Confidence Intervals

Jacques Carabain (1834-1933) was a Dutch-Belgian painter.

We have sections

\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}

Confidence Intervals for Means.

Given a random sampel $X_{1},X_{2},\cdots ,X_{n}$ from a normal distribution $N(\mu ,\sigma^{2})$, we shall now consider the closeness of $\bar{X}$ to the unknown mean $\mu$. For the probability $1-\alpha$, we can find a number $z_{\alpha /2}$ from the table satisfying

\[\mathbb{P}\left (-z_{\alpha /2}\leq\frac{\bar{X}-\mu}{\sigma /\sqrt{n}}\leq z_{\alpha /2}\right )=1-\alpha.\]

For example, if $1-\alpha=0.95$, then $z_{\alpha /2}=z_{0.025}=1.96$ and if $1-\alpha =0.9$, then $z_{\alpha /2}=z_{0.05}=1.645$. Now, we have

\[\mathbb{P}\left (\bar{X}-z_{\alpha /2}\cdot\frac{\sigma}{\sqrt{n}}\leq\mu\leq\bar{X}+z_{\alpha /2}\cdot\frac{\sigma}{\sqrt{n}}\right )=1-\alpha .\]

It means that the probability that the random interval

\[\left [\bar{X}-z_{\alpha /2}\cdot\frac{\sigma}{\sqrt{n}},\bar{X}+z_{\alpha /2}\cdot\frac{\sigma}{\sqrt{n}}\right ]\]

includes the unonkwn mean $\mu$ is $1-\alpha$. Once the sample is observed and the sample mean computed to equal to $\bar{x}$, the interval

\[\left [\bar{x}-z_{\alpha /2}\cdot\frac{\sigma}{\sqrt{n}},\bar{x}+z_{\alpha /2}\cdot\frac{\sigma}{\sqrt{n}}\right ]\]

is a known interval. Since the probability that the random interval covers $\mu$ before the sample is drawn is equal to $1-\alpha$, we now call the computed interval $\bar{x}\pm z_{\alpha /2}\cdot (\sigma /\sqrt{n})$ as a $100(1-\alpha )\%$ confidence interval}for the unknown mean $\mu$. For illustration, $\bar{x}\pm 1.96(\sigma /\sqrt{n})$ is a $95\%$ confidence interval for $\mu$. The number $100(1-\alpha )\%$, or equivalently, $1-\alpha$, is called the confidence coefficient. It is clear to see that, as $n$ increases, the value $z_{\alpha /2}(\sigma /\sqrt{n})$ decreases, resulting in a shorter confidence interval with the same confidence coefficient $1-\alpha$. A shorter confidence interval indicates that we have more reliance in $\bar{x}$ as an estimate of $\mu$. Statisticians who are not restricted by time, money, effort, or availability of observations can obviously make the confidence interval as short as they like by increasing the sample size $n$. For a fixed sample size $n$, the length of the confidence interval can also be shortened by decreasing the
confidence coefficient $1-\alpha$. But if this is done, we achieve a shorter confidence interval by losing some confidence.

Example. Let $X$ equal the length of life of a $60$-watt light bulb marketed by a certain manufacturer of light bulbs. Assume that the distribution of $X$ is $N(\mu ,1296)$. If a random sample of $n=27$ bulbs were tested until they are burned out, yielding a sample mean of $\bar{x}=1478$ hours, then a $95\%$ confidence interval for $\mu$ is given by

\begin{align*} & \left [\bar{x}-z_{0.025}\cdot\frac{\sigma}{\sqrt{n}},\bar{x}+z_{0.025}\cdot\frac{\sigma}{\sqrt{n}}\right ]\\ & \quad =\left [1478-1.96\cdot\frac{36}{\sqrt{27}},1478-1.96\cdot\frac{36}{\sqrt{27}}\right ]\\ & \quad =[1464.42,1491.58].\end{align*}

Example. Let $\bar{x}$ be the observed sample mean of the five items of a random sample from the normal distribution $N(\mu, 16)$. A $90\%$ confidence interval for the unknown mean $\mu$ is given by

\[\left [\bar{x}-1.645\sqrt{\frac{16}{5}},\bar{x}+1.645\sqrt{\frac{16}{5}}\right ].\]

For a particular sample, this interval either does or does not contain the mean $\mu$. However, if many such intervals were calculated, it should be trust that about $90\%$ of them contain the mean $\mu$. Fifty random samples of size five from the normal distribution $N(50,16)$ were simulated in a computer. A $90\%$ condifence interval was calculated from each random sample, as if the mean is unknown. For these $50$ intervals, $45$ (or $90\%$) of them contain the mean $\mu =50$. $\sharp$

If we cannot assume that the distribution from which the sample arose is normal, we can still obtain an approximate confidence interval for $\mu$. By the central limit theorem, the ratio

\[\frac{\bar{X}-\mu}{\sigma /\sqrt{n}}\]

has the approximate normal distribution $N(0,1)$ provided that $n$ is large enough when the underlying distribution is not normal. In this case, we have

\[\mathbb{P}\left (-z_{\alpha /2}\leq\frac{\bar{X}-\mu}{\sigma /\sqrt{n}}\leq z_{\alpha /2}\right )\approx 1-\alpha. \]

Therefore, we see that

\[\left [\bar{x}-z_{\alpha /2}\cdot\frac{\sigma}{\sqrt{n}},\bar{x}+z_{\alpha /2}\cdot\frac{\sigma}{\sqrt{n}}\right ]\]

is an approximate $100(1-\alpha )\%$ confidence interval for $\mu$.

Example. Let $X$ equal the amount of orange juice (in grams per day) consumed by American. Suppose it is known that the standard deviation of $X$ is $\sigma =96$. To estimate the mean $\mu$ of $X$, an orange growers’ association took a random sample of $n=576$ Americans and found that they consumed, on the average, $x=133$ grams of orange juice per day. Therefore, an approximate $90\%$ confidence interval for $\mu$ is given by

\[[133\pm 1.645\cdot\frac{96}{\sqrt{576}}\mbox{ or }[126.42,139.58].\]

However, in many applications, the sample sizes are small and we do not know the value of the standard deviation. Suppose that the underlying distribution is normal and that $\sigma^{2}$ is unknown. We again consider the ratio

\[T=\frac{\bar{X}-\mu}{S/\sqrt{n}}.\]

This ratio can be rewritten a follows

\begin{align*} T & =\frac{\bar{X}-\mu}{S\sqrt{n}}\\ & =\frac{\frac{\bar{X}-\mu}{\sigma \sqrt{n}}}{\sqrt{\frac{(n-1)S^{2}}{\sigma^{2}}/(n-1)}}.\end{align*}

From the previous discussions, we know that

\[Z=\frac{\bar{X}-\mu}{\sigma /\sqrt{n}}\]

is $N(0,1)$ and

\[U=\frac{(n-1)S^{2}}{\sigma^{2}}\]

is $\chi^{2}(n-1)$, and that these two are independent. That is, the random variable

\[T=\frac{Z}{\sqrt{U/(n-1)}}\]

has a $t$-distribution with $n-1$ degrees of freedom. For notational purposes, let $t_{\alpha}(r)$ denote the constant satisfying

\[\mathbb{P}(T\geq t_{\alpha}(r))=\alpha.\]

Let $T$ have a $t$-distribution with seven degrees of freedom. Then, consulting the table, we have

\begin{align*} \mathbb{P}(T\leq 1.415) & =0.9\\ \mathbb{P}(T\leq -1.415) & =1-\mathbb{P}(T<1.415)=0.1\\ \mathbb{P}(-1.895<T<1.415) & =0.9-0.05=0.85.\end{align*}

We also have $t_{0.01}(7)=1.415$, $t_{0.9}(7)=-t_{0.1}(7)=-1.415$, and $t_{0.025}(7)=2.365$.

Now we use the fact that

\[T=\frac{\bar{X}-\mu}{S/\sqrt{n}}\]

has a $t$-distribution with $n-1$ degrees of freedom, where $S^{2}$ is the usual unbiased estimator of $\sigma^{2}$. We take $t_{\alpha /2}$ satisfying

\[\mathbb{P}(T\geq t_{\alpha /2}(n-1))=\alpha /2.\]

Then, we have

\begin{align*} 1-\alpha & =\mathbb{P}\left (-t_{\alpha /2}(n-1)\leq\frac{\bar{X}-\mu}{S\sqrt{n}}\leq t_{\alpha /2}(n-1)\right )\\
& =\mathbb{P}\left (\bar{X}-t_{\alpha /2}(n-1)\cdot\frac{S}{\sqrt{n}}\leq\mu\leq\bar{X}+t_{\alpha /2}(n-1)\cdot\frac{S}{\sqrt{n}}\right ).\end{align*}

The observations of a random sample provide $\bar{x}$ and $s^{2}$. Then, we have that

\[\left [\bar{x}-t_{\alpha /2}(n-1)\cdot\frac{s}{\sqrt{n}},\bar{x}+t_{\alpha /2}(n-1)\cdot\frac{s}{\sqrt{n}}\right ]\]

is a $100(1-\alpha )\%$ confience interval for $\mu$.

Example. Let $X$ equal the amount of butterfat in pounds produced by a typical cow during a $305$-day milk production period between her first and second calves. Assume that the distribution of $X$ is $N(\mu ,\sigma^{2})$. To estimate $\mu$, a farmer measure the butterfat production for $n=20$ cows, yielding the values $\bar{x}=507.5$ and $s=89.75$. Therefore, a point estimate of $\mu$ is $\bar{x}=507.5$. Since $t_{0.05}(19)=1,729$, a $90\%$ confidence interval for $\mu$ is given by

\[507.5\pm 1.729\cdot\frac{89.75}{\sqrt{20}}\mbox{ or }[472.8,542.2].\]

When we are not able to assume that the underlying distribution is normal; that is, $\mu$ and $\sigma$ are both unknown, the approximate confidence intervals for $\mu$ can still be constructed using

\[T=\frac{\bar{X}-\mu}{S/\sqrt{n}},\]

which now only has an approximate $t$ distribution with $n-1$ degrees of freedom.

\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}

Confidence Intervals for Difference of Two Means.

Suppose that we are interested in comparing the means of two normal distributions. Let $X_{1},X_{2},\cdots ,X_{n}$ and $Y_{1},Y_{2},\cdots ,Y_{m}$ be two random samples of sizes $n$ and $m$ from two normal distributions $N(\mu_{X},\sigma_{X}^{2})$ and $N(\mu_{Y},\sigma_{Y}^{2})$, respectively. Suppose, for now, that $\sigma_{X}^{2}$ and $\sigma_{Y}^{2}$ are known. Since the random samples are independent, the respective sample means $\bar{X}$ and $\bar{Y}$ are also independent with distributions $N(\mu_{X},\sigma_{X}^{2}/n)$ and $N(\mu_{Y},\sigma_{Y}^{2}/m)$. Therefore, we know that the distribution of $W=\bar{X}-\bar{Y}$ is $N(\mu_{X}-\mu_{Y},\sigma_{X}^{2}/n+\sigma_{Y}^{2}/m)$ and

\[\mathbb{P}\left (-z_{\alpha /2}\leq\frac{(\bar{X}-\bar{Y})-(\mu_{X}-\mu_{Y})}{\sqrt{\sigma_{X}^{2}/n+\sigma_{Y}^{2}/m}}\leq z_{\alpha /2}\right )=1-\alpha ,\]

which can be rewritten as

\[\mathbb{P}[(\bar{X}-\bar{Y})-z_{\alpha /2}\cdot\sigma_{W}\leq\mu_{X}-\mu_{Y}\leq (\bar{X}-\bar{Y}+z_{\alpha /2}\cdot\sigma_{W}]=1-\alpha ,\]

where $\sigma_{W}=\sqrt{\sigma_{X}^{2}/n+\sigma_{Y}^{2}/m}$ is the standard deviation of $\bar{X}-\bar{Y}$. Once the experiments have been performed and the means $\bar{x}$ and $\bar{y}$ computed, it says that

\[\left [\bar{x}-\bar{y}-z_{\alpha /2}\cdot\sigma_{W},\bar{x}-\bar{y}+z_{\alpha /2}\cdot\sigma_{W}\right ]\]

provides a $100(1-\alpha )\%$ confidence interval for $\mu_{X}-\mu_{Y}$.

Example. In the preceding discussion, let $n=15$, $m=8$, $\bar{x}=70.1$, $\bar{y}=75.3$, $\sigma_{X}=60$, $\sigma_{Y}=40$, and $1-\alpha =0.9$. Then, we have

\[1-\alpha /2=0.95=\Phi (1.645).\]

Therefore, we obtain that

\[\left [70.1-75.3-1.645\cdot\sqrt{\frac{60}{15}+\frac{40}{8}},70.1-75.3+1.645\cdot\sqrt{\frac{60}{15}+\frac{40}{8}}\right ]=[-10.135,-0.265]\]

is a $90\%$ confidence interval for $\mu_{X}-\mu_{Y}$. Since the confidence interval does not contain zero, we suspect that $\mu_{Y}$ is greater than $\mu_{X}$. $\sharp$

If the sample sizes are large and $\sigma_{X}$ and $\sigma_{Y}$ are unknown, we can replace $\sigma_{X}^{2}$ and $\sigma_{Y}^{2}$ with $s_{X}^{2}$ and $s_{Y}^{2}$, where $s_{X}^{2}$ and $s_{Y}^{2}$ are the values of the respective unbiased estimates of the variances. This means that

\[\bar{x}-\bar{y}\pm z_{\alpha /2}\cdot\sqrt{\frac{s_{X}^{2}}{n}+\frac{s_{Y}^{2}}{m}}\]

serves as an approximate $100(1-\alpha )\%$ confidence interval for $\mu_{X}-\mu_{Y}$.

Now, we consider the problem of constructing confidence intervals for the difference of the means of two normal distributions when the variances are unknown but the sample sizes are small. Let $X_{1},X_{2},\cdots ,X_{n}$ and $Y_{1},Y_{2},\cdots ,Y_{m}$ be two independent random samples from the normal distributions $N(\mu_{X},\sigma_{X}^{2})$ and $N(\mu_{Y},\sigma_{Y}^{2})$, respectively. If the sample sizes are not large (e.g. smaller than $30$), this problem can be a difficult one. However, if we can assume common and unknown variances with $\sigma_{X}^{2}=\sigma_{Y}^{2}=\sigma^{2}$, there is a way out of this difficulty. We know that

\[Z=\frac{(\bar{X}-\bar{Y})-(\mu_{X}-\mu_{Y})}{\sqrt{\sigma^{2}/n+\sigma^{2}/m}}\]

is $N(0,1)$. Since the random samples are independent, it follows that

\[U=\frac{(n-1)S_{X}^{2}}{\sigma^{2}}+\frac{(m-1)S_{Y}^{2}}{\sigma^{2}}\]

is the sum of two independent chi-square random variables, which also says that $U$ is $\chi^{2}(n+m-2)$. In addition, the independence of the sample means and sample variances implies that $Z$ and $U$ are independent. Therefore, the random variable

\[T=\frac{Z}{\sqrt{U/(n+m-2)}}\]

has a $t$-distribution with $n+m-2$ degrees of freedom. Also, the random variable

\begin{align*} T & =\frac{\frac{(\bar{X}-\bar{Y})-(\mu_{X}-\mu_{Y})}{\sqrt{\sigma^{2}/n+\sigma^{2}/m}}}{\sqrt{\left (\frac{(n-1)S_{X}^{2}+(m-1)S_{Y}^{2}}{\sigma^{2}}\right )/(n+m-2)}}\\ & =\frac{(\bar{X}-\bar{Y})-(\mu_{X}-\mu_{Y})}{\sqrt{\left (\frac{(n-1)S_{X}^{2}+(m-1)S_{Y}^{2}}{n+m-2}\right )\left (\frac{1}{n}+\frac{1}{m}\right )}}\end{align*}

has a $t$-distribution with $n+m-2$ degrees of freedom. Therefore, we have

\[\mathbb{P}(-t_{\alpha /2}(n+m-2)\leq T\leq t_{\alpha /2}(n+m-2))=1-\alpha .\]

Solving the inequality for $\mu_{X}-\mu_{Y}$ yields

\[\mathbb{P}\left (\bar{X}-\bar{Y}-t_{\alpha /2}(n+m-2)\cdot S_{p}\cdot\sqrt{\frac{1}{n}+\frac{1}{m}}\leq \mu_{X}-\mu_{Y}\leq
\bar{X}-\bar{Y}+t_{\alpha /2}(n+m-2)\cdot S_{p}\cdot\sqrt{\frac{1}{n}+\frac{1}{m}}\right )=1-\alpha ,\]

where the pooled estimator of the common standard deviation is

\[S_{p}=\sqrt{\frac{(n-1)S_{X}^{2}+(m-1)S_{Y}^{2}}{n+m-2}}.\]

Let $\bar{x}$, $\bar{y}$, and $s_{p}$ be the observed values of $\bar{X}$, $\bar{Y}$, and $S_{p}$. Then

\[\left [\bar{x}-\bar{x}-t_{\alpha /2}(n+m-2)\cdot s_{p}\cdot\sqrt{\frac{1}{n}+\frac{1}{m}},\bar{x}-\bar{y}+t_{\alpha /2}(n+m-2)\cdot s_{p}\cdot\sqrt{\frac{1}{n}+\frac{1}{m}}\right ]\]

is a $100(1-\alpha )\%$ confidence interval for $\mu_{X}-\mu_{Y}$.

Example. Suppose that scores on a stnadardized test in mathematics taken by students from large and small high schools are $N(\mu_{X},\sigma^{2})$ and $N(\mu_{Y},\sigma^{2})$, respectively, where $\sigma^{2}$ is unknown. If a random sample of $n=9$ students from large high schools yielded $\bar{x}=81.31$ and $s_{X}^{2}=60.76$, and a random sample of $m=15$ students from small high schools yielded $\bar{y}=78.61$ and $s_{y}^{2}=48.24$, the endpoints for a $95\%$ confidence interval for $mu_{X}-\mu_{Y}$ are given by

\[81.31-78.61\pm 2.074\sqrt{\frac{8\cdot 60.76+14\cdot 48.24}{22}}\cdot\sqrt{\frac{1}{9}+\frac{1}{15}}.\]

because $t_{0.025}(22)=2.074$. The $95\%$ confidence interval is $[-3.65,9.05]$. $\sharp$

\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}

Confidence Intervals for Variances.

We have known that the distribution of $(n-1)S^{2}/\sigma^{2}$ is $\chi^{2}(n-1)$. Select constants $a$ and $b$ from table with $n-1$ degrees freedom satisfying

\[\mathbb{P}\left (a\leq\frac{(n-1)S^{2}}{\sigma^{2}}\leq b\right )=1-\alpha .\]

One way to do this is by selecting $a$ and $b$ satisfying $a=\chi_{1-\alpha /2}^{2}(n-1)$ and $b=\chi_{\alpha /2}^{2}(n-1)$. In other words, select $a$ and $b$ satisfying the probabilities in the two tails are equal. Then, solving the inequalities, we have

\begin{align*} 1-\alpha & =\mathbb{P}\left (\frac{a}{(n-1)S^{2}}\leq\frac{1}{\sigma^{2}}\leq\frac{b}{(n-1)S^{2}}\right )\\ & =\mathbb{P}\left (\frac{(n-1)S^{2}}{b}\leq\sigma^{2}\leq
\frac{(n-1)S^{2}}{a}\right ).\end{align*}

Therefore, the probability that the random interval $[(n-1)S^{2}/b,(n-1)S^{2}/a]$ contains the unknown $\sigma^{2}$ is $1-\alpha$. Once the values of $X_{1},X_{2},\cdots ,X_{n}$ are observed to be $x_{1},x_{2},\cdots ,x_{n}$ and $s^{2}$ computed, he interval $[(n-1)s^{2}/b,(n-1)s^{2}/a]$ is a $100(1-\alpha )\%$ confidence interval for $\sigma^{2}$. It follows that a $100(1-\alpha )\%$ confidnece interval for $\sigma$ (the standard deviation) is given by

\[\left [\sqrt{\frac{(n-1)s^{2}}{b}},\sqrt{\frac{(n-1)s^{2}}{a}}\right ]=\left [\sqrt{\frac{n-1}{b}}s,\sqrt{\frac{n-1}{a}}s\right ].\]

\begin{equation}{\label{ex1}}\tag{1}\mbox{}\end{equation}

Example \ref{ex1}. Assume that the time in days required for maturation of seeds of a species of guardiola, a flowering plant found in Mexico, is $N(\mu ,\sigma^{2})$. A radnom sample of $n=13$ seeds, both parents having narrow leaves, yielded $\bar{x}=18.97$ days and

\[12s^{2}=\sum_{i=1}^{13} (x_{i}-\bar{x})^{2}=128.41.\]

A $90\%$ confidence interval for $\sigma^{2}$ is given by

\[\left [\frac{128.41}{21.03},\frac{128.41}{5.226}\right ]=[6.11,24.57]\]

since $\chi_{0.95}^{2}(12)=5.226$ and $\chi_{0.05}^{2}(12)=21.03$. The corresponding $90\%$ confidence interval for $\sigma$ is given by

\[[\sqrt{6.11},\sqrt{24.57}]=[2.47,4.96].\]

There is an opportunity to compare the variances of two normal distributions. We do this by finding a confidence interval for $\sigma_{X}^{2}/\sigma_{Y}^{2}$ using the ratio of $S_{X}^{2}/\sigma_{X}^{2}$ and $S_{Y}^{2}/\sigma_{Y}^{2}$, where $S_{X}^{2}$ and $S_{Y}^{2}$ are two sample variances based on two independent samples of sizes $n$ and $m$ from $N(\mu_{X},\sigma_{X}^{2})$ and $N(\mu_{Y},\sigma_{Y}^{2})$, respectively. Howeve,r the reciprocal of that ratio can be rewritten as follows

\[\frac{\frac{S_{Y}^{2}}{\sigma_{Y}^{2}}}{\frac{S_{X}^{2}}{\sigma_{X}^{2}}}=\frac{\frac{(m-1)S_{Y}^{2}}{\sigma_{Y}^{2}}/(m-1)}{\frac{(n-1)S_{X}^{2}}{\sigma_{X}^{2}}/(n-1)}.\]

Since $(m-1)S_{Y}^{2}/\sigma_{Y}^{2}$ and $(n-1)S_{X}^{2}/\sigma_{X}^{2}$ are independent chi-square random variables with $m-1$ and $n-1$ degrees of freedom, respectively, this suggest that we define a new random variable, which is usually denoted by $F$. Assume that $F$ has an $F$-distribution with $r_{1}$ and $r_{2}$ degrees of freedom. Let $F_{\alpha}(r_{1},r_{2})$ denote the constant for which

\[\mathbb{P}(F\geq F_{\alpha}(r_{1},r_{2}))=\alpha .\]

Assume that the distribution of $F$ is $F(r_{1},r_{2})$.

When $r_{1}=7$ and $r_{2}=8$, we have $\mathbb{P}(F\geq 3.5)=0.95$, i.e., $F_{0.05}(7,8)=3.5$
When $r_{1}=9$ and $r_{2}=4$, we have $\mathbb{P}(F\geq 14.66)=0.99$, i.e., $F_{0.01}(9,4)=14.66$.

Since $F=(U/r_{1})/(V/r_{2})$, where $U$ and $V$ are independent and $\chi^{2}(r_{1})$ and $chi^{2}(r_{2})$, respectively, we see that $1/F=(V/r_{2})/ (U/r_{1})$ has a distribution $F(r_{2},r_{1})$. If the distribution of $F$ is $F(r_{1},r_{2})$ then

\[\mathbb{P}(F\leq F_{1-\alpha}(r_{1},r_{2}))=\alpha\]

and

\[\mathbb{P}\left (\frac{1}{F}\geq\frac{1}{F_{1-\alpha}}(r_{1},r_{2})\right )=\alpha .\]

The complement of $\{1/F\geq 1/F_{1-\alpha}(r_{1},r_{2})\}$ is $\{1/F<1/F_{1-\alpha}(r_{1},r_{2})\}$. Therefore, we have

\[\mathbb{P}\left (\frac{1}{F}<\frac{1}{F_{1-\alpha}}(r_{1},r_{2})\right )=1-\alpha .\]

Since the distribution of $1/F$ is $F(r_{2},r_{1})$, by definition of $F_{\alpha}(r_{2},r_{1})$, we have

\[P\left (\frac{1}{F}<F_{\alpha}(r_{2},r_{1})\right )=1-\alpha ,\]

which implies

\[\frac{1}{F_{1-\alpha}}(r_{1},r_{2})=F_{\alpha}(r_{2},r_{1}).\]

Now, we present how the $F$-distribution can be used to find a confidence interval for $\sigma_{X}^{2}/\sigma_{Y}^{2}$. Let $X_{1},X_{2},\cdots .X_{n}$ and $Y_{1},Y_{2},\cdots .Y_{n}$ be independent random samples of sizes $n$ and $m$ from two normal distributions $N(\mu_{X}^{2},\sigma_{X}^{2})$ and $N(\mu_{Y},\sigma_{Y}^{2})$, respectively. We know that $(n-1)S_{X}^{2}/\sigma_{X}^{2}$ is $\chi^{2}(n-1$ and $(m-1)S_{Y}^{2}/\sigma_{Y}^{2}$ is $\chi^{2}(m-1)$. The independence of the samples implies that $S_{X}$ and $S_{Y}$ are independent such that

\[F=\frac{(m-1)S_{Y}^{2}}{\sigma_{Y}^{2}(m-1)}/\frac{(n-1)S_{X}^{2}}{\sigma_{X}^{2}(n-1)}=\frac{S_{Y}^{2}}{\sigma_{Y}^{2}}/\frac{S_{X}^{2}}{\sigma_{X}^{2}}\]

has an $F$-distribution with $r_{1}=m-1$ and $r_{2}=n-1$ degrees of freedom. To form the confidence interval, select constants $c$ and $d$ from table satisfying

\begin{align*} 1-\alpha & =\mathbb{P}\left (c\leq\frac{S_{Y}^{2}/\sigma_{Y}^{2}}{S_{X}^{2}/\sigma_{X}^{2}}\leq d\right )\\ & =\mathbb{P}\left (c\cdot\frac{S_{X}^{2}}{S_{Y}^{2}}\leq\frac{\sigma_{X}^{2}}{\sigma_{Y}^{2}}\leqd\cdot\frac{S_{X}^{2}}{S_{Y}^{2}}\right ).\end{align*}

Let

\[c=F_{1-\alpha /2}(m-1,n-1)=1/F_{\alpha /2}(n-1,m-1)\mbox{ and }d=F_{\alpha /2}(m-1,n-1)$.\]

If $s_{X}^{2}$ and $s_{Y}^{2}$ are the observed values of $S_{X}^{2}$ and $S_{Y}^{2}$, respectively, then

\[\left [\frac{1}{F_{\alpha /2}}(n-1,m-1)\cdot\frac{s_{X}^{2}}{s_{Y}^{2}},F_{\alpha /2}(m-1,n-1)\cdot\frac{s_{X}^{2}}{s_{Y}^{2}}\right ]\]

is a $100(1-\alpha )\%$ confidence interval for $\sigma_{X}^{2}/\sigma_{Y}^{2}$. By taking square roots of both endpoints, we would obtain a $100(1-\alpha )\%$ confidence interval for $\sigma_{X}/\sigma_{Y}$.

Example. Continued from Example \ref{ex1}, we have $(n-1)s_{X}^{2}=12s^{2}=128.41$. Assume that the time in days required for maturation of seeds of species of guardiola, both parents having broad leaves, is $N(\mu_{Y},\sigma_{Y}^{2})$. A random sample of size $m=9$ seeds yielded $\bar{y}=23.2$ and

\[8s_{y}^{2}=\sum_{i=1}^{8} (y_{i}-\bar{y})^{2}=36.72.\]

A $98\%$ confidence interval for $\sigma_{X}^{2}/\sigma_{Y}^{2}$ is given by

\[\left [\frac{1}{5.67}\cdot\frac{128.41/12}{36.72/8}\right ]\]

since $F_{0.01}(12,8)=5.67$ and $F_{0.01}(8,12)=4.5$. It follows that a $98\%$ confidence interval for $\sigma_{X}\sigma_{Y}$ is given by

\[\left [\frac{1}{5.67}\cdot\frac{128.41/12}{36.72/8},4.5\cdot\frac{128.41/12}{36.72/8}\right ]=[0.41,10.49]\]

since $F_{0.01}(12,8)=5.67$ and $F_{0.01}(8,12)=4.5$. It follows that a $98\%$ confidence interval for $\sigma_{X}\sigma_{Y}$ is

\[[\sqrt{0.41},\sqrt{10.49}]=[0.64,3.24].\]

A random interval is a finite or infinite interval, where at least one of the end-points is a random variable. Let $L(X_{1},\cdots,X_{n})$ and $U(X_{1},\cdots ,X_{n})$ be two statistics such that $L(X_{1},\cdots ,X_{n})\leq U(X_{1},\cdots ,X_{n})$. We say that the random interval $[L(X_{1},\cdots ,X_{n},U(X_{1},\cdots ,X_{n})]$ is a confidence interval for $\theta$ with confidence coefficient $1-\alpha$ $(0<\alpha <1)$ when

\[\mathbb{P}_{\theta}\left (L(X_{1},\cdots ,X_{n})\leq\theta\leq U(X_{1},\cdots ,X_{n})\right )\geq 1-\alpha\mbox{ for all }\theta\in\Omega.\]

The interpretation of the confidence interval is as follows. Suppose that the random experiment under consideration is carried out independently $n$ times, and if $x_{j}$ is the observed value of $X_{j}$ for $j=1,\cdots ,j$, construct the interval $[L(x_{1},\cdots,x_{n}),U(x_{1},\cdots ,x_{n}]$. Suppose now that this process is repeated independently $N$ times such that we obtain $N$ intervals. Then, as $N$ gets larger and larger, at least $(1-\alpha )N$ of the $N$ intervals will cover the true parameter $\theta$.

\begin{equation}{\label{e2}}\tag{2}\mbox{}\end{equation}

Example \ref{e2}. Let $X_{1},\cdots ,X_{n}$ be i.i.d. random variables from $N(\mu ,\sigma^{2})$. Suppose that $\sigma$ is known such that $\mu$ is a parameter. Consider the random variables $T_{n}(\mu)=\sqrt{\mu} (\bar{X}-\mu )/\sigma$. Then $T_{n}(\mu )$ depends on the $X$’s only through the sufficient statistic $\bar{X}$ of $\mu$ and its distribution is $N(0,1)$ for all $\mu$. We want to determine two numbers $a$ and $b$ with $a<b$ satisfying

\[\mathbb{P}(a\leq N(0,1)\leq b)=1-\alpha.\]

Then, we have

\[\mathbb{P}_{\mu}\left(a\leq\frac{\sqrt{\mu}(\bar{X}-\mu )}{\sigma}\leq b\right)=1-\alpha,\]

which is equivalent to

\[\mathbb{P}_{\mu}\left{\bar{X}-b\frac{\sigma}{\sqrt{n}}\leq\mu\leq\bar{X}-a\frac{\sigma}{\sqrt{n}}\right}=1-\alpha .\]

Therefore

\[\left [\bar{X}-b\frac{\sigma}{\sqrt{n}},\bar{X}-a\frac{\sigma}{\sqrt{n}}\right ]\]

is a confidence interval for $\mu$ with confidence coefficient $1-\alpha$. Let $z_{\alpha /2}$ denote the upper $\alpha /2$ quantile of the $N(0,1)$ distribution. Then, we can take
$b=-a=z_{\alpha /2}$. Therefore,

\[\left [\bar{X}-z_{\alpha /2}\frac{\sigma}{\sqrt{n}},\bar{X}+z_{\alpha /2}\frac{\sigma}{\sqrt{n}}\right ]\]

is also a confidence interval for $\mu$ with confidence coefficient $1-\alpha$. Next, we assume that $\mu$ is known such that $\sigma^{2}$ is the parameter. Consider the random variable $U_{n}(\sigma^{2})=\frac{nS_{n}^{2}}{\sigma^{2}}$, where

\[S_{n}^{2}=\frac{1}{n}\sum_{j=1}^{n} (X_{j}-\mu )^{2}.\]

Then, the distribution of $U_{n}(\sigma^{2})$ is $\chi_{n}^{2}$ for all $\sigma^{2}$. We are going to determine two numbers $a$ and $b$ with $a<b$ satisfying

\[\mathbb{P}(a\leq\chi_{n}^{2}\leq b)=1-\alpha .\]

Then, we have

\[\mathbb{P}_{\sigma^{2}}\left(a\leq\frac{nS_{n}^{2}}{\sigma^{2}}\leq b\right)=1-\alpha,\]

which is equivalent to

\[\mathbb{P}_{\sigma^{2}}\left(\frac{nS_{n}^{2}}{b}\leq\sigma^{2}\leq\frac{nS_{n}^{2}}{a}\right)=1-\alpha .\]

Therefore

\[\left [\frac{nS_{n}^{2}}{b},\frac{nS_{n}^{2}}{a}\right ]\]

is a confidence interval for $\sigma^{2}$ with confidence coefficient $1-\alpha$. Let $a=\chi_{n,\alpha /2}^{2}$ and $b=\chi_{n,1-\alpha /2}^{2}$. Then

\[\left [\frac{nS_{n}^{2}}{\chi_{n,\alpha /2}^{2}},\frac{nS_{n}^{2}}{\chi_{n,1-\alpha /2}^{2}}\right ]\]

is a confidence interval for $\sigma^{2}$ with confidence coefficient. $\sharp$

\begin{equation}{\label{e3}}\tag{3}\mbox{}\end{equation}

Example \ref{e3}. Referring to Example \ref{e2}, assume that both $\mu$ and $\sigma$ are unknown. We are interested in constructing a confidence interval for $\mu$. For this purpose, we consider the random variable $T_{n}(\mu )$ of Example \ref{e2} and replace $\sigma$ by its usual estimator

\[S_{n-1}^{2}=\frac{1}{n-1}\sum_{j=1}^{n} (X_{j}-\bar{X}_{n})^{2}.\]

Therefore, we obtain the new random variable

\[U_{n}(\mu )=\frac{\sqrt{n}(\bar{X}_{n}-\mu )}{S_{n-1}}.\]

Since ${\displaystyle \sum_{j=1}^{n}\left ( \frac{X_{j}-\bar{X}_{n}}{\sigma}\right )^{2}}$ is $\chi_{n-1}^{2}$, it follows that ${\displaystyle \frac{(n-1)S_{n-1}^{2}}{\sigma^{2}}}$ is $\chi_{n-1}^{2}$. Since ${\displaystyle \frac{\sqrt{n}(\bar{X}_{n}-\mu )}{\sigma}}$ is $N(0,1)$, we see that $U_{n}(\mu )$ is $t_{n-1}$ distribution. As in Example \ref{e2}, we obtain a confidence interval of the form

\[\left [\bar{X}_{n}-b\frac{S_{n-1}}{\sqrt{n}},\bar{X}_{n}-a\frac{S_{n-1}}{\sqrt{n}}\right ].\]

Let $t_{n-1;\alpha /2}$ be the upper $\alpha /2$ quantile of the $t_{n-1}$ distribution. Then, we can get the confidence interval with confidence coefficient $1-\alpha$ by setting $b=-a=t_{n-1;\alpha/2}$. Suppose now that we wish to construct a confidence interval for $\sigma^{2}$. We set

\[V_{n}(\sigma^{2})=\frac{(n-1)S_{n-1}^{2}}{\sigma^{2}}\]

such that $V_{n}(\sigma^{2})$ is $\chi_{n-1}^{2}$ distributed. Then, we have that

\[\left [\frac{(n-1)S_{n-1}^{2}}{\chi_{n-1;1-\alpha /2}},\frac{(n-1)S_{n-1}^{2}}{\chi_{n-1;\alpha /2}}\right ]\]

is a confidence interval for $\sigma^{2}$ with confidence coefficient $1-\alpha$. $\sharp$

Example. Let $X_{1},\cdots ,X_{n}$ be i.i.d. random variables from the Gamma distribution with parameter $\beta$ and $\alpha$ a known positive integer, call it $r$. Then $\sum_{j=1}^{n} X_{j}$ is a sufficient statistic for $\beta$. Furthermore, for each $j=1,\cdots ,n$, the random variable $2X_{j}/\beta$ is $\chi_{2r}^{2}$ since

\[\phi_{2X_{j}/\beta}(t)=\phi_{X_{j}}\left (\frac{2t}{\beta}\right )=\frac{1}{(1-2it )^{2r/2}}.\]

Therefore, we see that

\[T_{n}(\beta )=\frac{2}{\beta}\sum_{j=1}^{n} X_{j}\]

is $\chi_{2rn}^{2}$ for all $\beta >0$. We want to determine $a$ and $b$ with $a<b$ satisfying

\[\mathbb{P}(a\leq\chi_{2rn}^{2}\leq b)=1-\alpha .\]

Then, we have

\[\mathbb{P}_{\beta}\left (a\leq\frac{2}{\beta}\sum_{j=1}^{n} X_{j}\leq b\right)=1-\alpha,\]

which is equivalently to

\[\mathbb{P}_{\beta}\left(2\sum_{j=1}^{n} X_{j}/b\leq\beta\leq 2\sum_{j=1}^{n} X_{j}/a\right)=1-\alpha .\]

Therefore, a confidence interval for $\beta$ with confidence coefficient $1-\alpha$ is given by

\[\left [\frac{2\sum_{j=1}^{n} X_{j}}{b},\frac{2\sum_{j=1}^{n} X_{j}}{a}\right ].\]

Similarly, as in the above example, we can set $a=\chi_{2rn,\alpha/2}$ and $b=\chi_{2rn,1-\alpha /2}$ to get the confidence interval for $\beta$ with confidence coefficient $1-\alpha$.$\sharp$

Example. Consider the independent random samples $X_{1},\cdots ,X_{m}$ from $N(\mu_{1},\sigma_{1}^{2})$ and $Y_{1},\cdots ,Y_{n}$ from $N(\mu_{2}, \sigma_{2}^{2})$, where all $\mu_{1},\mu_{2},\sigma_{1}$ and $\sigma_{2}$ are unknown. Assume that a confidence interval for $\mu_{1}- \mu_{2}$ is desired. For this purpose, we also need to assume $\sigma_{1}= \sigma_{2}=\sigma$. Consider the random variable

\[T_{m,n}(\mu_{1}-\mu_{2})=\frac{(\bar{X}_{m}-\bar{Y}{n})-(\mu_{1}-\mu_{2})}{{\displaystyle \sqrt{\frac{(m-1)S_{m-1}^{2}+(n-1)S_{n-1}^{2}}{m+n-2}\left (\frac{1}{m}+\frac{1}{n}\right )}}}.\]

Then $T_{m,n}$ is distributed as $t_{m+n-2}$. As in Example \ref{e2}, the confidence interval for $\mu_{1}-\mu_{2}$ with confidence coefficient $1-\alpha$ is given by

\[\begin{array}{l}
\left [(\bar{X}_{m}-\bar{Y}_{n})-t_{m+n-2;\alpha /2}{\displaystyle \sqrt{\frac{(m-1)S_{m-1}^{2}+(n-1)S_{n-1}^{2}}{m+n-2}\left (\frac{1}{m}+\frac{1}{n}\right )}},\right .\\
\left .(\bar{X}_{m}-\bar{Y}_{n})+t_{m+n-2;\alpha /2}{\displaystyle \sqrt{\frac{(m-1)S_{m-1}^{2}+(n-1)S_{n-1}^{2}}{m+n-2}\left (\frac{1}{m}+\frac{1}{n}\right )}}\right ].
\end{array}\]

Next, we are interested in constructing a confidence interval for $\sigma_{1}^{2}/\sigma_{2}^{2}$. We consider the random variable

\[U_{m,n}\left (\frac{\sigma_{1}}{\sigma_{2}}\right )=\frac{\sigma_{1}^{2}}{\sigma_{2}^{2}}\cdot\frac{S_{n-1}^{2}}{S_{m-1}^{2}}\]

which is distributed as $F_{n-1,m-1}$. We want to determine two numbers $a$ and $b$ with $0<a<b$ satisfying

\[\mathbb{P}(a\leq F_{n-1,m-1}\leq b)=1-\alpha .\]

Then, we have

\[\mathbb{P}\left(a\leq\frac{\sigma_{1}^{2}}{\sigma_{2}^{2}}\cdot\frac{S_{n-1}^{2}}{S_{m-1}^{2}}\leq b\right)=1-\alpha\]

\[\mathbb{P}\left(a\frac{S_{m-1}^{2}}{S_{n-1}^{2}}\leq\frac{\sigma_{1}^{2}}{\sigma_{2}^{2}}\leq b\frac{S_{n-1}^{2}}{S_{m-1}^{2}}\right)=1-\alpha .\]

In particular, the equal-tails confidence interval is provided by

\[\left [\frac{S_{m-1}^{2}}{S_{n-1}^{2}}F’_{n-1,m-1;\alpha /2}, \frac{S_{m-1}^{2}}{S_{n-1}^{2}}F_{n-1,m-1;\alpha /2}\right ],\]

where $F’_{n-1,m-1;\alpha /2}$ and $F_{n-1,m-1;\alpha /2}$ are the lower and the upper $\alpha /2$-qunatiles of $F_{n-1,m-1}$. We also have

\[F’_{n-1.m-1;\alpha /2}=\frac{1}{F_{m-1,n-1;\alpha /2}}=F_{m-1,n-1;\alpha /2}.\]

\begin{equation}{\label{t11}\tag{4}\mbox{}\end{equation}

Theorem \ref{t11}. Let $X_{1},\cdots ,X_{n}$ be i.i.d $N(\mu ,\sigma^{2})$. Then $\bar{X}_{n}$ and $S_{n}^{2}$ are independent.

Example. Refer to Example \ref{e3}, we are going to construct a confidence region in $\mathbb{R}^{2}$ for $(\mu ,\sigma^{2})$. Consider the random variables

\[\frac{\sqrt{n}(\bar{X}_{n}-\mu )}{\sigma}\mbox{ and }\frac{(n-1)S_{n-1}^{2}}{\sigma^{2}},\]

which are independent by Theorem \ref{t11} and distributed as $N(0,1)$ and $\chi_{n-1}^{2}$, respectively. We want to determine the constants $c>0$, $a$ and $b$ with $0<a<b$ satisfying

\[\mathbb{P}(-c\leq N(0,1)\leq c)=\sqrt{1-\alpha}\]

and

\[\mathbb{P}(a\leq\chi_{n-1}^{2}\leq b)=\sqrt{1-\alpha}.\]

From these relationship, we obtain

\begin{align*} & \mathbb{P}_{\mu,\sigma}\left(-c\leq\frac{\sqrt{n}(\bar{X}_{n}-\mu )}{\sigma}\leq c,a\leq\frac{(n-1)S_{n-1}^{2}}{\sigma^{2}}\leq b\right)\\
& \quad=\mathbb{P}_{\mu,\sigma}\left(-c\leq\frac{\sqrt{n}(\bar{X}_{n}-\mu )}{\sigma}\leq c\right)\times\mathbb{P}_{\mu ,\sigma}\left(a\leq\frac{(n-1)S_{n-1}^{2}}
{\sigma^{2}}\leq b\right)\mbox{ (by independence)}\\
& \quad= 1-\alpha.\end{align*}

Equivalently, we have

\[\mathbb{P}_{\mu ,\sigma}\left((\mu -\bar{X}_{n})^{2}\leq\frac{c^{2}\sigma^{2}}{n}, \frac{(n-1)S_{n-1}^{2}}{b}\leq\sigma^{2}\leq\frac{(n-1)S_{n-1}^{2}}{a}\right)=1-\alpha.\]

For the observed values of the $X$’s, we can get the confidence region. $\sharp$

Let $X_{1},\cdots ,X_{n}$ be i.i.d. random variables. Assume that $\sqrt{n}(\bar{X}_{n}-\mu )/\sigma$ is approximately $N(0,1)$. Therefore, when $\sigma$ is known, a confidence interval for $\mu$ with approximate confidence coefficient $1-\alpha$ is given by

\[\left [\bar{X}_{n}-z_{\alpha /2}\frac{\sigma}{\sqrt{n}},\bar{X}_{n}+z_{\alpha /2}\frac{\sigma}{\sqrt{n}}\right ],\]

provided $n$ is sufficiently large. Suppose now that $\sigma$ is also unknown. Since

\[S_{n}^{2}=\frac{1}{n}\sum_{j=1}^{n} (X_{j}-\bar{X}_{n})^{2}\rightarrow\sigma^{2}\]

as $n\rightarrow\infty$ in probability, we have that $\sqrt{n}(\bar{X}_{n}-\mu )S_{n}$ is again approximately $N(0,1)$.