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Chapter 10: Problem 104

An experimenter wishes to obtain a CI for the difference between true average breaking strength for cables manufactured by company I and by company II. Suppose breaking strength is normally distributed for both types of cable with \(\sigma_{1}=30 \mathrm{psi}\) and \(\sigma_{2}=20 \mathrm{psi}\). a. If costs dictate that the sample size for the type I cable should be three times the sample size for the type II cable, how many observations are required if the \(99 \%\) CI is to be no wider than \(20 \mathrm{psi}\) ? b. Suppose a total of 400 observations is to be made. How many of the observations should be made on type I cable samples if the width of the resulting interval is to be a minimum?

Short Answer

Expert verified
a) 54 type I and 18 type II observations; b) 300 type I observations.

Step by step solution

01

Define Variables and Concepts

We need to determine the sample sizes for both types of cables. Let \( n_1 \) be the sample size for type I and \( n_2 \) for type II. We know from the problem that \( n_1 = 3n_2 \) and we need to find the confidence interval (CI) width for the difference in means.
02

Confidence Interval Formula

The formula for a confidence interval (CI) for the difference of two population means is given by: \[ CI = \left( \bar{x}_1 - \bar{x}_2 \right) \pm z_{\alpha/2} \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \]where \( z_{\alpha/2} \) is the Z-value for the 99% confidence level.
03

Set CI Width Equal to Desired Width

The problem states that the CI width should not be wider than 20 psi. Therefore, we set:\[ 2z_{\alpha/2} \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \leq 20 \]For 99% CI, \( z_{\alpha/2} \approx 2.576 \) (from Z-table).
04

Simplify the Equation

Substitute the given \( \sigma_1 = 30 \) psi, \( \sigma_2 = 20 \) psi, and \( z_{\alpha/2} = 2.576 \):\[ 2(2.576) \sqrt{\frac{30^2}{3n_2} + \frac{20^2}{n_2}} \leq 20 \]Simplify the inequality to find \( n_2 \):\[ \sqrt{\frac{900}{3n_2} + \frac{400}{n_2}} \leq \frac{10}{2.576} \]
05

Solve for Sample Size for Type II

Calculate \( n_2 \) from the simplified inequality:\[ 900/(3n_2) + 400/n_2 \leq (\frac{10}{2.576})^2 \]Multiply through by \( 3n_2 \) to clear the fractions and solve for \( n_2 \). This results in a quadratic equation in \( n_2 \). Solving, we find \( n_2 \approx 18 \), so \( n_1 = 3n_2 = 54 \).
06

Total Observation Allocation for Part B

Given a total of 400 observations (\( n_1 + n_2 = 400 \)), express \( n_1 \) in terms of \( n_2 \):From \( n_1 = 3n_2 \), substitute into total:\[ 3n_2 + n_2 = 400 \]Solve for \( n_2 = 100 \). Therefore, \( n_1 = 300 \).

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Key Concepts

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

Sample Size Calculation
Determining the sample size is crucial to ensure that the results of a study are both reliable and valid. When a confidence interval is required for the difference in means between two groups, like breaking strengths of cables, the sample size calculation becomes even more important.

A larger sample size reduces variability, making the confidence interval narrower and ensuring results that truly reflect the population. In the original exercise, sample size for type I cables ( _1) is set as three times that of type II ( _2). By using the confidence interval formula for two means and setting the desired maximum width, we solve for these sample sizes. Adjusting the sample sizes according to constraints (such as cost), allows for optimal yet cost-effective sampling.

Understanding how to calculate sample size helps in planning resources efficiently, leading to better experimental outcomes.
Mean Difference
In experiments, especially those involving comparative studies like the cable strengths, the mean difference represents the average variation in a measured variable between two groups. In this case, it's the difference in breaking strength between cables from different companies.

Mean difference is central to understanding the result of an experiment, allowing researchers to determine if a significant difference exists between the two groups. It plays an integral role in building the confidence interval, which provides a range of values within which the true mean difference likely falls.

Taking mean differences into account guides decision-making processes by providing insight into the effectiveness or superiority of methods or products under study.
Normal Distribution
Normal distribution, often depicted as a bell curve, is a crucial concept in statistics. It suggests that data collected under normal conditions will have most values concentrated around the mean, with symmetric tails extending on either side.

In the exercise, it's assumed that the breaking strength follows a normal distribution for both cable types. This assumption allows us to use standard statistical methods like Z-tests or t-tests when analyzing the data. Normal distribution makes it easier to predict probabilities and calculate statistics such as variance and standard deviation.

Recognizing the normal distribution helps in choosing the right statistical tools and understanding data variability and reliability.
Z-value
The Z-value is a standard score that indicates how many standard deviations away from the mean a data point is. When constructing confidence intervals, the Z-value corresponds to the desired confidence level.

For instance, a 99% confidence interval will use a Z-value of approximately 2.576. This relationship enables us to find the range that covers a specific proportion of the data, creating a reliable confidence interval.

In the exercise, the Z-value is utilized to calculate the maximum allowable width of the confidence interval for the difference in means. Understanding Z-values in normal distribution contexts helps interpret and convey the reliability and significance of findings.

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Most popular questions from this chapter

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