/*! 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 64 A random sample of 50 suspension... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A random sample of 50 suspension helmets used by motorcycle riders and automobile race-car drivers was subjected to an impact test, and some damage was observed on 18 of these helmets. (a) Find a \(95 \%\) two-sided confidence interval on the true proportion of helmets that would show damage from this test. (b) Using the point estimate of \(p\) from the 50 helmets, how many helmets must be tested to be \(95 \%\) confident that the error in estimating \(p\) is less than \(0.02 ?\) (c) How large must the sample be if we wish to be at least \(95 \%\) confident that the error in estimating \(p\) is less than 0.02 regardless of the true value of \(p ?\)

Short Answer

Expert verified
(a) Confidence interval: 0.2259 to 0.4941. (b) Test at least 2215 helmets. (c) Test at least 2401 helmets.

Step by step solution

01

Find Point Estimate of Proportion

The point estimate \( \hat{p} \) of the true proportion \( p \) is calculated by dividing the number of damaged helmets by the total number of helmets tested. Since 18 helmets were damaged out of 50 tested, \[ \hat{p} = \frac{18}{50} = 0.36 \].
02

Calculate Standard Error for Proportion

The standard error (SE) of \( \hat{p} \) is found using the formula \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \], where \( n \) is the sample size. Plugging in the values, \[ SE = \sqrt{\frac{0.36 \times 0.64}{50}} \approx 0.0689 \].
03

Find Z-Score for 95% Confidence Interval

For a 95% confidence level, the Z-score \( Z \) is approximately 1.96 (from standard normal distribution tables).
04

Calculate Confidence Interval

The confidence interval (CI) is calculated by \[ \hat{p} \pm Z \times SE \]. Substituting the values: \[ 0.36 \pm 1.96 \times 0.0689 \]. This gives \[ (0.2259, 0.4941) \].
05

Calculate Sample Size for Desired Margin of Error

Using the formula \[ n = \left( \frac{Z}{E} \right)^2 \hat{p}(1-\hat{p}) \] to find the required sample size where the margin of error \( E = 0.02 \) and \( Z = 1.96 \). Substituting the values, \[ n = \left( \frac{1.96}{0.02} \right)^2 \times 0.36 \times 0.64 \approx 2214.56 \]. Rounding up, we require 2215 helmets.
06

Calculate Sample Size for Maximum Variability of p

To find the worst-case sample size (when \( p \) is unknown), use \( \hat{p} = 0.5 \) for maximum variability. Substituting into the formula, \[ n = \left( \frac{1.96}{0.02} \right)^2 \times 0.5 \times 0.5 \]. This results in \( n \approx 2401 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Point Estimate
The point estimate is a crucial concept in statistics as it provides a single best guess of a population parameter based on sample data. In this context, the point estimate, denoted as \( \hat{p} \), represents the estimated proportion of helmets that incur damage during impact testing.
For the given problem, there are 18 damaged helmets out of 50 tested, resulting in a point estimate of\[ \hat{p} = \frac{18}{50} = 0.36. \] This means we estimate that 36% of all helmets would be damaged under similar conditions.
Point estimates provide an initial assessment but are limited by not accounting for variability or uncertainty inherent in sampling; hence, they are often coupled with interval estimates like confidence intervals.
Sample Size Determination
Determining the right sample size is vital to ensure the accuracy and reliability of statistical estimates. Sample size affects the precision of confidence intervals and the margin of error you are willing to accept.
In our exercise, we calculated the needed sample size to achieve a specific confidence level of 95% and a desired margin of error less than 0.02 for our point estimate. Using the formula: \[ n = \left( \frac{Z}{E} \right)^2 \hat{p}(1-\hat{p}), \]
  • "\(Z\)" is the Z-score,
  • "\(E\)" is the margin of error,
  • "\(\hat{p}\)" is the point estimate.
By using \(\hat{p} = 0.36\) and applying the formula, the calculated sample size is 2215 helmets to achieve the desired accuracy. Furthermore, considering maximum variability by setting \(\hat{p} = 0.5\) for the worse-case scenario, about 2401 helmets would be needed, giving a broader assurance regardless of actual proportion values.
Standard Error
The standard error (SE) helps us understand the precision of the point estimate and how much it is expected to vary from one sample to another. It's a measure of the sampling distribution's spread.
In our case, the standard error of the proportion \( \hat{p} \) is calculated by the formula:\[SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\] where "\(n\)" is the sample size. Using the proportion \(\hat{p} = 0.36\) and sample size \(n = 50\), we find:\[ SE \approx 0.0689. \]
This indicates the extent to which the proportion estimate \(\hat{p}\) could fluctuate due to random sampling. By understanding SE, statisticians are better equipped to create reliable confidence intervals, acknowledging the inherent variability of sample estimates.
Z-Score
In statistical analysis, the Z-score is pivotal as it quantifies the number of standard deviations a data point is from the mean. It allows us to calculate confidence intervals with precision for normally distributed data.
For the confidence interval construction in our exercise, a Z-score of 1.96 was selected for a 95% confidence level, meaning we expect our interval to contain the true population parameter 95% of the time.
The Z-score is derived from standard normal distribution tables and varies according to the confidence level:
  • Common Z-scores include 1.96 for 95% confidence,
  • 2.58 for 99% confidence,
  • and 1.64 for 90% confidence.
Understanding and applying the Z-score correctly is instrumental in constructing reliable confidence intervals, allowing statisticians to account for variability in data points.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The diameter of holes for a cable harness is known to have a normal distribution with \(\sigma=0.01\) inch. A random sample of size 10 yields an average diameter of 1.5045 inch. Find a \(99 \%\) two-sided confidence interval on the mean hole diameter.

An article in the ASCE Journal of Energy Engineering ["Overview of Reservoir Release Improvements at 20 TVA Dams" (Vol. \(125,\) April \(1999,\) pp. \(1-17)]\) presents data on dissolved oxygen concentrations in streams below 20 dams in the Tennessee Valley Authority system. The observations are (in milligrams per liter): \(5.0,3.4,3.9,1.3,0.2,0.9,2.7,3.7,\) \(3.8,4.1,1.0,1.0,0.8,0.4,3.8,4.5,5.3,6.1,6.9,\) and 6.5. (a) Is there evidence to support the assumption that the dissolved oxygen concentration is normally distributed? (b) Find a \(95 \%\) CI on the mean dissolved oxygen concentration. (c) Find a \(95 \%\) prediction interval on the dissolved oxygen concentration for the next stream in the system that will be tested. (d) Find an interval that will contain \(95 \%\) of the values of the dissolved oxygen concentration with \(99 \%\) confidence. (e) Explain the difference in the three intervals computed in parts (b), (c), and (d).

Suppose that \(n=100\) random samples of water from a freshwater lake were taken and the calcium concentration (milligrams per liter) measured. A \(95 \%\) CI on the mean calcium concentration is \(0.49 \leq \mu \leq 0.82\). (a) Would a \(99 \%\) CI calculated from the same sample data be longer or shorter? (b) Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between 0.49 and \(0.82 .\) Is this statement correct? Explain your answer. (c) Consider the following statement: If \(n=100\) random samples of water from the lake were taken and the \(95 \% \mathrm{CI}\) on \(\mu\) computed, and this process were repeated 1000 times, 950 of the CIs would contain the true value of \(\mu\). Is this statement correct? Explain your answer.

The article "Mix Design for Optimal Strength Development of Fly Ash Concrete" (Cement and Concrete Research, 1989, Vol. \(19(4),\) pp. \(634-640\) ) investigates the compressive strength of concrete when mixed with fly ash (a mixture of silica, alumina, iron, magnesium oxide, and other ingredients). The compressive strength for nine samples in dry conditions on the 28 th day are as follows (in megapascals): \(\begin{array}{lllll}40.2 & 30.4 & 28.9 & 30.5 & 22.4 \\ 25.8 & 18.4 & 14.2 & 15.3 & \end{array}\) (a) Given the following probability plot of the data, what is a logical assumption about the underlying distribution of the data? (b) Find a \(99 \%\) lower one-sided confidence interval on mean compressive strength. Provide a practical interpretation of this interval. (c) Find a \(98 \%\) two-sided confidence interval on mean compressive strength. Provide a practical interpretation of this interval and explain why the lower end-point of the interval is or is not the same as in part (b). (d) Find a \(99 \%\) upper one-sided confidence interval on the variance of compressive strength. Provide a practical interpretation of this interval. (e) Find a \(98 \%\) two-sided confidence interval on the variance of compression strength. Provide a practical interpretation of this interval and explain why the upper end-point of the interval is or is not the same as in part (d). (f) Suppose that it was discovered that the largest observation 40.2 was misrecorded and should actually be 20.4 . Now the sample mean \(\bar{x}=23\) and the sample variance \(s^{2}=39.8\). Use these new values and repeat parts (c) and (e). Compare the original computed intervals and the newly computed intervals with the corrected observation value. How does this mistake affect the values of the sample mean, sample variance, and the width of the two-sided confidence intervals? (g) Suppose, instead, that it was discovered that the largest observation 40.2 is correct but that the observation 25.8 is incorrect and should actually be \(24.8 .\) Now the sample mean \(\bar{x}=25\) and the standard deviation \(s=8.41 .\) Use these new values and repeat parts (c) and (e). Compare the original computed intervals and the newly computed intervals with the corrected observation value. How does this mistake affect the values of the sample mean, the sample variance, and the width of the two-sided confidence intervals? (h) Use the results from parts (f) and (g) to explain the effect of mistakenly recorded values on sample estimates. Comment on the effect when the mistaken values are near the sample mean and when they are not.

An article in Technometrics (1999, Vol. 41, pp. 202- 211 ) studied the capability of a gauge by measuring the weight of paper. The data for repeated measurements of one sheet of paper are in the following table. Construct a \(95 \%\) one-sided upper confidence interval for the standard deviation of these measurements. Check the assumption of normality of the data and comment on the assumptions for the confidence interval. $$ \begin{array}{lllll} \hline &&{\text { Observations }} \\ \hline 3.481 & 3.448 & 3.485 & 3.475 & 3.472 \\ 3.477 & 3.472 & 3.464 & 3.472 & 3.470 \\ 3.470 & 3.470 & 3.477 & 3.473 & 3.474 \\ \hline \end{array} $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.