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Chapter 8: Problem 2

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?

Short Answer

Expert verified
a. The sample mean is 115.0 Hz. b. The interval (114.4, 115.6) has a 90% confidence level because it is narrower than (114.1, 115.9), which has a 99% confidence level.

Step by step solution

01

Identify Sample Mean from Confidence Intervals

The sample mean is the midpoint of a confidence interval. It can be calculated by taking the average of the lower and upper limits of any given confidence interval. Let's choose interval (114.4, 115.6):\[ \text{Sample mean} = \frac{114.4 + 115.6}{2} = 115.0 \text{ Hz} \]
02

Compare Interval Widths

Calculate the width of each confidence interval by subtracting the lower bound from the upper bound.\[ \text{Width of (114.4, 115.6)} = 115.6 - 114.4 = 1.2 \text{Width of (114.1, 115.9)} = 115.9 - 114.1 = 1.8 \]
03

Determine Confidence Levels

A confidence interval with a larger width has a higher confidence level because it accounts for more potential variability. The interval (114.1, 115.9) with a width of 1.8 is wider than the interval (114.4, 115.6) with a width of 1.2. Therefore, \( (114.4, 115.6) \) is the 90% confidence interval, and \( (114.1, 115.9) \) is the 99% confidence interval.

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Key Concepts

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

Sample Mean
The sample mean is a vital concept when discussing confidence intervals. It's often described as a central point around which your data or observations are concentrated. When we're given a confidence interval, calculating the sample mean is straightforward. You simply find the midpoint of the interval. This involves taking the average of the upper and lower boundaries of the confidence interval. For example, if the confidence interval is from 114.4 to 115.6, the sample mean is calculated as:
  • First, add the two boundaries: 114.4 + 115.6
  • Second, divide the sum by 2: \( \frac{114.4 + 115.6}{2} = 115.0 \text{ Hz} \)
The sample mean provides a single value estimate of the population mean, giving you a quick snapshot of what the average might be.
Confidence Level
Confidence level is a statistical term that reveals how certain you are that the true population parameter falls within the confidence interval you calculated. It's expressed as a percentage and reflects the probability that the interval includes the actual mean. If you have a 90% confidence level, it means you expect that 90 out of 100 such intervals would contain the true mean.
The relationship between confidence level and the width of the interval is inverse. A higher confidence level (like 99%) means the interval is wider because you need to account for more variation to be more certain. Conversely, a lower confidence level (like 90%) is narrower, showing less certainty.
  • 90% confidence interval: Less wide, quicker conclusions, less certainty.
  • 99% confidence interval: Wider, more detailed analysis, more certainty.
Interval Width
Interval width is a simple calculation but it carries significant insight into the precision of an estimate. It is the range covered by the confidence interval, calculated by subtracting the lower limit from the upper limit. A larger interval width reflects more variability and suggests that your estimate of the population mean is less precise.
A narrower interval width means you have a more precise estimate, assuming a fixed sample size and variability in the data.
  • Interval width of \((114.4, 115.6)\): \(115.6 - 114.4 = 1.2\)
  • Interval width of \((114.1, 115.9)\): \(115.9 - 114.1 = 1.8\)
Understanding interval width helps you interpret how confident you should be about the results and can guide decision-making in data analysis.

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Most popular questions from this chapter

A random sample of \(n=15\) heat pumps of a certain type yielded the following observations on lifetime (in years): \(\begin{array}{rrrrrrrr}2.0 & 1.3 & 6.0 & 1.9 & 5.1 & .4 & 1.0 & 5.3 \\ 15.7 & .7 & 4.8 & .9 & 12.2 & 5.3 & .6 & \end{array}\) a. Assume that the lifetime distribution is exponential and use an argument parallel to that of Example \(8.5\) to obtain a \(95 \%\) CI for expected (true average) lifetime. b. How should the interval of part (a) be altered to achieve a confidence level of \(99 \%\) ? c. What is a \(95 \%\) CI for the standard deviation of the lifetime distribution? [Hint: What is the standard deviation of an exponential random variable?]

Even as traditional markets for sweetgum lumber have declined, large section solid timbers traditionally used for construction bridges and mats have become increasingly scarce. The article "Development of Novel Industrial Laminated Planks from Sweetgum Lumber" (J. of Bridge Engr., 2008: 64-66) described the manufacturing and testing of composite beams designed to add value to low- grade sweetgum lumber. Here is data on the modulus of rupture (psi; the article contained summary data expressed in MPa): \(\begin{array}{llllll}6807.99 & 7637.06 & 6663.28 & 6165.03 & 6991.41 & 6992.23 \\ 6981.46 & 7569.75 & 7437.88 & 6872.39 & 7663.18 & 6032.28 \\\ 6906.04 & 6617.17 & 6984.12 & 7093.71 & 7659.50 & 7378.61 \\ 7295.54 & 6702.76 & 7440.17 & 8053.26 & 8284.75 & 7347.95 \\ 7422.69 & 7886.87 & 6316.67 & 7713.65 & 7503.33 & 7674.99\end{array}\) a. Verify the plausibility of assuming a normal population distribution. b. Estimate the true average modulus of rupture in a way that conveys information about precision and reliability. c. Predict the modulus for a single beam in a way that conveys information about precision and reliability. How does the resulting prediction compare to the estimate in (b).

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.

A manufacturer of college textbooks is interested in estimating the strength of the bindings produced by a particular binding machine. Strength can be measured by recording the force required to pull the pages from the binding. If this force is measured in pounds, how many books should be tested to estimate the average force required to break the binding to within . \(1 \mathrm{lb}\) with \(95 \%\) confidence? Assume that \(\sigma\) is known to be .8.

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}
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