Chapter 19, Problem 29E

(a)

To determine

To find out how many of this younger age must we survey in order to estimate the proportion of non-grads to within $\text{6\% with 90\%}$ confidence.

Answer to Problem 29E

  $141$ .

Explanation of Solution

It is given in the question that it is believed that as many as $\text{25\%}$ of adults over $\text{50}$ never graduated from high school. They wish to see if the percentage is the same among the $\text{25 to 30}$ age group. Thus, to find the number of younger age must we survey in order to estimate the proportion of non-grads to within $\text{6\% with 90\%}$ confidence, we have,

  $\begin{array}{l}0.06=1.645\times \sqrt{\frac{0.25\times 0.75}{n}}\\ \text{Solve for }n\\ \Rightarrow 0.0365=\sqrt{\frac{0.25\times 0.75}{n}}\\ \Rightarrow n=141\end{array}$

Thus, $141$ younger age must we survey in order to estimate the proportion of non-grads to within $\text{6\% with 90\%}$ confidence.

(b)

To determine

To explain what is the necessary sample size.

Answer to Problem 29E

  $318$ .

Explanation of Solution

It is given in the question that it is believed that as many as $\text{25\%}$ of adults over $\text{50}$ never graduated from high school. They wish to see if the percentage is the same among the $\text{25 to 30}$ age group. Thus, to find the necessary sample size, we have,

  $\begin{array}{l}0.04=1.645\times \sqrt{\frac{0.25\times 0.75}{n}}\\ \text{Solve for }n\\ \Rightarrow 0.0243=\sqrt{\frac{0.25\times 0.75}{n}}\\ \Rightarrow n=318\end{array}$

Thus, the necessary sample size is $318$ .

(c)

To determine

To find out what sample size would produce a margin of error of $\text{3\%}$ .

Answer to Problem 29E

  $564$ .

Explanation of Solution

It is given in the question that it is believed that as many as $\text{25\%}$ of adults over $\text{50}$ never graduated from high school. They wish to see if the percentage is the same among the $\text{25 to 30}$ age group. Thus, to find sample size that would produce a margin of error of $\text{3\%}$ , we have,

  $\begin{array}{l}0.03=1.645\times \sqrt{\frac{0.25\times 0.75}{n}}\\ \text{Solve for }n\\ \Rightarrow 0.0182=\sqrt{\frac{0.25\times 0.75}{n}}\\ \Rightarrow n=564\end{array}$

Thus, $564$ sample size we would produce for a margin of error of $\text{3\%}$ .