Chapter 8.3, Problem 37E

(a)

To determine

Appropriate words to fill in the blanks.

Answer to Problem 37E

We are $98\%$ confident that the population proportion is between $0.732082$ and $0.870128$.

Explanation of Solution

Given information:

A MINITAB output for a $98\%$ confidence interval for a proportion is given as below.

$\begin{array}{l}\text{Sample X N Sample p }\text{ 98}\%\text{CI}\\ 11451810.801105\left(0.732082,0.870128\right)\end{array}$

A statement is given as:

We are ______________ confident that the population proportion is between___________ and ______.

Calculation:

A MINITAB output for a $98\%$ confidence interval for a proportion is given as below.

$\begin{array}{l}\text{Sample X N Sample p }\text{ 98}\%\text{CI}\\ 11451810.801105\left(0.732082,0.870128\right)\end{array}$

From above output it is clear that the $98\%$ confident interval for the given population proportion is between $(0.732082,0.870128)$.

This mean -

We are $98\%$ confident that the population proportion is between $0.732082$ and $0.870128$.

(b)

To determine

The $90\%$ confidence interval for $p$.

Answer to Problem 37E

$90\%$ confidence interval for population proportion would be $0.751\text{ to 0}\text{.849}$.

Explanation of Solution

Given information:

A MINITAB output for a $98\%$ confidence interval for a proportion is given as below.

$\begin{array}{l}\text{Sample X N Sample p }\text{ 98}\%\text{CI}\\ 11451810.801105\left(0.732082,0.870128\right)\end{array}$

Calculation:

$\begin{array}{l}\text{Sample X N Sample p }\text{ 98}\%\text{CI}\\ 11451810.801105\left(0.732082,0.870128\right)\end{array}$

From the standard normal table, the z-critical value at $90\%$ confidence is $1.645$.

So, $90\%$ confidence interval for population proportion can be calculated as below -

$\begin{array}{c}90\%CI.=\overline{p}\pm z_C\times \sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}\\ =0.80\pm 1.645\times \sqrt{\frac{0.80\left(1-0.80\right)}{181}}\\ =0.80\pm 0.049\left(\text{Margin of Error}\right)\\ =0.751\text{ to 0}\text{.849}\end{array}$

Hence, a $90\%$ confidence interval for population proportion would be $0.751\text{ to 0}\text{.849}$.