Chapter 7, Problem 59SE

a.

To determine

Show that $P\left({\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}<\frac{Y}{\theta }<{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}\right)=1-\alpha$.

Derive the $100\left(1-\alpha \right)\%$ confidence interval for $\theta$ based on given probability statement.

Answer to Problem 59SE

It is verified that $P\left({\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}<\frac{Y}{\theta }<{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}\right)=1-\alpha$

The $100\left(1-\alpha \right)\%$ confidence interval for $\theta$ based on given probability statement is $\underset{\_}{CI=\left(\frac{\max \left(X_i\right)}{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}},\frac{\max \left(X_i\right)}{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}\right)}$.

Explanation of Solution

Given info:

A random sample $X_1,X_2,\mathrm{...},X_n$ is from a uniform distribution on the interval $\left[0,\theta \right]$ and the probability density function of the random variable $U=\frac{Y}{\theta }=\frac{\max \left(X_i\right)}{\theta }$ is,

$f_U\left(u\right)=\{\begin{array}{l}n{u}^{n-1},0\le u\le 1\\ 0,\text{otherwise}\end{array}$.

Calculation:

Let $X_1,X_2,\mathrm{...},X_n$ be a random sample from a uniform distribution on the interval $\left[0,\theta \right]$.

The probability density function of the random variable $U=\frac{Y}{\theta }=\frac{\max \left(X_i\right)}{\theta }$ is,

$f_U\left(u\right)=\{\begin{array}{l}n{u}^{n-1},0\le u\le 1\\ 0,\text{otherwise}\end{array}$

Probability value:

$\begin{array}{c}P\left({\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}<\frac{Y}{\theta }<{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}\right)=P\left({\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}<U<{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}\right)\\ ={\displaystyle \underset{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}{\overset{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}}{\int }}{f}_U\left(u\right)du}\\ ={\displaystyle \underset{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}{\overset{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}}{\int }}\left(n{u}^{n-1}\right)du}\\ =n\times {\displaystyle \underset{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}{\overset{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}}{\int }}{u}^{n-1}du}\end{array}$

$\begin{array}{l}=n\times {\displaystyle \underset{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}{\overset{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}}{\int }}{u}^{n-1}du}\\ =n\times {\left[\frac{u^{n-1+1}}{n-1+1}\right]}_{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}^{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}}\\ ={\left[u^n\right]}_{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}^{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}}\\ ={\left({\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}\right)}^n-{\left({\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}\right)}^n\end{array}$

$\begin{array}{l}=1-\frac{\alpha }{2}-\frac{\alpha }{2}\\ =1-\alpha \end{array}$

Therefore, it is verified that $P\left({\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}<\frac{Y}{\theta }<{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}\right)=1-\alpha$.

Confidence interval based on given probability statement:

The confidence interval for $\theta$ is,

$\begin{array}{l}P\left({\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}<\frac{Y}{\theta }<{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}\right)=1-\alpha \\ P\left(\frac{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}{Y}<\frac{1}{\theta }<\frac{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}}{Y}\right)=1-\alpha \\ P\left(\frac{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}{\max \left(X_i\right)}<\frac{1}{\theta }<\frac{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}}{\max \left(X_i\right)}\right)=1-\alpha \\ P\left(\frac{\max \left(X_i\right)}{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}}<\theta <\frac{\max \left(X_i\right)}{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}\right)=1-\alpha \end{array}$

Thus, the $100\left(1-\alpha \right)\%$ confidence interval for $\theta$ based on given probability statement is $\underset{\_}{CI=\left(\frac{\max \left(X_i\right)}{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}},\frac{\max \left(X_i\right)}{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}\right)}$.

b.

To determine

Show that $P\left({\alpha }^{\frac{1}{n}}<\frac{Y}{\theta }<1\right)=1-\alpha$.

Derive the $100\left(1-\alpha \right)\%$ confidence interval for $\theta$ based on given probability statement.

Answer to Problem 59SE

It is verified that $P\left({\alpha }^{\frac{1}{n}}<\frac{Y}{\theta }<1\right)=1-\alpha$

The $100\left(1-\alpha \right)\%$ confidence interval for $\theta$ based on given probability statement is $\underset{\_}{CI=\left(\max \left(X_i\right),\frac{\max \left(X_i\right)}{{\alpha }^{\frac{1}{n}}}\right)}$.

Explanation of Solution

Calculation:

Let $X_1,X_2,\mathrm{...},X_n$ be a random sample from a uniform distribution on the interval $\left[0,\theta \right]$.

The probability density function of the random variable $U=\frac{Y}{\theta }=\frac{\max \left(X_i\right)}{\theta }$ is,

$f_U\left(u\right)=\{\begin{array}{l}n{u}^{n-1},0\le u\le 1\\ 0,\text{otherwise}\end{array}$

Probability value:

$\begin{array}{c}P\left({\alpha }^{\frac{1}{n}}<\frac{Y}{\theta }<1\right)=1-\alpha =P\left({\alpha }^{\frac{1}{n}}<U<1\right)\\ ={\displaystyle \underset{{\alpha }^{\frac{1}{n}}}{\overset{1}{\int }}{f}_U\left(u\right)du}\\ ={\displaystyle \underset{{\alpha }^{\frac{1}{n}}}{\overset{1}{\int }}\left(n{u}^{n-1}\right)du}\\ =n\times {\displaystyle \underset{{\alpha }^{\frac{1}{n}}}{\overset{1}{\int }}{u}^{n-1}du}\end{array}$

                       $\begin{array}{l}=n\times {\displaystyle \underset{{\alpha }^{\frac{1}{n}}}{\overset{1}{\int }}{u}^{n-1}du}\\ =n\times {\left[\frac{u^{n-1+1}}{n-1+1}\right]}_{{\alpha }^{\frac{1}{n}}}^1\\ ={\left[u^n\right]}_{{\alpha }^{\frac{1}{n}}}^1\\ ={\left(1\right)}^n-{\left({\alpha }^{\frac{1}{n}}\right)}^n\end{array}$

                       $=1-\alpha$

Therefore, it is verified that $P\left({\alpha }^{\frac{1}{n}}<\frac{Y}{\theta }<1\right)=1-\alpha$.

Confidence interval:

The confidence interval for $\theta$ is,

$\begin{array}{l}P\left({\alpha }^{\frac{1}{n}}<\frac{Y}{\theta }<1\right)=1-\alpha \\ P\left(\frac{{\alpha }^{\frac{1}{n}}}{Y}<\frac{1}{\theta }<\frac{1}{Y}\right)=1-\alpha \\ P\left(\frac{{\alpha }^{\frac{1}{n}}}{\max \left(X_i\right)}<\frac{1}{\theta }<\frac{1}{\max \left(X_i\right)}\right)=1-\alpha \\ P\left(\max \left(X_i\right)<\theta <\frac{\max \left(X_i\right)}{{\alpha }^{\frac{1}{n}}}\right)=1-\alpha \end{array}$

Thus, the $100\left(1-\alpha \right)\%$ confidence interval for $\theta$ is $\underset{\_}{CI=\left(\max \left(X_i\right),\frac{\max \left(X_i\right)}{{\alpha }^{\frac{1}{n}}}\right)}$.

c.

To determine

Find the shorter interval among the obtained two intervals.

Find the 95% confidence interval for $\theta$ using the shorter interval among the obtained two intervals.

Answer to Problem 59SE

The confidence interval obtained in part (b) is shorter than the interval obtained in part (a).

The 95% confidence interval for $\theta$ using the shorter interval is $\underset{\_}{\left(4.2,7.65\right)}$.

Explanation of Solution

Given info:

The data represents the sample of 5 waiting times of a morning bus. The variable waiting time for a morning bus is uniformly distributed.

Calculation:

The confidence interval for $\theta$ obtained in part (a) is $CI=\left(\frac{\max \left(X_i\right)}{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}},\frac{\max \left(X_i\right)}{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}\right)$

Width of the confidence interval obtained in part (b):

$\begin{array}{c}{\text{Width}}_1=\frac{\max \left(X_i\right)}{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}-\frac{\max \left(X_i\right)}{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}}\\ =\max \left(X_i\right)\left(\frac{1}{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}-\frac{1}{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}}\right)\\ ==\max \left(X_i\right)\left(\frac{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}-{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}\times {\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}}\right)\end{array}$

The confidence interval for $\theta$ obtained in part (b) is $CI=\left(\max \left(X_i\right),\frac{\max \left(X_i\right)}{{\alpha }^{\frac{1}{n}}}\right)$

Width of the confidence interval obtained in part (b):

$\begin{array}{c}{\text{Width}}_2=\frac{\max \left(X_i\right)}{{\alpha }^{\frac{1}{n}}}-\max \left(X_i\right)\\ =\max \left(X_i\right)\left(\frac{1}{{\alpha }^{\frac{1}{n}}}-1\right)\end{array}$

Here, the value of $\alpha$ will be always between 0 and 1.

That is, $0<\alpha <1$.

And the sample size n will be always greater than 1.

That is, $n>1$.

From this it can be said that, $\left(\frac{{\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}-{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}}{{\left(\frac{\alpha }{2}\right)}^{\frac{1}{n}}\times {\left(1-\frac{\alpha }{2}\right)}^{\frac{1}{n}}}\right)$ will be greater than $\left(\frac{1}{{\alpha }^{\frac{1}{n}}}-1\right)$.

Therefore, the width of the interval in part (b) will be narrower than the width of the interval in part (a).

Thus, the shorter confidence interval for $\theta$ is $CI=\left(\max \left(X_i\right),\frac{\max \left(X_i\right)}{{\alpha }^{\frac{1}{n}}}\right)$.

95% confidence interval:

For 95% confidence level,

$\begin{array}{c}1-\alpha =0.95\\ \alpha =0.05\end{array}$

Thus, the level of significance is $\alpha =0.05$.

The maximum waiting time for a sample of 5 waiting times is $\max \left(X_i\right)=4.2$.

The 95% confidence interval for $\theta$ is,

$\begin{array}{c}CI=\left(\max \left(X_i\right),\frac{\max \left(X_i\right)}{{\alpha }^{\frac{1}{n}}}\right)\\ =\left(4.2,\frac{4.2}{{0.05}^{\frac{1}{5}}}\right)\\ =\left(4.2,\frac{4.2}{{0.05}^{0.2}}\right)\\ =\left(4.2,7.65\right)\end{array}$

Thus, the 95% confidence interval for $\theta$ is $\underset{\_}{\left(4.2,7.65\right)}$.