Chapter 9.4, Problem 6E

Find p̅ and q̅.

a. X₁ = 6, n₁ = 15, X₂ = 9, n₂ = 15

b. X₁ = 21, n₁ = 100, X₂ = 43, n₂ = 150

c. X₁ = 20, n₁ = 80, X₂ = 65, n₂ = 120

d. X₁ = 15, n₁ = 50, X₂ = 3, n₂ = 12

e. X₁ = 24, n₁ = 40, X₂ = 18, n₂ = 36

To determine

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

Answer to Problem 6E

The value of $\overline{p}$ and $\overline{q}$ is 0.5 and 0.5, respectively.

Explanation of Solution

Given info:

$X_1=6$, $n_1=15$, $X_2=9$ and $n_2=15$.

Calculation:

The value of $\overline{p}$ is,

$\begin{array}{c}\overline{p}=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{6+9}{15+15}\\ =\frac{\text{15}}{\text{30}}\\ =0.5\end{array}$

The value of $\overline{q}$ is,

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

Thus, the value of $\overline{p}$ and $\overline{q}$ is 0.5 and 0.5, respectively.

To determine

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

Answer to Problem 6E

The value of $\overline{p}$ and $\overline{q}$ is 0.26 and 0.74, respectively.

Explanation of Solution

Given info:

$X_1=21$, $n_1=100$, $X_2=43$ and $n_2=150$.

Calculation:

The value of $\overline{p}$ is,

$\begin{array}{c}\overline{p}=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{21+43}{100+150}\\ =\frac{\text{64}}{\text{250}}\\ =0.26\end{array}$

The value of $\overline{q}$ is,

$\begin{array}{c}\overline{q}=1-\overline{p}\\ =1-0.26\\ =0.74\end{array}$

Thus, the value of $\overline{p}$ and $\overline{q}$ is 0.26 and 0.74, respectively.

To determine

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

Answer to Problem 6E

The value of $\overline{p}$ and $\overline{q}$ is 0.43 and 0.57, respectively.

Explanation of Solution

Given info:

$X_1=20$, $n_1=80$, $X_2=65$ and $n_2=120$.

Calculation:

The value of $\overline{p}$ is,

$\begin{array}{c}\overline{p}=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{20+65}{80+120}\\ =\frac{\text{85}}{\text{200}}\\ =0.43\end{array}$

The value of $\overline{q}$ is,

$\begin{array}{c}\overline{q}=1-\overline{p}\\ =1-0.43\\ =0.57\end{array}$

Thus, the value of $\overline{p}$ and $\overline{q}$ is 0.43 and 0.57, respectively.

To determine

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

Answer to Problem 6E

The value of $\overline{p}$ and $\overline{q}$ is 0.29 and 0.71, respectively.

Explanation of Solution

Given info:

$X_1=15$, $n_1=50$, $X_2=3$ and $n_2=12$.

Calculation:

The value of $\overline{p}$ is,

$\begin{array}{c}\overline{p}=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{15+3}{50+12}\\ =\frac{\text{18}}{\text{62}}\\ =0.29\end{array}$

The value of $\overline{q}$ is,

$\begin{array}{c}\overline{q}=1-\overline{p}\\ =1-0.29\\ =0.71\end{array}$

Thus, the value of $\overline{p}$ and $\overline{q}$ is 0.29 and 0.71, respectively.

To determine

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

Answer to Problem 6E

The value of $\overline{p}$ and $\overline{q}$ is 0.55 and 0.45, respectively.

Explanation of Solution

Given info:

$X_1=24$, $n_1=40$, $X_2=18$ and $n_2=36$.

Calculation:

The value of $\overline{p}$ is,

$\begin{array}{c}\overline{p}=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{24+18}{40+36}\\ =\frac{\text{42}}{\text{76}}\\ =0.55\end{array}$

The value of $\overline{q}$ is,

$\begin{array}{c}\overline{q}=1-\overline{p}\\ =1-0.55\\ =0.45\end{array}$

Thus, the value of $\overline{p}$ and $\overline{q}$ is 0.55 and 0.45, respectively.