Chapter 9.4, Problem 5E

(a)

To determine

To find: The $\widehat{p}$ and $\widehat{q}$.

Answer to Problem 5E

The value of ${\widehat{p}}_1$, ${\widehat{p}}_2$, ${\widehat{q}}_1$ and ${\widehat{q}}_2$ is 0.5, 0.44, 0.5 and 0.56, respectively.

Explanation of Solution

Given info:

$X_1=25$, $n_1=75$, $X_2=40$ and $n_2=90$.

Calculation:

The value of ${\widehat{p}}_1$ is,

$\begin{array}{c}{\widehat{p}}_1=\frac{X_1}{n_1}\\ =\frac{25}{75}\\ =0.5\end{array}$

The value of ${\widehat{q}}_1$ is,

$\begin{array}{c}{\widehat{q}}_1=1-{\widehat{p}}_1\\ =1-0.5\\ =0.5\end{array}$

The value of ${\widehat{p}}_2$ is,

$\begin{array}{c}{\widehat{p}}_2=\frac{X_2}{n_2}\\ =\frac{40}{90}\\ =0.44\end{array}$

The value of ${\widehat{q}}_2$ is,

$\begin{array}{c}{\widehat{q}}_2=1-{\widehat{p}}_2\\ =1-0.44\\ =0.56\end{array}$

Thus, the value of ${\widehat{p}}_1$, ${\widehat{p}}_2$, ${\widehat{q}}_1$ and ${\widehat{q}}_2$ is 0.5, 0.44, 0.5 and 0.56, respectively.

(b)

To determine

To find: The $\widehat{p}$ and $\widehat{q}$.

Answer to Problem 5E

The value of ${\widehat{p}}_1$, ${\widehat{p}}_2$, ${\widehat{q}}_1$ and ${\widehat{q}}_2$ is 0.6, 0.4, 0.35 and 0.65, respectively.

Explanation of Solution

Given info:

$X_1=9$, $n_1=15$, $X_2=7$ and $n_2=20$.

Calculation:

The value of ${\widehat{p}}_1$ is,

$\begin{array}{c}{\widehat{p}}_1=\frac{X_1}{n_1}\\ =\frac{9}{15}\\ =0.6\end{array}$

The value of ${\widehat{q}}_1$ is,

$\begin{array}{c}{\widehat{q}}_1=1-{\widehat{p}}_1\\ =1-0.6\\ =0.4\end{array}$

The value of ${\widehat{p}}_2$ is,

$\begin{array}{c}{\widehat{p}}_2=\frac{X_2}{n_2}\\ =\frac{7}{20}\\ =0.35\end{array}$

The value of ${\widehat{q}}_2$ is,

$\begin{array}{c}{\widehat{q}}_2=1-{\widehat{p}}_2\\ =1-0.35\\ =0.65\end{array}$

Thus, the value of ${\widehat{p}}_1$, ${\widehat{p}}_2$, ${\widehat{q}}_1$ and ${\widehat{q}}_2$ is 0.6, 0.4, 0.35 and 0.65, respectively.

(c)

To determine

To find: The $\widehat{p}$ and $\widehat{q}$.

Answer to Problem 5E

The value of ${\widehat{p}}_1$, ${\widehat{p}}_2$, ${\widehat{q}}_1$ and ${\widehat{q}}_2$ is 0.15, 0.85, 0.125 and 0.875, respectively.

Explanation of Solution

Given info:

$X_1=3$, $n_1=20$, $X_2=5$ and $n_2=40$.

Calculation:

The value of ${\widehat{p}}_1$ is,

$\begin{array}{c}{\widehat{p}}_1=\frac{X_1}{n_1}\\ =\frac{3}{20}\\ =0.15\end{array}$

The value of ${\widehat{q}}_1$ is,

$\begin{array}{c}{\widehat{q}}_1=1-{\widehat{p}}_1\\ =1-0.15\\ =0.85\end{array}$

The value of ${\widehat{p}}_2$ is,

$\begin{array}{c}{\widehat{p}}_2=\frac{X_2}{n_2}\\ =\frac{5}{40}\\ =0.125\end{array}$

The value of ${\widehat{q}}_2$ is,

$\begin{array}{c}{\widehat{q}}_2=1-{\widehat{p}}_2\\ =1-0.125\\ =0.875\end{array}$

Thus, the value of ${\widehat{p}}_1$, ${\widehat{p}}_2$, ${\widehat{q}}_1$ and ${\widehat{q}}_2$ is 0.15, 0.85, 0.125 and 0.875, respectively.

(d)

To determine

To find: The $\widehat{p}$ and $\widehat{q}$.

Answer to Problem 5E

The value of ${\widehat{p}}_1$, ${\widehat{p}}_2$, ${\widehat{q}}_1$ and ${\widehat{q}}_2$ is 0.42, 0.58, 0.64 and 0.36, respectively.

Explanation of Solution

Given info:

$X_1=21$, $n_1=50$, $X_2=32$ and $n_2=50$.

Calculation:

The value of ${\widehat{p}}_1$ is,

$\begin{array}{c}{\widehat{p}}_1=\frac{X_1}{n_1}\\ =\frac{21}{50}\\ =0.42\end{array}$

The value of ${\widehat{q}}_1$ is,

$\begin{array}{c}{\widehat{q}}_1=1-{\widehat{p}}_1\\ =1-0.42\\ =0.58\end{array}$

The value of ${\widehat{p}}_2$ is,

$\begin{array}{c}{\widehat{p}}_2=\frac{X_2}{n_2}\\ =\frac{32}{50}\\ =0.64\end{array}$

The value of ${\widehat{q}}_2$ is,

$\begin{array}{c}{\widehat{q}}_2=1-{\widehat{p}}_2\\ =1-0.64\\ =0.36\end{array}$

Thus, the value of ${\widehat{p}}_1$, ${\widehat{p}}_2$, ${\widehat{q}}_1$ and ${\widehat{q}}_2$ is 0.42, 0.58, 0.64 and 0.36, respectively.

(e)

To determine

To find: The $\widehat{p}$ and $\widehat{q}$.

Answer to Problem 5E

The value of ${\widehat{p}}_1$, ${\widehat{p}}_2$, ${\widehat{q}}_1$ and ${\widehat{q}}_2$ is 0.42, 0.58, 0.64 and 0.36, respectively.

Explanation of Solution

Given info:

$X_1=20$, $n_1=150$, $X_2=30$ and $n_2=50$.

Calculation:

The value of ${\widehat{p}}_1$ is,

$\begin{array}{c}{\widehat{p}}_1=\frac{X_1}{n_1}\\ =\frac{20}{150}\\ =0.13\end{array}$

The value of ${\widehat{q}}_1$ is,

$\begin{array}{c}{\widehat{q}}_1=1-{\widehat{p}}_1\\ =1-0.13\\ =0.87\end{array}$

The value of ${\widehat{p}}_2$ is,

$\begin{array}{c}{\widehat{p}}_2=\frac{X_2}{n_2}\\ =\frac{30}{50}\\ =0.6\end{array}$

The value of ${\widehat{q}}_2$ is,

$\begin{array}{c}{\widehat{q}}_2=1-{\widehat{p}}_2\\ =1-0.6\\ =0.4\end{array}$

Thus, the value of ${\widehat{p}}_1$, ${\widehat{p}}_2$, ${\widehat{q}}_1$ and ${\widehat{q}}_2$ is 0.13, 0.87, 0.64 and 0.4, respectively.