/*! 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 60 The result in Exercise 7.58 hold... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The result in Exercise 7.58 holds even if the sample sizes differ. That is, if \(X_{1}, X_{2}, \ldots, X_{n_{1}}\) and \(Y_{1}, Y_{2}, \ldots, Y_{n_{2}}\) constitute independent random samples from populations with means \(\mu_{1}\) and \(\mu_{2}\) and variances \(\sigma_{1}^{2}\) and \(\sigma_{2}^{2}\), respectively, then \(\bar{X}-\bar{Y}\) will be approximately normally distributed, for large \(n_{1}\) and \(n_{2},\) with mean \(\mu_{1}-\mu_{2}\) and variance \(\left(\sigma_{1}^{2} / n_{1}\right)+\left(\sigma_{2}^{2} / n_{2}\right)\).The flow of water through soil depends on, among other things, the porosity (volume proportion of voids) of the soil. To compare two types of sandy soil, \(n_{1}=50\) measurements are to be taken on the porosity of soil \(\mathrm{A}\) and \(n_{2}=100\) measurements are to be taken on soil \(\mathrm{B}\). Assume that \(\sigma_{1}^{2}=.01\) and \(\sigma_{2}^{2}=.02 .\) Find the probability that the difference between the sample means will be within. 05 unit of the difference between the population means \(\mu_{1}-\mu_{2}\).

Short Answer

Expert verified
The probability is approximately 0.9876.

Step by step solution

01

Determine Distribution and Parameters

The difference \( \bar{X} - \bar{Y} \) follows a normal distribution approximately, with mean \(\mu_1 - \mu_2\) and variance \(\left(\frac{\sigma_1^2}{n_1}\right) + \left(\frac{\sigma_2^2}{n_2}\right)\). Let's calculate the variance. Given \(\sigma_1^2 = 0.01\), \(\sigma_2^2 = 0.02\), \(n_1 = 50\), and \(n_2 = 100\): \[\text{Variance} = \left(\frac{0.01}{50}\right) + \left(\frac{0.02}{100}\right) = 0.0002 + 0.0002 = 0.0004.\]
02

Determine the Standard Deviation

The standard deviation of \(\bar{X}-\bar{Y}\) is the square root of the variance. Let's calculate it:\[\text{Standard Deviation} = \sqrt{0.0004} = 0.02.\]
03

Define the Range of Interest

We are interested in the probability that the difference \(\bar{X} - \bar{Y}\) is within 0.05 units of \(\mu_1 - \mu_2\). This means we are looking at the interval \((\mu_1 - \mu_2) - 0.05\) to \((\mu_1 - \mu_2) + 0.05\).
04

Use the Standard Normal Distribution

Standardize the problem by using a Z-score transformation. We calculate the probability that a standard normal variable is within\[\frac{0.05}{0.02} = 2.5\] units of the mean (0). This implies calculating:\[P(-2.5 < Z < 2.5)\] where \(Z\) is a standard normal variable.
05

Calculate the Probability

Use the standard normal distribution table or a calculator to find:\[P(-2.5 < Z < 2.5) = P(Z < 2.5) - P(Z < -2.5)\]Since the normal distribution is symmetric, \(P(Z < -2.5) = 1 - P(Z < 2.5)\).\[P(Z < 2.5) \approx 0.9938\]\[P(Z < -2.5) \approx 0.0062\]So:\[P(-2.5 < Z < 2.5) = 0.9938 - 0.0062 = 0.9876\]
06

Interpret the Result

The probability that the difference between the sample means \(\bar{X} - \bar{Y}\) is within 0.05 units of the difference between the population means \(\mu_1 - \mu_2\) is approximately 0.9876.

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.

Normal Distribution
The normal distribution is one of the most important concepts in statistics. It's a type of continuous probability distribution that is symmetrical around its mean. Picture it like a bell-shaped curve, where most of the observations cluster around the central peak and the probabilities for values taper off equally on both sides. This is why it's also known as the "bell curve." A key feature of a normal distribution is that it is defined by two parameters: the mean (average) and the standard deviation (spread or width of the curve).

In statistics, many real-world variables, like test scores or measurement errors, tend to follow a normal distribution. The significance of the normal distribution also arises because of the Central Limit Theorem. This theorem states that, given a large enough sample size, the sampling distribution of the sample mean will be approximately normal, no matter the distribution of the original population. This creates a foundation for making inferences about population parameters using sample data.

Understanding the normal distribution helps in predicting outcomes and verifying assumptions in various statistical analyses.
Sample Means
The sample mean is a statistical measure representing the average or central value of a set of data from a sample. It's denoted as \( \bar{X} \) for the mean of the X-values and comes into play critically in statistics when making inferences about a population. The sample mean offers a snapshot that helps in estimating the population mean.

Imagine you've measured the porosity of soil from two different samples; the average value of these measurements is your sample mean. It's an essential step since measuring the entire population might be impractical, so a sample helps predict and understand the overall behavior of the population.

A notable property of sample means, as inferred by the Central Limit Theorem, is that the distribution of these sample means tends to be normally distributed as the sample size becomes large enough. This characteristic is crucial when calculating probabilities and constructing confidence intervals.
Variance Calculation
Variance is a measure of how data points differ from the mean. It signifies the degree to which each number in a dataset varies around the mean. The formula for variance when dealing with sample data is:
  • Population variance: \( \sigma^2 = \frac{\sum (X_i - \mu)^2}{N} \)
  • Sample variance: \( s^2 = \frac{\sum (X_i - \bar{X})^2}{n-1} \)
Here, \( X_i \) is each data point, \( \mu \) is the population mean, \( \bar{X} \) is the sample mean, \( N \) is the number of data points in the population, and \( n \) is the sample size.

In our context of comparing two soil types, variance tells us how spread out the porosity measurements are within each sample. By comparing these variances, we can understand differences between the two types of soil more effectively.

Lower variance means data points tend to be very close to the mean, whereas higher variance indicates data points are spread out over a wider range. Understanding variance is pivotal when normalizing data or evaluating the reliability of sample mean estimates.
Z-score Transformation
The Z-score transformation is a statistical technique that allows data points to be expressed in terms of standard deviations from the mean. It's calculated using the formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
where \( X \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

Z-scores allow us to determine how unusual or typical a data point is within a distribution. In our exercise on soil porosity, the Z-score helps standardize the differences between sample means so that these can be evaluated using the standard normal distribution.

By transforming the values into Z-scores, we can easily compute the probability of a random sample mean difference falling within a particular range, facilitating decisions about the significance or likelihood of observations. Because of the symmetry of the normal distribution, Z-scores also allow us to understand percentiles and probabilities with ease, aiding in making data-driven inferences.

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 times that a cashier spends processing individual customer's order are independent random variables with mean 2.5 minutes and standard deviation 2 minutes. What is the approximate probability that it will take more than 4 hours to process the orders of 100 people?

Suppose that \(Z\) has a standard normal distribution and that \(Y\) is an independent \(\chi^{2}\) -distributed random variable with \(\nu\) df. Then, according to Definition 7.2, $$T=\frac{Z}{\sqrt{Y / \nu}}$$ has a \(t\) distribution with \(\nu\) df. \(^{*}\) a. If \(Z\) has a standard normal distribution, give \(E(Z)\) and \(E\left(Z^{2}\right)\). [Hint: For any random variable, \(\left.E\left(Z^{2}\right)=V(Z)+(E(Z))^{2} \cdot\right]\) b. According to the result derived in Exercise \(4.112(\mathrm{a}),\) if \(Y\) has a \(\chi^{2}\) distribution with \(\nu\) df, then $$E\left(Y^{a}\right)=\frac{\Gamma([\nu / 2]+a)}{\Gamma(\nu / 2)} 2^{a}, \quad \text { if } \nu>-2 a$$. Use this result, the result from part (a), and the structure of \(T\) to show the following. [Hint: Recall the independence of \(Z \text { and } Y \)] i. \(E(T)=0,\) if \(\nu>1\) ii. \(V(T)=\nu /(\nu-2),\) if \(\nu>2\)

Refer to Exercise \(7.34 .\) Suppose that \(F\) has an \(F\) distribution with \(\nu_{1}=50\) numerator degrees of freedom and \(\nu_{2}=70\) denominator degrees of freedom. Notice that Table 7 , Appendix 3 , does not contain entries for 50 numerator degrees of freedom and 70 denominator degrees of freedom. a. What is \(E(F) ?\) b. Give \(V(F)\) c. Is it likely that \(F\) will exceed \(3 ?[\text { Hint: Use Tchebysheff's theorem. }]\)

The quality of computer disks is measured by the number of missing pulses. Brand X is such that \(80 \%\) of the disks have no missing pulses. If 100 disks of brand \(X\) are inspected, what is the probability that 15 or more contain missing pulses?

Suppose that independent samples (of sizes \(n_{i}\) ) are taken from each of \(k\) populations and that population \(i\) is normally distributed with mean \(\mu_{i}\) and variance \(\sigma^{2}, i=1,2, \ldots, k\). That is, all populations are normally distributed with the same variance but with (possibly) different means. Let \(\bar{X}_{i}\) and \(S_{i}^{2}, i=1,2, \ldots, k\) be the respective sample means and variances. Let \(\theta=c_{1} \mu_{1}+c_{2} \mu_{2}+\cdots+c_{k} \mu_{k},\) where \(c_{1}, c_{2}, \ldots, c_{k}\) are given constants. a. Give the distribution of \(\hat{\theta}=c_{1} \bar{X}_{1}+c_{2} \bar{X}_{2}+\cdots+c_{k} \bar{X}_{k}\). Provide reasons for any claims that you make. b. Give the distribution of $$\frac{\mathrm{SSE}}{\sigma^{2}}, \quad \text { whereSSE }=\sum_{i=1}^{k}\left(n_{i}-1\right) S_{i}^{2}$$. Provide reasons for any claims that you make. c. Give the distribution of $$\frac{\hat{\theta}-\theta}{\sqrt{\left(\frac{c_{1}^{2}}{n_{1}}+\frac{c_{2}^{2}}{n_{2}}+\cdots+\frac{c_{k}^{2}}{n_{k}}\right) \mathrm{MSE}}}, \quad \text { whereMSE }=\frac{\mathrm{SSE}}{n_{1}+n_{2}+\cdots+n_{k}-k}$$ Provide reasons for any claims that you make.

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.