Suppose that we take a sample of size n1 from a normally...

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 $(\mathrm{d}) ?$

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 $(\mathrm{d}) ?$

Key Concepts

Known Variance Ratio

This concept arises when there is a known proportional relationship between the variances of the two populations (e.g., one variance is a constant multiple of the other). Knowing the ratio simplifies the adjustment of the standard error in the test statistics and allows for the exact derivation of the sampling distributions (such as the standard normal, chi-square, or t distributions) used for valid statistical inference.

Pooled Variance Estimate

When dealing with two independent samples, the pooled variance estimate is a weighted average of the individual sample variances. It is particularly useful when the variances are either equal or proportionally related, as it produces a more reliable estimate of the common variance, which is then used in test statistics for drawing inferences about the difference in means.

t Distribution

The t distribution emerges when a standard normal variable is divided by the square root of an independent chi-square variable divided by its degrees of freedom. It is a cornerstone of inference when population variances are unknown and must be estimated, enabling the formulation of confidence intervals and hypothesis tests for the difference between two means.

Standard Normal Distribution

In two-sample comparisons, standardizing the difference between sample means, when scaled appropriately, results in a variable that follows a standard normal distribution. This property is essential for deriving exact test statistics when certain population parameters (or their proportional relationships) are known, and it underpins many inferential procedures.

Two-sample Inference

This concept involves comparing two independent samples drawn from normally distributed populations to make inferences about the difference between their means. It encompasses the construction of test statistics and confidence intervals based on the sampling distributions of the sample means and variances, taking into account underlying assumptions about the populations.

Chi-square Distribution

The chi-square distribution is vital in characterizing the distribution of scaled sample variances derived from normally distributed data. In two-sample settings, combining variances (after suitable scaling) yields a chi-square distributed variable, which plays a critical role in constructing composite test statistics and determining degrees of freedom for further inference.

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