Chapter 19, Problem 14E

(a)

To determine

To find the margin of error for this poll if we want $95\%$ confidence in our estimate of the percent of American adults who approve cloning humans.

Answer to Problem 14E

  $0.019$ .

Explanation of Solution

It is given in the question that a May Gallup poll found that only $11\%$ of a random sample of $1003$ adults approved of attempts to clone a human. Thus, to find the margin of error for this poll if we want $95\%$ confidence in our estimate of the percent of American adults who approve cloning humans, we have,

First find the critical value as:

  $\begin{array}{l}{z}^{*}=invNorm(0.95,0,1)\\ =1.960\end{array}$

Thus, the margins of error will be:

  $\begin{array}{l}ME=1.960\times \sqrt{\frac{0.11\times 0.89}{1003}}\\ =0.019\end{array}$

Thus, $0.019$ margin of error for this poll is needed if we want $95\%$ confidence in our estimate of the percent of American adults who approve cloning humans.

(b)

To determine

To explain what that margin of error means.

Explanation of Solution

It is given in the question that a May Gallup poll found that only $11\%$ of a random sample of $1003$ adults approved of attempts to clone a human. Thus, $0.019$ margin of error for this poll is needed if we want $95\%$ confidence in our estimate of the percent of American adults who approve cloning humans. Thus, the margin of error means the error region about the point estimate.

(c)

To determine

To explain will the margin of error be larger or smaller.

Answer to Problem 14E

The margin of error be smaller.

Explanation of Solution

It is given in the question that a May Gallup poll found that only $11\%$ of a random sample of $1003$ adults approved of attempts to clone a human. Thus, $0.019$ margin of error for this poll is needed if we want $95\%$ confidence in our estimate of the percent of American adults who approve cloning humans. Thus, the margin of error will be smaller because a $90\%$ confidence level would contain a smaller range which means that the margin of error must be smaller.

(d)

To determine

To find that margin of error.

Answer to Problem 14E

  $0.016$ .

Explanation of Solution

It is given in the question that a May Gallup poll found that only $11\%$ of a random sample of $1003$ adults approved of attempts to clone a human. Thus, to find the margin of error, we have,

First find the critical value as:

  $\begin{array}{l}{z}^{*}=invNorm(0.995,0,1)\\ =1.645\end{array}$

Thus, the margins of error will be:

  $\begin{array}{l}ME=1.645\times \sqrt{\frac{0.11\times 0.89}{1003}}\\ =0.016\end{array}$

(e)

To determine

To explain would smaller samples produce smaller or larger margins of error.

Answer to Problem 14E

The smaller samples produce larger margins of error.

Explanation of Solution

It is given in the question that a May Gallup poll found that only $11\%$ of a random sample of $1003$ adults approved of attempts to clone a human. Thus, smaller samples produce larger margins of error because the smaller the sample, the more variability it will have. This means a larger margin or error.