91Ó°ÊÓ

Suppose that \(X_{1}, X_{2}, \ldots, X_{m}\) and \(Y_{1}, Y_{2}, \ldots, Y_{n}\) are independent random samples, with the variables \(X_{i}\) normally distributed with mean \(\mu_{1}\) and variance \(\sigma_{1}^{2}\) and the variables \(Y_{i}\) normally distributed with mean \(\mu_{2}\) and variance \(\sigma_{2}^{2} .\) The difference between the sample means, \(\bar{X}-\bar{Y},\) is then a linear combination of \(m+n\) normally distributed random variables and, by Theorem \(6.3,\) is itself normally distributed. a. Find \(E(\bar{X}-\bar{Y})\). b. Find \(V(\bar{X}-\bar{Y})\). c. Suppose that \(\sigma_{1}^{2}=2, \sigma_{2}^{2}=2.5,\) and \(m=n .\) Find the sample sizes so that \((\bar{X}-\bar{Y})\) will be within 1 unit of \(\left(\mu_{1}-\mu_{2}\right)\) with probability .95

Step by step solution

01

Calculate E(XÌ„)

The expected value of the sample mean \( \bar{X} \) is the same as the mean of the population from which the samples are drawn. Therefore, \( E(\bar{X}) = \mu_1 \).

02

Calculate E(YÌ„)

Similarly, the expected value of the sample mean \( \bar{Y} \) is the mean of the population from which the samples \( Y_i \) are drawn. Thus, \( E(\bar{Y}) = \mu_2 \).

03

Find E(XÌ„ - YÌ„)

The expectation of the difference \( \bar{X} - \bar{Y} \) is given by \( E(\bar{X} - \bar{Y}) = E(\bar{X}) - E(\bar{Y}) = \mu_1 - \mu_2 \).

04

Compute V(XÌ„)

The variance of \( \bar{X} \) for a sample of size \( m \) is \( V(\bar{X}) = \frac{\sigma_1^2}{m} \).

05

Compute V(YÌ„)

The variance of \( \bar{Y} \) for a sample of size \( n \) is \( V(\bar{Y}) = \frac{\sigma_2^2}{n} \).

06

Find V(XÌ„ - YÌ„)

The variance of the difference \( \bar{X} - \bar{Y} \) is \( V(\bar{X} - \bar{Y}) = V(\bar{X}) + V(\bar{Y}) = \frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n} \).

07

Introduce given variances and condition m = n

Given \( \sigma_1^2 = 2 \), \( \sigma_2^2 = 2.5 \), and \( m = n \), substitute these into the variance formula: \( V(\bar{X} - \bar{Y}) = \frac{2}{m} + \frac{2.5}{m} = \frac{4.5}{m} \).

08

Determine sample size for desired precision

We want \( \bar{X} - \bar{Y} \) to be within 1 unit of the true mean difference \( \mu_1 - \mu_2 \) with probability 0.95. Using a normal distribution, this is equivalent to setting the margin of error to 1: \( Z_{0.025} \cdot \sqrt{\frac{4.5}{m}} = 1 \), where \( Z_{0.025} \approx 1.96 \).

09

Solve for the sample size

Rearrange and solve: \( 1.96 \cdot \sqrt{\frac{4.5}{m}} = 1 \). Squaring both sides gives: \( 1.96^2 \cdot \frac{4.5}{m} = 1 \). Thus, \( m = 1.96^2 \cdot 4.5 \approx 17.64 \). Round up to the nearest whole number \( m = n = 18 \).

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Key Concepts

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

Normal Distribution

The normal distribution is a fundamental concept in statistics. It describes how data points are distributed and is often graphically represented as a bell-shaped curve. This distribution is symmetric around the mean, which means most of the observations tend to cluster around the central peak. The width of the bell curve is determined by the standard deviation, which affects how data is spread. The normal distribution is particularly important because many statistical tests and methods assume that the data follows this distribution. These methods rely on properties such as:

• A single mode with a shape symmetric around the mean
• A probability that decreases as you move away from the mean
• An empirical rule, where approximately 68% of data falls within one standard deviation of the mean, 95% within two, and 99.7% within three

Understanding how your data fits into this distribution can help in accurately interpreting results and predictions.

Sample Variance

Sample variance measures the spread or dispersion of a set of sample observations. It is calculated by averaging the squared deviations of each observation from the sample mean. Mathematically, if you have a sample of size \( n \) with observations \( x_1, x_2, \ldots, x_n \), the sample variance \( s^2 \) is given by:\[s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2\]where \( \bar{x} \) is the sample mean.

• Sample variance is crucial because it tells us about the data's variability.
• Unlike the variance of the population \( \sigma^2 \), which overestimates real-world data due to N in the denominator, sample variance corrects this bias with \(n-1\).

This measure is foundational, helping in hypothesis testing and the construction of confidence intervals. It allows one to understand the stability and reliability of the sample data.

Sample Size Determination

Deciding the appropriate sample size is crucial in statistical analysis. A proper sample size ensures that results are statistically significant and can be generalized to a larger population. Several factors influence sample size determination, including the desired level of precision, the population's variability, and the acceptable error margin. In the context of the original exercise, when determining sample size to ensure \( \bar{X} - \bar{Y} \) is within 1 unit of \( \mu_1 - \mu_2 \) with 95% probability, the following considerations were made:

• The variability of each population, represented by \( \sigma_1^2 \) and \( \sigma_2^2 \).
• The symmetry and probabilities of the normal distribution, using Z-scores to translate confidence levels into sample size requirements.

This is how we solved for \( m = n = 18 \), an optimal sample size necessary to achieve desired accuracy.

Expected Value

The expected value is a key concept in probability and statistics, representing the average value one would expect to obtain if an experiment were repeated indefinitely. For a random variable \(X\), the expected value \(E(X)\) is the weighted average of all possible values, with weights being their probabilities. In the realm of sample means, such as \( \bar{X} \) and \( \bar{Y} \), their expected values equal the population means they come from, i.e., \(E(\bar{X}) = \mu_1\) and \(E(\bar{Y}) = \mu_2\).

• In our exercise, the expected value of the difference \( \bar{X} - \bar{Y} \) became \(\mu_1 - \mu_2\).
• This situation makes it much easier to understand how the results of different samples reflect the true population parameters.

The expected value assists researchers in making predictions and informed decisions based on existing data patterns.