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Chapter 9: Problem 45

Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\) are independent random samples from normal distributions with means \(\mu_{X}\) and \(\mu_{Y}\) and known standard deviations \(\sigma_{X}\) and \(\sigma_{Y}\), respectively. Derive a \(100(1-\alpha) \%\) confidence interval for \(\mu_{X}-\mu_{Y}\).

Short Answer

Expert verified
The \(100(1-\alpha) \%\) confidence interval for \(D = \mu_{X} - \mu_{Y}\) is given by \((\overline{X} - \overline{Y}) \pm Z_{\alpha/2}\sqrt{\sigma_{X}^{2}/n + \sigma_{Y}^{2}/m}\).

Step by step solution

01

Identify known variables

We know that the random samples \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\) are independent and they are taken from normal distributions with means \(\mu_{X}\) and \(\mu_{Y}\) and known standard deviations \(\sigma_{X}\) and \(\sigma_{Y}\), respectively.
02

Calculate sample means

We need to calculate the sample means for both sets of samples. They are given by
03

Define the difference in means

Define \(D = \mu_{X} - \mu_{Y}\), which is the difference in the population means we are interested in.
04

Calculate standard error of the difference

The standard deviation of the difference in the sample means is given by \(\sigma_{D} = \sqrt{\sigma_{X}^{2}/n + \sigma_{Y}^{2}/m}\), where n and m are the sizes of the two samples.
05

Derive confidence interval

The \(100(1-\alpha) \%\) confidence interval for \(D = \mu_{X} - \mu_{Y}\) is given by \((\overline{X} - \overline{Y}) \pm Z_{\alpha/2}\sigma_{D}\), where \(Z_{\alpha/2}\) is the z-value that captures the middle \(100(1-\alpha) \%\) area under the standard normal curve.

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Key Concepts

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

Understanding Normal Distribution
A normal distribution is a common way to describe data that clusters around a central mean value. Here's why it's important:

  • The normal distribution is symmetric, meaning it looks the same on both sides of the mean.
  • Most of the data tends to be close to the mean, with fewer cases appearing as you move away.
  • It follows a bell-shaped curve, known as the Gaussian curve.
In many natural phenomena, data is expected to cluster in this way. For example, heights, test scores, and other measurable traits often show a normal distribution.

This distribution makes statistical analysis simpler and more predictable, especially when making inferences about a population based on a sample.
The Importance of Sample Mean
The sample mean is a key value in statistics, representing the average of a set of data points collected from a larger population.

  • It is calculated by summing all the data values and dividing by the number of values.
  • The sample mean serves as an unbiased estimator of the population mean.
This value allows researchers to make educated guesses about the overall population. In the context of the problem, finding the sample mean for each dataset (\(\overline{X}\) for the sample of \(X\)s and \(\overline{Y}\) for the sample of \(Y\)s) helps compare their central tendencies.

The closer the sample mean is to the true population mean, the more accurate the estimate will be.
Exploring Standard Error
Standard error measures the precision of the sample mean when estimating the population mean.

  • It is calculated as the standard deviation of the sampling distribution.
  • For the difference in means in our problem, it is found using: \(\sigma_{D} = \sqrt{\sigma_{X}^{2}/n + \sigma_{Y}^{2}/m}\).
The standard error tells us how much we can expect the sample mean to fluctuate from the population mean. A smaller standard error indicates more reliable, precise estimates.

In practical terms, it helps establish the boundaries of our confidence interval, which we use to make predictions about the population difference \(D = \mu_{X} - \mu_{Y}\).
Understanding the Z-value
The Z-value is a critical part of calculating confidence intervals. It corresponds to a position under the standard normal distribution curve.

  • For a given confidence level, the Z-value determines how far the sample mean is from the population mean.
  • For a 95% confidence interval, \(Z_{0.025}\) is typically used, as it leaves 2.5% in each tail of the distribution.
This value helps quantify the certainty of the interval estimate, indicating how far we must look under the standard normal curve to enclose the desired coverage.

Thus, using the Z-value, we can derive bounds that reflect our confidence in how close the sample statistics are to the actual population parameters.

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Most popular questions from this chapter

If \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\) are independent random samples from normal distributions with the same \(\sigma^{2}\), prove that their pooled sample variance, \(S_{p}^{2}\), is an unbiased estimator for \(\sigma^{2}\).

Suppose that \(H_{0}: \mu_{X}=\mu_{Y}\) is being tested against \(H_{1}\) : \(\mu_{X} \neq \mu_{Y}\), where \(\sigma_{X}^{2}\) and \(\sigma_{Y}^{2}\) are known to be \(17.6\) and \(22.9\), respectively. If \(n=10, m=20, \bar{x}=81.6\), and \(\bar{y}=79.9\), what \(P\)-value would be associated with the observed \(Z\) ratio?

A company markets two brands of latex paint regular and a more expensive brand that claims to dry an hour faster. A consumer magazine decides to test this claim by painting ten panels with each product. The average drying time of the regular brand is \(2.1\) hours with a sample standard deviation of 12 minutes. The fast-drying version has an average of \(1.6\) hours with a sample standard deviation of 16 minutes. Test the null hypothesis that the more expensive brand dries an hour quicker. Use a onesided \(H_{1}\). Let \(\alpha=0.05\).

In a study designed to investigate the effects of a strong magnetic field on the early development of mice (7), ten cages, each containing three 30 -day- old albino female mice, were subjected for a period of 12 days to a magnetic field having an average strength of \(80 \mathrm{Oe} / \mathrm{cm}\). Thirty other mice, housed in ten similar cages, were not put in the magnetic field and served as controls. Listed in the table are the weight gains, in grams, for each of the twenty sets of mice. Test whether the variances of the two sets of weight gains are significantly different. Let \(\alpha=0.05\). For the mice in the magnetic field, \(s_{X}=5.67\); for the other mice, \(s_{Y}=3.18\).

Water witching, the practice of using the movements of a forked twig to locate underground water (or minerals), dates back over four hundred years. Its first detailed description appears in Agricola's De re Metallica, published in 1556. That water witching works remains a belief widely held among rural people in Europe and throughout the Americas. [In 1960 the number of "active" water witches in the United States was estimated to be more than 20,000 (205).] Reliable evidence supporting or refuting water witching is hard to find. Personal accounts of isolated successes or failures tend to be strongly biased by the attitude of the observer. Of all the wells dug in Fence Lake, New Mexico, twenty-nine "witched" wells and thirty-two "nonwitched" wells were sunk. Of the "witched"' wells, twenty-four were successful. For the "nonwitched"' wells, there were twentyseven successes. What would you conclude?
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