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Chapter 5: Problem 73

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
\( \bar{X} - \bar{Y} \sim N(5, 2.6286) \). \( P(-1 \leq \bar{X}-\bar{Y} \leq 1) \approx 0.0065 \). If \( \bar{X}-\bar{Y} \geq 10 \), \( \mu_1-\mu_2=5 \) might be incorrect.

Step by step solution

01

Identify the distributions of \( \bar{X} \) and \( \bar{Y} \)

The Central Limit Theorem (CLT) states that the distribution of the sample mean \( \bar{X} \) for sufficiently large sample sizes will be approximately normal. Here, the sample size for type-A steel is 40, which is large enough. Therefore, \( \bar{X} \sim N(\mu_X, \sigma_X^2/n)\), where \(\mu_X = 105\ ksi\) and \(\sigma_X = 8\ ksi\). For type-B steel with a sample size of 35, \( \bar{Y} \sim N(\mu_Y, \sigma_Y^2/m)\), where \(\mu_Y = 100\ ksi\) and \(\sigma_Y = 6\ ksi\).
02

Calculate variances and standard deviations of \( \bar{X} \) and \( \bar{Y} \)

The variance of \( \bar{X} \) is given by \( \frac{\sigma_X^2}{n} = \frac{8^2}{40} = 1.6 \ ksi^2\). Thus, the standard deviation is \( \sqrt{1.6} \approx 1.265 \ ksi\). For \( \bar{Y} \), the variance is \( \frac{\sigma_Y^2}{m} = \frac{6^2}{35} \approx 1.0286 \ ksi^2\) and the standard deviation is \( \sqrt{1.0286} \approx 1.014 \ ksi\).
03

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

Since \( \bar{X} \) and \( \bar{Y} \) are both normally distributed, their difference \( \bar{X} - \bar{Y} \) is also normally distributed. The mean is \( \mu_X - \mu_Y = 105 - 100 = 5 \ ksi\) and the variance is the sum of the variances: \( \frac{8^2}{40} + \frac{6^2}{35} = 1.6 + 1.0286 = 2.6286 \ ksi^2\). The standard deviation is \( \sqrt{2.6286} \approx 1.621 \ ksi\). Thus, \( \bar{X} - \bar{Y} \sim N(5, 2.6286)\).
04

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

This probability can be found by standardizing the interval and referring to the standard normal distribution table: \[ P\left(-1 \leq \bar{X}-\bar{Y} \leq 1\right) = P\left( \frac{-1-5}{1.621} \leq Z \leq \frac{1-5}{1.621} \right) \approx P(-3.700 \leq Z \leq -2.469) \]. Using standard normal tables or computational tools, find that \( P(-1 \leq \bar{X}-\bar{Y} \leq 1) \approx 0.0065 \).
05

Calculate \( P(\bar{X}-\bar{Y} \geq 10) \) and assess \( \mu_1-\mu_2=5 \)

We convert the inequality to a standard normal probability: \[ P(\bar{X}-\bar{Y} \geq 10) = P\left( Z \geq \frac{10-5}{1.621} \right) = P(Z \geq 3.084) \approx 0.0010 \]. If \( \bar{X} - \bar{Y} \geq 10 \) is observed, this event is very unlikely under the assumption \( \mu_1-\mu_2=5 \), suggesting that this assumption might be incorrect.

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Key Concepts

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

Normal Distribution
A normal distribution is a continuous probability distribution that is symmetrical around its mean. This specific shape is known as a bell curve. It's one of the most important concepts in statistics because it describes how the values of a variable are distributed. Many natural phenomena tend to follow a normal distribution.

Key properties of the normal distribution include:
  • The mean, median, and mode are all equal.
  • It is defined by two parameters: mean (\( \mu \)) and standard deviation (\( \sigma \)).
  • It asymptotically approaches the x-axis; that is, it never actually touches the axis.
In the context of the given exercise, the Central Limit Theorem assures us that the sample means for both types of steel are normally distributed, even if the tensile strength itself isn't. This is because the sample sizes are large enough.
Sample Mean
When dealing with sample data, the sample mean is a crucial statistic. It represents the average of a set of observations or data points from a sample.

The formula for calculating a sample mean is simple: sum up all the sample values and then divide by the number of observations:\[\bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i\]where \( n \) is the number of observations in the sample.

In the exercise, we compute the sample means \( \bar{X} \) and \( \bar{Y} \) which refer to the average tensile strengths of samples from type-A and type-B steel respectively. These means help us understand the typical tensile strength values we might expect from these types of steel.
Hypothesis Testing
Hypothesis testing is a statistical method used for making decisions about a parameter such as the mean of a population. It involves formulating a null hypothesis (\( H_0 \) ), which is a statement of no effect or no difference. Then, an alternative hypothesis (\( H_a \) ), which is what you aim to support or prove, is compared to decide the validity of \( H_0 \).

In our exercise, if we observe a difference of at least 10 ksi between the sample means, we must reconsider our assumption that the mean tensile strengths differ by only 5 ksi (\( \mu_1 - \mu_2 = 5 \)). Given the improbability (\( P \approx 0.001 \)) of observing such a large difference under the null hypothesis, we might conclude that the assumption about the population means could be incorrect.
Probability Calculation
Probability calculations involve determining the likelihood of an event occurring. In the context of a normal distribution, this is often done by standardizing the values, which helps to use standard normal distribution tables or computational tools.

For instance, to calculate \( P(-1 \leq \bar{X}-\bar{Y} \leq 1) \) in the exercise, we convert the expression into a standard normal form: \[ Z = \frac{x-\mu}{\sigma} \] where \( x \) is the value from the data, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. This transformation allows us to find probabilities using a standard normal distribution table. We determined that \( P(-1 \leq \bar{X}-\bar{Y} \leq 1) \approx 0.0065 \), indicating a very low probability for this range of differences between \( \bar{X} \) and \( \bar{Y} \).

This kind of calculation is crucial for understanding how often certain outcomes can occur under given assumptions.

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Most popular questions from this chapter

In cost estimation, the total cost of a project is the sum of component task costs. Each of these costs is a random variable with a probability distribution. It is customary to obtain information about the total cost distribution by adding together characteristics of the individual component cost distributions-this is called the "roll-up" procedure. For example, \(E\left(X_{1}+\cdots+X_{n}\right)=E\left(X_{1}\right)+\cdots+E\left(X_{n}\right)\), so the roll-up procedure is valid for mean cost. Suppose that there are two component tasks and that \(X_{1}\) and \(X_{2}\) are independent, normally distributed random variables. Is the roll-up procedure valid for the 75 th percentile? That is, is the 75 th percentile of the distribution of \(X_{1}+X_{2}\) the same as the sum of the 75 th percentiles of the two individual distributions? If not, what is the relationship between the percentile of the sum and the sum of percentiles? For what percentiles is the roll-up procedure valid in this case?

Manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal distribution, and the three times are independent of one another. The mean values are 15,30 , and 20 min, respectively, and the standard deviations are 1,2 , and \(1.5\) min, respectively. What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component?

You have two lightbulbs for a particular lamp. Let \(X=\) the lifetime of the first bulb and \(Y=\) the lifetime of the second bulb (both in \(1000 \mathrm{~s}\) of hours). Suppose that \(X\) and \(Y\) are independent and that each has an exponential distribution with parameter \(\lambda=1\). a. What is the joint pdf of \(X\) and \(Y\) ? b. What is the probability that each bulb lasts at most 1000 hours (i.e., \(X \leq 1\) and \(Y \leq 1)\) ? c. What is the probability that the total lifetime of the two bulbs is at most 2 ? [Hint: Draw a picture of the region \(A=\\{(x, y): x \geq 0, y \geq 0, x+y \leq 2\\}\) before integrating.] d. What is the probability that the total lifetime is between 1 and \(2 ?\)

A particular brand of dishwasher soap is sold in three sizes: \(25 \mathrm{oz}, 40 \mathrm{oz}\), and \(65 \mathrm{oz}\). Twenty percent of all purchasers select a 25 -oz box, \(50 \%\) select a 40 -oz box, and the remaining \(30 \%\) choose a \(65-\mathrm{oz}\) box. Let \(X_{1}\) and \(X_{2}\) denote the package sizes selected by two independently selected purchasers. a. Determine the sampling distribution of \(\bar{X}\), calculate \(E(\bar{X})\), and compare to \(\mu\). b. Determine the sampling distribution of the sample variance \(S^{2}\), calculate \(E\left(S^{2}\right)\), and compare to \(\sigma^{2}\).

The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of \(500 \mathrm{psi}\). a. What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200 ? b. If the sample size had been 15 rather than 40 , could the probability requested in part (a) be calculated from the given information?
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