasets.
Margin of Error
The margin of error (ME) quantifies the range around the sample mean in which the true population mean is expected to fall for a given confidence level. It is crucial for interpreting confidence intervals.

In the equation, \[ ME = t_{\alpha/2} \times \frac{s}{\sqrt{n}} \] the t-value represents the critical value for our chosen confidence level based on the t-distribution, \( s \) is the sample standard deviation, and \( n \) is the sample size. Essentially, this formula tells you how far the estimated mean could deviate from the actual mean.

The margin of error helps us evaluate the precision of our estimate. A smaller margin suggests a more precise estimate of the true average fracture strength. In the exercise, the margin of error calculated as 0.56 implies that our estimate is reasonably precise—suggesting a narrow range around the sample mean for the true mean.

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Most popular questions from this chapter

The one-sample \(t\) CI for \(\mu\) is also a confidence interval for the population median \(\tilde{\mu}\) when the population distribution is normal. We now develop a CI for \(\tilde{\mu}\) that is valid whatever the shape of the population distribution as long as it is continuous. Let \(X_{1}, \ldots, X_{n}\) be a random sample from the distribution and \(Y_{1}, \ldots, Y_{n}\) denote the corresponding order statistics (smallest observation, second smallest, and so on). a. What is \(P\left(X_{1}<\tilde{\mu}\right) ?\) What is \(P\left(\left\\{X_{1}<\tilde{\mu}\right\\} \cap\right.\) \(\left.\left\\{X_{2}<\tilde{\mu}\right\\}\right) ?\) b. What is \(P\left(Y_{n}<\tilde{\mu}\right)\) ? What is \(P\left(Y_{1}>\tilde{\mu}\right)\) ? [Hint: What condition involving all of the \(X_{i}\) 's is equivalent to the largest being smaller than the population median?] c. What is \(P\left(Y_{1}<\tilde{\mu}

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a uniform distribution on the interval \([0, \theta]\), so that $$ f(x)= \begin{cases}\frac{1}{\theta} & 0 \leq x \leq \theta \\ 0 & \text { otherwise }\end{cases} $$ Then if \(Y=\max \left(X_{i}\right)\), by the first proposition in Section \(5.5, U=Y / \theta\) has density function $$ f_{U}(u)=\left\\{\begin{array}{cl} n u^{n-1} & 0 \leq u \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Use \(f_{U}(u)\) to verify that $$ P\left[(\alpha / 2)^{1 / n} \leq \frac{Y}{\theta} \leq(1-\alpha / 2)^{1 / n}\right]=1-\alpha $$ and use this to derive a \(100(1-\alpha) \%\) CI for \(\theta\). b. Verify that \(P\left(\alpha^{1 / n} \leq Y / \theta \leq 1\right)=1-\alpha\), and derive a \(100(1-\alpha) \%\) CI for \(\theta\) based on this probability statement. c. Which of the two intervals derived previously is shorter? If your waiting time for a morning bus is uniformly distributed and observed waiting times are \(x_{1}=4.2, x_{2}=3.5, x_{3}=1.7\), \(x_{4}=1.2\), and \(x_{5}=2.4\), derive a 95\% CI for \(\theta\) by using the shorter of the two intervals.

Determine the \(t\) critical value that will capture the desired \(t\) curve area in each of the following cases: a. Central area \(=.95\), df \(=10\) b. Central area \(=.95, \mathrm{df}=20\) c. Central area \(=99\), df \(=20\) d. Central area \(=.99, \mathrm{df}=50\) e. Upper-tail area \(=.01, \mathrm{df}=25\) f. Lower-tail area \(=.025\), df \(=5\)

If you go to a major league baseball game, how long do you expect the game to be? From the 2,429 games played in 2001 , here is a random sample of 25 times in minutes: \(\begin{array}{llllllllll}352 & 150 & 164 & 167 & 225 & 159 & 142 & 182 & 229 & 163 \\ 188 & 197 & 189 & 235 & 161 & 195 & 177 & 166 & 195 & 160 \\ 154 & 130 & 189 & 188 & 225 & & & & & \end{array}\) This is one of those rare instances in which we can do a confidence interval and compare with the true population mean. The mean of all 2,429 lengths is \(178.29\) (almost \(3 \mathrm{~h}\) ). a. Compute the \(t\)-based confidence interval of Section \(8.3 .\) b. Use a normal plot to see if part (a) is valid. c. Generate a bootstrap sample of 999 means. d. Use the standard deviation for part (c) to get a \(95 \%\) confidence interval for the population mean. e. Investigate the distribution of the bootstrap means to see if the CI of part (d) is valid. f. Use part (c) to form the \(95 \%\) confidence interval using the percentile method. g. Say which interval should be used and explain why. Does your interval include the true value, \(178.29\) ?

Each of the following is a confidence interval for \(\mu=\) true average (i.c., population mean) resonance frequency \((\mathrm{Hz})\) for all tennis rackets of a certain type: \((114.4,115.6) \quad(114.1,115.9)\) a. What is the value of the sample mean resonance frequency? b. Both intervals were calculated from the same sample data. The confidence level for one of these intervals is \(90 \%\) and for the other is \(99 \%\). Which of the intervals has the \(90 \%\) confidence level, and why?

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