Chapter 8.4, Problem 37E

One of the questions on a survey of 1000 adults asked if today’s children will be better off than their parents (Rasmussen Reports website, October 26, 2012). Representative data are shown in the DATAfile named ChildOutlook. A response of Yes indicates that the adult surveyed did think today’s children will be better off than their parents. A response of No indicates that the adult surveyed did not think today’s children will be better off than their parents. A response of Not Sure was given by 23% of the adults surveyed.

1. a. What is the point estimate of the proportion of the population of adults who do think that today’s children will be better off than their parents?
2. b. At 95% confidence, what is the margin of error?
3. c. What is the 95% confidence interval for the proportion of adults who do think that today’s children will be better off than their parents?
4. d. What is the 95% confidence interval for the proportion of adults who do not think that today’s children will be better off than their parents?
5. e. Which of the confidence intervals in parts (c) and (d) has the smaller margin of error? Why?

To determine

Find the point estimate of the proportion of the population of adults who think that today’s children will be better than their parents.

Answer to Problem 37E

The point estimate of the proportion of the population of adults who think that today’s children will be better than their parents is 0.24.

Explanation of Solution

Calculation:

The survey is on the today’s children will be better than their parents or not. The sample of 1,000 adults gave responses on the question. The responses of the survey is yes, no and not sure.

From the data, the response of yes is 240, the response of no is 530 and the not sure is 230.

The sampling distribution of $\overline{p}$ is the population proportion p.

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

$\begin{array}{c}\overline{p}=\frac{\text{Numbe of possible cases}}{\text{Total number of cases}}\\ =\frac{240}{1,000}\\ =0.24\end{array}$

Thus, the point estimate of the proportion is 0.24.

To determine

Find the margin of error at 95% confidence.

Answer to Problem 37E

The margin of error at 95% confidence is 0.0265.

Explanation of Solution

Calculation:

The formula for margin of error is,

$\text{Margin of error}=z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}$

From the results of part (a),

The value of $\overline{p}$ is 0.24.

From the “Table 8.1 value of $z_{\frac{\alpha }{2}}$”, the value of $z_{\frac{\alpha }{2}}$ for 95% is 1.96.

The value of margin of error is,

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.24\left(1-0.24\right)}{1,000}}\\ =1.96\sqrt{\frac{0.1824}{1,000}}\\ =1.96\times 0.0135\\ =0.0265\end{array}$

Thus, the value of margin of error is 0.0265.

To determine

Find the 95% confidence interval for the population proportion of the population of adults who think that today’s children will be better than their parents.

Answer to Problem 37E

The 95% confidence interval for the population proportion of the population of adults who think that today’s children will be better than their parents is (0.2135, 0.2665).

Explanation of Solution

Calculation:

From the results of parts (a) and (b), the point estimate of the proportion $\overline{p}$ is 0.24 and the value Margin of error is 0.0265.

The 95% confidence interval for population proportion is,

$\begin{array}{c}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}=0.24\pm 0.0265\\ =\left(0.24-0.0265,0.24+0.0265\right)\\ =\left(0.2135,0.2665\right)\end{array}$

Thus, the 95% confidence interval for population proportion is (0.2135, 0.2665).

To determine

Find the 95% confidence interval for the population proportion of the population of adults who do not think that today’s children will be better than their parents.

Answer to Problem 37E

The 95% confidence interval for the population proportion of the population of adults who do not think that today’s children will be better than their parents is (0.4991, 0.5609).

Explanation of Solution

Calculation:

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

$\begin{array}{c}\overline{p}=\frac{\text{Numbe of possible cases}}{\text{Total number of cases}}\\ =\frac{530}{1,000}\\ =0.53\end{array}$

Thus, the point estimate of the proportion is 0.53.

From the “Table 8.1 value of $z_{\frac{\alpha }{2}}$”, the value of $z_{\frac{\alpha }{2}}$ for 95% is 1.96.

The value of margin of error is,

$\begin{array}{c}\text{Margin of error}=1.96\sqrt{\frac{0.53\left(1-0.53\right)}{1,000}}\\ =1.96\sqrt{\frac{0.2491}{1,000}}\\ =1.96\times 0.0158\\ =0.0309\end{array}$

Thus, the value of margin of error is 0.0309.

The 95% confidence interval for population proportion is,

$\begin{array}{c}\overline{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\overline{p}\left(1-\overline{p}\right)}{n}}=0.53\pm 0.0309\\ =\left(0.53-0.0309,0.53+0.0309\right)\\ =\left(0.4991,0.5609\right)\end{array}$

Thus, the 95% confidence interval for population proportion is (0.4991, 0.5609).

To determine

Find the smallest margin of error in part (c) and (d).

Explain the reason for smaller margin of error

Answer to Problem 37E

The smaller margin of error contains in part (c) confidence interval and the smaller margin of error because $\underset{\_}{\overline{p}}$ is smaller.

Explanation of Solution

Calculation:

From the part (c), the margin of error is 0.0265.

From the part (d), the margin of error is 0.0309.

The smaller margin of error contains in part (c) confidence interval.

In part (c), the point estimator $\overline{p}$ is smaller compare with part (d).

Thus, the smaller margin of error because $\underset{\_}{\overline{p}}$ is smaller.