/*! 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 4 A CI is desired for the true ave... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A CI is desired for the true average stray-load loss \(\mu\) (watts) for a certain type of induction motor when the line current is held at 10 amps for a speed of \(1,500 \mathrm{rpm}\). Assume that stray-load loss is normally distributed with \(\sigma=3.0\). a. Compute a \(95 \%\) CI for \(\mu\) when \(n=25\) and \(\bar{x}=58.3 .\) b. Compute a \(95 \%\) CI for \(\mu\) when \(n=100\) and \(\pi=58.3\). c. Compute a \(99 \%\) CI for \(\mu\) when \(n=100\) and \(\pi=58.3\) d. Compute an \(82 \%\) CI for \(\mu\) when \(n=100\) and \(\bar{x}=58.3 .\) e. How large must \(n\) be if the width of the \(99 \%\) interval for \(\mu\) is to be \(1.0\) ?

Short Answer

Expert verified
The CIs are: (57.124, 59.476), (57.712, 58.888), (57.527, 59.073), and (57.898, 58.702). For part e, n must be at least 240.

Step by step solution

01

Understanding the Confidence Interval Formula

To find a confidence interval (CI) for the mean, use the formula: \[ \text{CI} = \bar{x} \pm z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right) \] where \( \bar{x} \) is the sample mean, \( \sigma \) is the population standard deviation, \( n \) is the sample size, and \( z_{\alpha/2} \) is the z-score corresponding to the desired level of confidence.
02

Computing 95% CI for n=25

For part (a), the confidence level is 95% and \( z_{0.025} \approx 1.96 \). Use \( n = 25 \), \( \bar{x} = 58.3 \) and \( \sigma = 3.0 \). Calculate: \[ \text{CI} = 58.3 \pm 1.96 \left(\frac{3.0}{\sqrt{25}}\right) = 58.3 \pm 1.176 \] This gives the interval \( (57.124, 59.476) \).
03

Computing 95% CI for n=100

For part (b), use the same confidence level of 95% and \( z_{0.025} \approx 1.96 \). Now \( n = 100 \). Calculate: \[ \text{CI} = 58.3 \pm 1.96 \left(\frac{3.0}{\sqrt{100}}\right) = 58.3 \pm 0.588 \] This gives the interval \( (57.712, 58.888) \).
04

Computing 99% CI for n=100

For part (c), the confidence level is 99% and \( z_{0.005} \approx 2.576 \). Use the same \( n = 100 \). Calculate: \[ \text{CI} = 58.3 \pm 2.576 \left(\frac{3.0}{\sqrt{100}}\right) = 58.3 \pm 0.773 \] This gives the interval \( (57.527, 59.073) \).
05

Computing 82% CI for n=100

For part (d), the confidence level is 82% and \( z_{0.09} \approx 1.34 \). Use \( n = 100 \). Calculate: \[ \text{CI} = 58.3 \pm 1.34 \left(\frac{3.0}{\sqrt{100}}\right) = 58.3 \pm 0.402 \] This gives the interval \( (57.898, 58.702) \).
06

Determining Sample Size for Narrow CI

For part (e), the desired width of a 99% CI is 1, so we calculate \( n \). Use \( \text{width} = 2 \times z_{0.005} \left(\frac{\sigma}{\sqrt{n}}\right) = 1 \). Rearrange: \[ n = \left(\frac{2 \times z_{0.005} \times \sigma}{\text{width}}\right)^2 = \left(\frac{2 \times 2.576 \times 3.0}{1}\right)^2 \] Calculate: \[ n = \left(15.456\right)^2 = 239.82 \] Since \( n \) must be a whole number, round up to 240.

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.

Sample Size Determination
Sample size determination is an essential part of statistical analysis, especially when constructing confidence intervals. It's crucial to understand how the size of a sample affects the reliability of your estimates. In the provided exercise, part (e) demonstrates how to calculate the required sample size for a given confidence interval width. This is particularly important when you need a more precise estimate of the mean under a specific confidence level.
  • The formula for determining sample size, given a desired confidence interval width, is: \[ n = \left(\frac{2 \times z_{\alpha/2} \times \sigma}{\text{width}}\right)^2 \]
  • This formula allows us to ensure that the confidence interval is narrow enough to be useful for precise decision-making.
  • In our example, the desire was for a 99% confidence interval with a width of 1. By rearranging the formula, we calculated that the necessary sample size is 240. This assures that the interval is accurate and reflects the real average stray-load loss closely, assuming the sample follows a normal distribution.
Sample size determination often depends on several factors:
  • Desired width of the confidence interval.
  • The confidence level, which dictates the critical z-value.
  • The population standard deviation, which influences how narrow the interval can become with a given sample size.
Understanding the right sample size helps in both designing experiments and interpreting the results effectively.
Normal Distribution
The concept of normal distribution is pivotal in creating confidence intervals. The normal distribution, with its characteristic bell-shaped curve, is defined by two parameters: the mean and the standard deviation. A key feature of the normal distribution is that it is symmetric around the mean.
In the context of the exercise, we assume that the stray-load loss is normally distributed, which means we can use properties of this distribution to extrapolate data about the population from our sample.
  • With a known standard deviation, we use the central limit theorem, which posits that, given a sufficiently large sample size, the sampling distribution of the sample mean will be approximately normal regardless of the distribution of the population.
  • This allows us to apply the z-score in conjunction with our sample mean to calculate confidence intervals.
The normal distribution is foundational in statistics due to:
  • Its natural occurrence in many real-world phenomena.
  • The ease with which it lends itself to analysis using various statistical tools.
  • Providing a framework for making inferences about a population from sample data.
This model makes it easier to compute probabilities and percentages within various segments of the data distribution, which is why it’s such a powerful tool in statistical analysis. Understanding how data conforms to this distribution enables us to trust these intervals and make informed decisions.
z-Score Calculation
The z-score calculation is a fundamental part of determining confidence intervals. A z-score indicates how many standard deviations an element is from the mean. When calculating confidence intervals, the z-score helps quantify the margin of error around the sample mean, allowing us to understand the range within which the population mean likely falls.
In the exercise, the z-score reflects the confidence level chosen for the interval:
  • A 95% confidence level uses a z-score of approximately 1.96.
  • A 99% confidence level requires a higher z-score of about 2.576 to increase the range and thus the likelihood that the population parameter falls within it.
The formula to find a confidence interval for the mean relies on this calculation:
  • \[ \text{CI} = \bar{x} \pm z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right) \]
  • Where \(\bar{x}\) is the sample mean, \(\sigma\) is the standard deviation, and \(n\) is the sample size.
To effectively use z-scores:
  • Understand that they are crucial for translating a confidence level into a margin of error.
  • Adjust the z-score to account for the desired level of confidence and the variability in your data.
  • Use it in conjunction with the sample and population parameters to calculate your confidence intervals accurately.
Z-scores make it simple to compute how many standard deviations away our interval limits are from the estimated mean, thus helping in accurately determining the probable range for the population mean based on the sampled data.

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

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.

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.

Here is a sample of ACT scores (average of the Math, English, Social Science, and Natural Science scores) for students taking college freshman calculus: \(\begin{array}{lllllll}24.00 & 28.00 & 27.75 & 27.00 & 24.25 & 23.50 & 26.25 \\\ 24.00 & 25.00 & 30.00 & 23.25 & 26.25 & 21.50 & 26.00 \\ 28.00 & 24.50 & 22.50 & 28.25 & 21.25 & 19.75 & \end{array}\) a. Using an appropriate graph, see if it is plausible that the observations were selected from a normal distribution. b. Calculate a two-sided \(95 \%\) confidence interval for the population mean. c. The university ACT average for entering freshmen that year was about 21. Are the calculus students better than average, as measured by the ACT?

The article "An Evaluation of Football Helmets Under Impact Conditions" (Amer.J. Sports Med., 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 \(\hat{p}\) ?

The superintendent of a large school district, having once had a course in probability and statistics, believes that the number of teachers absent on any given day has a Poisson distribution with parameter \(\lambda\). Use the accompanying data on absences for 50 days to derive a large-sample CI for \(\lambda\). [Hint: The mean and variance of a Poisson variable both cqual \(\lambda, s 0\) $$ Z=\frac{X-\lambda}{\sqrt{\lambda / n}} $$ has approximately a standard normal distribution. Now proceed as in the derivation of the interval for \(p\) by making a probability statement (with probability \(1-\alpha\) ) and solving the resulting inequalities for \(\lambda\) (see the argument just after \((8.10))\) ]. \begin{tabular}{l|lll} Number of absences & 0 & 1 & 2 \\ \hline Frequency & 1 & 4 & 8 \end{tabular}

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.