91Ó°ÊÓ

Suppose that we take a sample of size \(n_{1}\) from a normally distributed population with mean and variance \(\mu_{1}\) and \(\sigma_{1}^{2}\) and an independent of sample size \(n_{2}\) from a normally distributed population with mean and variance \(\mu_{2}\) and \(\sigma_{2}^{2} .\) If it is reasonable to assume that \(\sigma_{1}^{2}=\sigma_{2}^{2},\) then the results given in Section 8.8 apply. What can be done if we cannot assume that the unknown variances are equal but are fortunate enough to know that \(\sigma_{2}^{2}=k \sigma_{1}^{2}\) for some known constant \(k \neq 1 ?\) Suppose, as previously, that the sample means are given by \(\bar{Y}_{1}\) and \(\bar{Y}_{2}\) and the sample variances by \(S_{1}^{2}\) and \(S_{2}^{2}\), respectively. a. Show that \(Z^{\star}\) given below has a standard normal distribution. $$Z^{*}=\frac{\left(\bar{Y}_{1}-\bar{Y}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{\sigma_{1} \sqrt{\frac{1}{n_{1}}+\frac{k}{n_{2}}}}$$ b. Show that \(W^{\star}\) given below has a \(\chi^{2}\) distribution with \(n_{1}+n_{2}-2\) df. $$W^{*}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2} / k}{\sigma_{1}^{2}}$$ c. Notice that \(Z^{\star}\) and \(W^{\star}\) from parts (a) and (b) are independent. Finally, show that $$T^{*}=\frac{\left(\bar{Y}_{1}-\bar{Y}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{S_{p}^{*} \sqrt{\frac{1}{n_{1}}+\frac{k}{n_{2}}}}, \quad \text { where } S_{p}^{2 *}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2} / k}{n_{1}+n_{2}-2}$$ has a \(t\) distribution with \(n_{1}+n_{2}-2\) df. d. Use the result in part (c) to give a \(100(1-\alpha) \%\) confidence interval for \(\mu_{1}-\mu_{2},\) assuming that \(\sigma_{2}^{2}=k \sigma_{1}^{2}\) e. What happens if \(k=1\) in parts \((\mathrm{a}),(\mathrm{b}),(\mathrm{c}),\) and \((\mathrm{d}) ?\)

Step by step solution

01

Derive Distribution for $Z^*$

Given:\[Z^{*} = \frac{(\bar{Y}_1 - \bar{Y}_2) - (\mu_1 - \mu_2)}{\sigma_1 \sqrt{\frac{1}{n_1} + \frac{k}{n_2}}}\]The numerator \((\bar{Y}_1 - \bar{Y}_2) - (\mu_1 - \mu_2)\) represents the difference between two independent sample means, which follows a normal distribution with mean 0 and variance \(\sigma_1^2 \left(\frac{1}{n_1} + \frac{k}{n_2}\right)\) after scaling since \(\text{Var}(\bar{Y}_1) = \sigma_1^2/n_1\) and \(\text{Var}(\bar{Y}_2) = \sigma_2^2/n_2 = k\sigma_1^2/n_2\). Thus, dividing by its standard deviation suggests that \(Z^*\) follows a Standard Normal Distribution \(N(0,1)\).

02

Show $W^*$ has a Chi-Squared Distribution

Given:\[W^{*} = \frac{(n_1 - 1) S_{1}^{2} + (n_2 - 1) S_{2}^{2} / k}{\sigma_{1}^{2}}\]Each sample variance, \(S_1^2\) and \(S_2^2\), represents a scaled chi-squared distribution because they are estimated based on sample sizes \(n_1\) and \(n_2\), respectively. The chi-square degrees of freedom are \((n_1-1)\) for \(S_1^2\) and \((n_2-1)\) for \(S_2^2/k\). Thus, scaling these appropriately, the entire expression \(W^*\) follows a \(\chi^2(n_1 + n_2 - 2)\) distribution.

03

Show $T^*$ follows a $t$ distribution

Given:\[T^{*} = \frac{(\bar{Y}_1 - \bar{Y}_2) - (\mu_1 - \mu_2)}{S_{p}^{*} \sqrt{\frac{1}{n_1} + \frac{k}{n_2}}}\]Where:\[S_{p}^{2 *} = \frac{(n_1 - 1) S_1^{2} + (n_2 - 1) S_2^{2} / k}{n_1 + n_2 - 2}\]Since \( S_p^{2*} \) is a weighted average of sample variances, \(T^*\) represents a ratio of independent standard normal distribution \(Z^*\) and a chi-square distribution divided by its degrees of freedom. Consequently, \(T^*\) follows a \(t\) distribution with \(n_1 + n_2 - 2\) degrees of freedom.

04

Derive Confidence Interval for $\mu_1 - \mu_2$

Given the result in part (c), the \(100(1-\alpha)\%\) confidence interval for \(\mu_1 - \mu_2\) is given by:\[(\bar{Y}_1 - \bar{Y}_2) \pm t_{n_1+n_2-2,\alpha/2} \times S_p^{*} \sqrt{\frac{1}{n_1} + \frac{k}{n_2}}\]Where \(t_{n_1+n_2-2,\alpha/2}\) is the critical value of the \(t\)-distribution with \(n_1 + n_2 - 2\) degrees of freedom.

05

Analyze the Case When $k=1$

When \(k = 1\):- Part (a): The expression for \(Z^*\) assumes both populations have the same variance, simplifying the denominator.- Part (b): The form of \(W^*\) becomes the classic pooled variance estimator with a simple \(\chi^2\) distribution.- Part (c): \(S_p^{*}\) becomes a straightforward pooled variance estimator, making \(T^*\) the classic two-sample \(t\)-statistic.- Part (d): The confidence interval becomes the traditional interval for the difference of means assuming equal variances.

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

In mathematical statistics, the normal distribution plays a pivotal role. It is a continuous probability distribution that is symmetric, forming a classic bell-shaped curve.
The mean \(\mu\) determines the center of the distribution, while the variance \(\sigma^2\) controls its spread.
The probability density function for a normal distribution is given by \[ f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \]
When dealing with sample data from a population assumed to be normally distributed, statisticians often analyze the sample mean.
In our exercise, the difference of the sample means \(\bar{Y}_1 - \bar{Y}_2\) is considered. Because these sample means are taken from independent normally distributed populations, they themselves form a distribution that is approximately normal.
Understanding the normal distribution is crucial to many statistical methods, as it allows for approximations and inferences to be made with known probabilities.

Sample Variance

Sample variance is a measure of the variability or dispersion of a sample dataset. It's calculated to understand how much the data in a sample vary from the sample mean.
Mathematically, the sample variance, denoted as \(S^2\), is calculated by the formula \[ S^2 = \frac{1}{n-1}\sum_{i=1}^{n}(X_i - \bar{X})^2 \] where \(X_i\) are the individual data points, and \(\bar{X}\) is the sample mean.
In the context of the exercise, we have two samples, each with its own sample variance, \(S_1^2\) and \(S_2^2\). These variances measure the spread of data in their respective samples.
Understanding sample variance is essential since it provides the basis for estimating the variability of the population, especially when the sample size is small. This estimate is used when constructing confidence intervals and performing hypothesis tests.

Chi-Square Distribution

The chi-square distribution is a critical concept in statistical analysis, especially for tests of variance. It is widely used as it describes the distribution of the sum of squared standard normal deviates.
The chi-square distribution with \(k\) degrees of freedom is given by \[ X^2 = \sum_{i=1}^{k}(Z_i)^2 \] where \(Z_i\) are independent standard normal variables.
In the given exercise, \(W^{*}\) is calculated to be a chi-square distribution with \(n_1 + n_2 - 2\) degrees of freedom. This results from the combination of scaled sample variances, which individually follow scaled chi-square distributions.
Understanding this concept helps in constructing tests like the chi-squared test for goodness of fit and the derivation of pooled variance methods, especially when dealing with normal distributions.

Confidence Interval

Confidence intervals provide a range of values for an unknown population parameter. This range is derived from the sample data and a specified confidence level, such as 95% or 99%.
For the difference between two population means where variance is not necessarily equal, a confidence interval can be constructed using the formula derived from a \(t\)-distribution.
The general formula is \[(\bar{Y}_1 - \bar{Y}_2) \pm t_{df, \alpha/2} \times S_p^{*} \sqrt{\frac{1}{n_1} + \frac{k}{n_2}}\] where \(t_{df, \alpha/2}\) is the critical \(t\)-value for the desired confidence level and degrees of freedom.
Such intervals are pivotal because they provide a method to express uncertainty and make informed conclusions about population differences, all while accounting for sample variability. This is essential for hypothesis testing and inference statistics.