/*! 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 73 Suppose the expected tensile str... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose the expected tensile strength of type-A steel is \(105 \mathrm{ksi}\) and the standard deviation of tensile strength is \(8 \mathrm{ksi}\). For type-B steel, suppose the expected tensile strength and standard deviation of tensile strength are \(100 \mathrm{ksi}\) and \(6 \mathrm{ksi}\), respectively. Let \(\bar{X}=\) the sample average tensile strength of a random sample of 40 type-A specimens, and let \(\bar{Y}=\) the sample average tensile strength of a random sample of 35 type-B specimens. a. What is the approximate distribution of \(\bar{X}\) ? Of \(\bar{Y}\) ? b. What is the approximate distribution of \(\bar{X}-\bar{Y}\) ? Justify your answer. c. Calculate (approximately) \(P(-1 \leq \bar{X}-\bar{Y} \leq 1)\). d. Calculate \(P(\bar{X}-\bar{Y} \geq 10\) ). If you actually observed \(\bar{X}-\bar{Y} \geq\) 10 , would you doubt that \(\mu_{1}-\mu_{2}=5\) ?

Short Answer

Expert verified
\( \bar{X} \) and \( \bar{Y} \) are normally distributed. \( \bar{X} - \bar{Y} \) is \( N(5, 1.62) \). Probabilities for questions (c) and (d) are very small.

Step by step solution

01

Identify the Distribution of Sample Means

For both types of steel, since we are dealing with sample means from large random samples (40 for type A and 35 for type B), the Central Limit Theorem tells us that the sample means \( \bar{X} \) and \( \bar{Y} \) will be approximately normally distributed. Specifically, \( \bar{X} \sim N(105, \frac{8}{\sqrt{40}}) \) and \( \bar{Y} \sim N(100, \frac{6}{\sqrt{35}}) \).
02

Find the Distribution of \( \bar{X} - \bar{Y} \)

To find the distribution of \( \bar{X} - \bar{Y} \), we need to consider the means and variances. Given that \( \bar{X} \sim N(105, \frac{8}{\sqrt{40}}) \) and \( \bar{Y} \sim N(100, \frac{6}{\sqrt{35}})\), we have \( \bar{X} - \bar{Y} \sim N(105 - 100, \sqrt{\frac{8^2}{40} + \frac{6^2}{35}})\). This simplifies to \( \bar{X} - \bar{Y} \sim N(5, \sqrt{1.6 + 1.03}) \).
03

Calculate the Variance and Standard Deviation of \( \bar{X} - \bar{Y} \)

Calculate the combined variance: \( 1.6 + 1.03 = 2.63 \). The standard deviation is \( \sqrt{2.63} \approx 1.62 \). Thus, \( \bar{X} - \bar{Y} \sim N(5, 1.62) \).
04

Calculate Probability \( P(-1 \leq \bar{X} - \bar{Y} \leq 1) \)

First, standardize the variable using \( Z = \frac{\bar{X} - \bar{Y} - 5}{1.62} \). For \( \bar{X} - \bar{Y} = -1 \), \( Z = \frac{-1 - 5}{1.62} \approx -3.70 \). For \( \bar{X} - \bar{Y} = 1 \), \( Z = \frac{1 - 5}{1.62} \approx -2.47 \). Use the standard normal distribution table to find \( P(-3.70 \leq Z \leq -2.47) \). This probability is very small, practically close to zero.
05

Calculate Probability \( P(\bar{X} - \bar{Y} \geq 10) \)

For \( \bar{X} - \bar{Y} = 10 \), standardize using \( Z = \frac{10 - 5}{1.62} = 3.09 \). Find \( P(Z \geq 3.09) \) using the standard normal table. This probability is also very small, suggesting observing \( \bar{X} - \bar{Y} \geq 10 \) is improbable.

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Key Concepts

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

Normal Distribution
In statistics, a normal distribution is a fundamental concept often referred to as a "bell curve" due to its distinctive shape. It is a continuous probability distribution characterized by its mean and standard deviation.
The mean determines the center of the distribution, while the standard deviation dictates the spread or width. Most values are clustered around the mean, with fewer values trailing off symmetrically on either side. The Central Limit Theorem (CLT) plays a crucial role in connecting normal distribution to real-world scenarios. It states that the distribution of sample means will tend to be normal or nearly normal as the sample size becomes larger, even if the underlying data distribution is not normal. This is why in our exercise, even with different sample sizes of 40 and 35, the sample means (\(\bar{X}\) and \(\bar{Y}\)) were approximately normally distributed.
Tensile Strength
Tensile strength is a measure of how much stress a material can withstand while being stretched or pulled before breaking. It is a crucial property for materials that are expected to endure forces that may cause stretching, like steel.
In the given problem, tensile strength has been measured for two types of steel: Type A and Type B.
  • Type A has an expected tensile strength of 105 ksi.
  • Type B has an expected tensile strength of 100 ksi.
Understanding tensile strength helps manufacturers ensure materials meet specifications needed for safety and performance, particularly in construction or manufacturing. The values are usually found by conducting experiments on samples under controlled conditions, resulting in statistical data like mean and standard deviation which describe the material's characteristics.
Sample Variance
Sample variance is a measure of how much values in a data set vary. It provides key insights into the distribution and reliability of data. In statistical terms, it is the square of the standard deviation and it indicates the extent to which each number in the dataset differs from the mean.
In our exercise, we estimate the variance of sample means to draw conclusions. Sample variance helps in analyzing how spread out the observations are. For instance, when comparing tensile strengths of steel samples, calculating each sample's variance gives us a broader picture of the structural consistency. To compute sample variance: - Determine the mean of the sample. - Subtract the mean from each data point and square the result. - Sum these squared differences. - Divide by the number of data points minus one.
Standard Deviation
Standard deviation is a statistical metric that indicates the amount of variation or dispersion in a set of values. A low standard deviation implies data points are clustered near the mean, while a high standard deviation indicates a wider spread.
In the exercise, a standard deviation of 8 ksi for Type A steel and 6 ksi for Type B provides insight into the consistency of tensile strengths. Calculating standard deviation involves: - Computing the mean of the data set. - Subtracting the mean from each data point, then squaring the result to remove negatives. - Finding the average of these squared differences. - Taking the square root of this average results in the standard deviation. Understanding standard deviation is essential as it helps with:
  • Assessing risk in fields like quality control.
  • Comparing consistency among different datasets.
  • Determining the impact of changes in individual measurements.

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