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Chapter 7: Problem 23

The article "An Evaluation of Football Helmets Under Impact Conditions" (Amer. J. Sports Medicine, 1984: 233-237) reports that when each football helmet in a random sample of 37 suspension-type helmets was subjected to a certain impact test, 24 showed damage. Let \(p\) denote the proportion of all helmets of this type that would show damage when tested in the prescribed manner. a. Calculate a \(99 \%\) CI for \(p\). b. What sample size would be required for the width of a \(99 \%\) CI to be at most .10, irrespective of \(\hat{p}\) ?

Short Answer

Expert verified
a. The 99% CI for \( p \) is (0.4478, 0.8494). b. The required sample size is 665.

Step by step solution

01

Identify Given Values

We have a sample size of 37 helmets and 24 of these showed damage. We denote the sample proportion of damaged helmets as \( \hat{p} \). First, calculate \( \hat{p} \): \( \hat{p} = \frac{24}{37} \approx 0.6486 \).
02

Calculate Standard Error

To find the confidence interval, the standard error (SE) for the sample proportion \( \hat{p} \) is needed. Calculate it using the formula: \( SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } = \sqrt{ \frac{0.6486 \times 0.3514}{37} } \approx 0.078 \).
03

Determine Z-score for 99% Confidence Level

For a 99% confidence interval, the Z-score for two-tailed approximations from the standard normal distribution is approximately 2.576. This can be found from Z-tables or standard normal distribution calculators.
04

Calculate Confidence Interval

The confidence interval for \( p \) is given by \( \hat{p} \pm Z \times SE \). Substitute the values: \( 0.6486 \pm 2.576 \times 0.078 \approx (0.4478, 0.8494) \). Thus, the 99% confidence interval for \( p \) is from 0.4478 to 0.8494.
05

Formulate Width Requirement Condition

For the required sample size calculation, we want the width of 99% CI for \( p \) to be at most 0.10. The width of the CI is twice the margin of error \( ME \), hence \( 2 \times (Z \times SE) \leq 0.10 \).
06

Equation for Sample Size

Reformulating margin of error condition, \( Z \times \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } = 0.05 \). Assume \( \hat{p} = 0.5 \) for maximum variability. Substitute \( Z = 2.576 \) and solve for \( n \): \( 2.576 \times \sqrt{ \frac{0.25}{n} } = 0.05 \).
07

Calculate Required Sample Size

Solve the equation \( 2.576 \times \sqrt{ \frac{0.25}{n} } = 0.05 \) for \( n \). This simplifies to \( 2.576 \times \sqrt{ 0.25 } = 0.05 \times \sqrt{ n} \). Thus, \( n = \left( \frac{2.576}{0.05} \right)^2 \times 0.25 \approx 665 \). Therefore, a sample size of at least 665 is needed.

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Key Concepts

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

Sample Size Calculation
Calculating the required sample size is crucial to ensure that our confidence intervals are precise and reliable. Here, we focus on adjusting the sample size to achieve a desired width for the confidence interval. For instance, to ensure that the confidence interval is no wider than 0.10 with 99% confidence, we need to calculate adequate sample size, typically larger than original samples.The width of the confidence interval is related to the margin of error, which is controlled by the formula: \(2 \times (Z \times \text{Standard Error})\). To adjust the margin of error to 0.05 (since 0.05 * 2 equals 0.10), the formula becomes:
  • \(Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = 0.05\)
Here, \(\hat{p}\) is assumed to be 0.5, enforcing maximum variability, making the formula setup appropriate for calculating the conservative sample size.By substituting the values, using the Z-score of 2.576 for a 99% confidence level, you equate the margin of error to 0.05 and solve for \(n\) by algebraically manipulating the equation. Ultimately, it results in requiring over 665 samples to ensure the interval's width is satisfactory.
Standard Error of Proportion
The standard error (SE) of a proportion specifies how much an observed sample proportion likely varies around the true population proportion. It's pivotal when building confidence intervals, providing insight into the variability of the sample rate.To compute the standard error of a sample proportion \(\hat{p}\), use the formula:
  • \(SE = \sqrt{\frac{\hat{p} \times (1 - \hat{p})}{n}}\)
Applying the formula with \(\hat{p} = 0.6486\) and \(n = 37\), you get:
  • \(SE = \sqrt{\frac{0.6486 \times 0.3514}{37}} \approx 0.078\)
This value reflects the variability expected in repeated sample collections of size 37 under similar conditions. A smaller SE indicates that the sample proportion reliably estimates the actual population proportion.
Z-score for Confidence Intervals
The Z-score is an essential component when constructing confidence intervals for proportions, specifically telling you how far away the limits of the interval should be from the sample mean to achieve your confidence level.For a 99% confidence level, the Z-score is approximately 2.576, derived or found from the Z-distribution table, typically representing the number of standard errors you expect the population proportion to fall within this range of the sample mean.The numerical representation of Z is crucial because it determines the width of the confidence interval in tandem with the standard error. Substituting this Z-score into the formula:
  • Confidence Interval = \(\hat{p} \pm Z \times SE\)
shows, with \(\hat{p} = 0.6486\) and \(SE\) at 0.078, the confidence interval spans from about 0.4478 to 0.8494. This Z-score confirms the interval is the correct size, which reflects our level of certainty for the estimated parameter \(p\) across repeated samples.

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Most popular questions from this chapter

A sample of 25 pieces of laminate used in the manufacture of circuit boards was selected and the amount of warpage (in.) under particular conditions was determined for each piece, resulting in a sample mean warpage of \(.0635\) and a sample standard deviation of .0065. a. Calculate a prediction for the amount of warpage of a single piece of laminate in a way that provides information about precision and reliability. b. Calculate an interval for which you can have a high degree of confidence that at least \(95 \%\) of all pieces of laminate result in amounts of warpage that are between the two limits of the interval.

The article "Gas Cooking, Kitchen Ventilation, and Exposure to Combustion Products" (Indoor Air, 2006: 65-73) reported that for a sample of 50 kitchens with gas cooking appliances monitored during a one-week period, the sample mean \(\mathrm{CO}_{2}\) level (ppm) was 654.16, and the sample standard deviation was \(164.43\). a. Calculate and interpret a \(95 \%\) (two-sided) confidence interval for true average \(\mathrm{CO}_{2}\) level in the population of all homes from which the sample was selected. b. Suppose the investigators had made a rough guess of 175 for the value of \(s\) before collecting data. What sample size would be necessary to obtain an interval width of \(50 \mathrm{ppm}\) for a confidence level of \(95 \%\) ?

The charge-to-tap time (min) for a carbon steel in one type of open hearth furnace was determined for each heat in a sample of size 46 , resulting in a sample mean time of \(382.1\) and a sample standard deviation of \(31.5\). Calculate a \(95 \%\) upper confidence bound for true average charge-to-tap time.

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 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).
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