/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 A random sample of \(n=8\) E-gla... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 32

A random sample of \(n=8\) E-glass fiber test specimens of a certain type yielded a sample mean interfacial shear yield stress of \(30.2\) and a sample standard deviation of \(3.1\) ("On Interfacial Failure in Notched Unidirectional Glass/Epoxy Composites," J. of Composite Materials, 1985: 276-286). Assuming that interfacial shear yield stress is normally distributed, compute a \(95 \%\) CI for true average stress (as did the authors of the cited article).

Short Answer

Expert verified
The 95% CI for the true average stress is approximately (27.609, 32.791).

Step by step solution

01

Identify the Given Values

We are given that the sample size is \( n = 8 \), the sample mean is \( \bar{x} = 30.2 \), and the sample standard deviation is \( s = 3.1 \). We need to compute a \(95\%\) confidence interval for the true average interfacial shear yield stress.
02

Determine the Critical t-value

Since the sample size \( n = 8 \) is small, we use the t-distribution. The degrees of freedom \( df \) is \( n - 1 = 7 \). For a \(95\%\) confidence interval, look up the t-value in a t-distribution table or use a calculator to find that \( t_{\alpha/2} = t_{0.025,7} \approx 2.3646 \).
03

Calculate the Standard Error

The standard error (SE) of the sample mean is calculated as \( SE = \frac{s}{\sqrt{n}} = \frac{3.1}{\sqrt{8}} \approx 1.096 \).
04

Compute the Confidence Interval

The formula for a confidence interval using the t-distribution is \( \bar{x} \pm t_{\alpha/2} \times SE \). Substituting in the values, we have:\[ 30.2 \pm 2.3646 \times 1.096 \approx 30.2 \pm 2.591 \]Thus, the confidence interval is approximately \( (27.609, 32.791) \).

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Key Concepts

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

t-distribution
When we have a small sample size, typically less than 30, and the population standard deviation is unknown, we use the t-distribution to calculate confidence intervals. It is a type of probability distribution that is symmetric and bell-shaped, similar to the normal distribution, but has thicker tails. These thick tails help account for the increased variability and uncertainty present in small samples.
  • Thicker tails mean that there is a higher probability of values farther from the mean.
  • The t-distribution approaches the normal distribution as sample sizes become larger.
In the confidence interval calculation, the critical value that we derive from the t-distribution helps us to create a range wherein we expect the true population parameter to lie. We rely on it especially when the sample data appears to diverge noticeably from the norm, which is often the case for small samples.
sample standard deviation
The sample standard deviation (often denoted as 's') is a measure that quantifies the amount of variation or dispersion in a set of sample data points. It's calculated as the square root of the variance, which itself is the average of the squared differences from the sample mean. Knowing the standard deviation is crucial as it gives us insight into how spread out the values of the sample data are.
  • A lower standard deviation indicates that the data points are close to the mean.
  • A higher standard deviation shows that the data points are spread out over a wider range of values.
In context to finding confidence intervals, the sample standard deviation plays a vital role in determining the standard error and thus affects the width of the confidence interval.
degrees of freedom
Degrees of freedom (df) in statistics refers to the number of independent values or quantities which can be assigned to a statistical distribution. It is related to the sample size and affects the shape of the t-distribution used in confidence interval calculations.
  • Degrees of freedom is calculated as the sample size minus one, or \( n - 1 \).
  • It provides context for determining the critical value from the t-distribution table or calculator.
Ultimately, degrees of freedom influence the shape and spread of your sampling distribution, ensuring that smaller samples account for additional variability by widening the confidence interval.
standard error
Standard error (SE) is a statistical measure that describes the accuracy with which a sample mean represents the entire population mean. It is the standard deviation of the sample mean's sampling distribution.
  • Calculated as \( SE = \frac{s}{\sqrt{n}} \), where \( s \) is the sample standard deviation and \( n \) is the sample size.
  • The SE decreases as the sample size increases, indicating more precise estimates of the population mean.
The SE is pivotal in the computation of confidence intervals because it determines how much the sample mean would be expected to vary from the true population mean. Smaller values of SE indicate less variability, resulting in a more narrow confidence interval.

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Most popular questions from this chapter

Let \(0 \leq \gamma \leq \alpha\). Then a \(100(1-\alpha)\) \% CI for \(\mu\) when \(n\) is large is $$ \left(\bar{x}-z_{\gamma} \cdot \frac{s}{\sqrt{n}}, \bar{x}+z_{\alpha-\gamma} \cdot \frac{s}{\sqrt{n}}\right) $$ The choice \(\gamma=\alpha / 2\) yields the usual interval derived in Section 7.2; if \(\gamma \neq \alpha / 2\), this interval is not symmetric about \(\bar{x}\). The width of this interval is \(w=s\left(z_{\gamma}+z_{\alpha-\gamma}\right) / \sqrt{n}\). Show that \(w\) is minimized for the choice \(\gamma=\alpha / 2\), so that the symmetric interval is the shortest. [Hints: (a) By definition of \(z_{a}, \Phi\left(z_{a}\right)=1-\alpha\), so that \(z_{a}=\Phi^{-1}(1-\alpha)\); (b) the relationship between the derivative of a function \(y=f(x)\) and the inverse function \(x=f^{-1}(y)\) is \((d / d y)\) \(\left.f^{-1}(y)=1 / f^{\prime}(x) .\right]\)

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)?

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a continuous probability distribution having median \(\tilde{\mu}\) (so that \(P\left(X_{i} \leq \tilde{\mu}\right)=\) \(\left.P\left(X_{i} \geq \tilde{\mu}\right)=.5\right)\) a. Show that $$ P\left(\min \left(X_{i}\right)<\tilde{\mu}<\max \left(X_{i}\right)\right)=1-\left(\frac{1}{2}\right)^{n-1} $$ so that \(\left(\min \left(x_{i}\right)\right.\), \(\left.\max \left(x_{i}\right)\right)\) is a \(100(1-\alpha) \%\) confidence interval for \(\tilde{\mu}\) with \(\alpha=\left(\frac{1}{2}\right)^{n-1}\). [Hint: The complement of the event \(\left\\{\min \left(X_{i}\right)<\tilde{\mu}<\max \left(X_{i}\right)\right\\}\) is \(\left\\{\max \left(X_{i}\right) \leq \tilde{\mu}\right\\} \cup\) \(\left\\{\min \left(X_{i}\right) \geq \tilde{\mu}\right\\}\). But \(\max \left(X_{i}\right) \leq \tilde{\mu}\) iff \(X_{i} \leq \tilde{\mu}\) for all \(i\).] b. For each of six normal male infants, the amount of the amino acid alanine \((\mathrm{mg} / 100 \mathrm{~mL})\) was determined while the infants were on an isoleucine-free diet, resulting in the following data: \(\begin{array}{llllll}2.84 & 3.54 & 2.80 & 1.44 & 2.94 & 2.70\end{array}\) Compute a \(97 \%\) CI for the true median amount of alanine for infants on such a diet ("The Essential Amino Acid Requirements of Infants," Amer. J. Nutrition, 1964 : \(322-330)\) c. Let \(x_{(2)}\) denote the second smallest of the \(x_{1} \mathrm{~s}\) and \(x_{(\mathrm{n}-1)}\) denote the second largest of the \(x_{i}\) s. What is the confidence coefficient of the interval \(\left(x_{(2)}, x_{(x-1)}\right)\) for \(\tilde{\mu}\) ?

A study of the ability of individuals to walk in a straight line ("Can We Really Walk Straight?" Amer. J. of Physical Anthro., 1992: 19-27) reported the accompanying data on cadence (strides per second) for a sample of \(n=20\) randomly selected healthy men. \(\begin{array}{rrrrrrrrrr}.95 & .85 & .92 & .95 & .93 & .86 & 1.00 & .92 & .85 & .81 \\ .78 & .93 & .93 & 1.05 & .93 & 1.06 & 1.06 & .96 & .81 & .96\end{array}\) A normal probability plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from MINITAB follows: \(\begin{array}{lccccc}\text { Variable } \mathbb{N} & \text { Mean } & \text { Median } & \text { TrMean } & \text { StDev } & \text { SEMean } \\ \text { cadence } 20 & 0.9255 & 0.9300 & 0.9261 & 0.0809 & 0.0181 \\ \text { Variable } & \text { Min } & \text { Max } & \text { Q1 } & \text { Q3 } & \\ \text { cadence } & 0.7800 & 1.0600 & 0.8525 & 0.9600 & \end{array}\) a. Calculate and interpret a \(95 \%\) confidence interval for population mean cadence. b. Calculate and interpret a \(95 \%\) prediction interval for the cadence of a single individual randomly selected from this population. c. Calculate an interval that includes at least \(99 \%\) of the cadences in the population distribution using a confidence level of \(95 \%\).

It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work in temperatures of \(200-500^{\circ} \mathrm{F}\). In a test of one type of mask, 11 of 55 masks had lenses pop out at \(250^{\circ}\). Construct a \(90 \%\) CI for the true proportion of masks of this type whose lenses would pop out at \(250^{\circ}\).
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