Chapter 21, Problem 16E

(a)

To determine

To find: The 95% confidence interval for the proportion of adults who feel that basic medical care in the United States was affordable when the sample size n is 750.

Answer to Problem 16E

Solution: The confidence interval is $\left(0.484,0.556\right)$.

Explanation of Solution

Calculation:

For a two-tailed test at 5% significance level, the tabulated value of $z$ statistic is 1.96 that is provided in Table 21.1 of the textbook.

Compute the confidence interval using the formula as follows:

$\text{Confidence interval}=\left[\widehat{p}\pm z\left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]$

where

$\begin{array}{l}\widehat{p}\text{ is the sample proportion}\text{.}\\ z\text{ is the tabulated value for desired confidence interval}\text{.}\\ n\text{ is the sample size}\text{.}\end{array}$

Now, compute the 95% confidence interval as follows:

$\begin{array}{c}\text{Confidence interval}=\left[\widehat{p}\pm z\left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]\\ =\left[0.52\pm 1.96\left(\sqrt{\frac{0.52\left(1-0.52\right)}{750}}\right)\right]\\ =\left(0.52\pm 0.036\right)\\ =\left(0.484,0.556\right)\end{array}$

Interpretation: It can be concluded with 95% confidence that proportion of adults who feel that basic medical care in the United States was affordable lies between approximately 48.4% and 55.6%.

(b)

To determine

To find: The 95% confidence interval for the proportion of adults who feel that basic medical care in the United States was affordable when the sample size n is 1500.

Answer to Problem 16E

Solution: The confidence interval is $\left(0.495,0.545\right)$.

Explanation of Solution

Calculation:

For a two-tailed test at 5% significance level, the tabulated value of $z$ statistic is 1.96 that is provided in Table 21.1 of the textbook.

Compute the confidence interval using the formula as follows:

$\text{Confidence interval}=\left[\widehat{p}\pm z\left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]$

where

$\begin{array}{l}\widehat{p}\text{ is the sample proportion}\text{.}\\ z\text{ is the tabulated value for desired confidence interval}\text{.}\\ n\text{ is the sample size}\text{.}\end{array}$

Now, compute the 95% confidence interval as follows:

$\begin{array}{c}\text{Confidence interval}=\left[\widehat{p}\pm z\left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]\\ =\left[0.52\pm 1.96\left(\sqrt{\frac{0.52\left(1-0.52\right)}{1500}}\right)\right]\\ =\left(0.52\pm 0.025\right)\\ =\left(0.495,0.545\right)\end{array}$

Interpretation: It can be concluded with 95% confidence that proportion of adults who feel that basic medical care in the United States was affordable lies between approximately 49.5% and 54.5%.

(c)

To determine

To find: The 95% confidence interval for the proportion of adults who feel that basic medical care in the United States was affordable when the sample size n is 3000.

Answer to Problem 16E

Solution: The confidence interval is $\left(0.502,0.538\right)$.

Explanation of Solution

Calculation:

For a two-tailed test at 5% significance level, the tabulated value of $z$ statistic is 1.96 that is provided in Table 21.1 of the textbook.

Compute the confidence interval using the formula as follows:

$\text{Confidence interval}=\left[\widehat{p}\pm z\left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]$

where

$\begin{array}{l}\widehat{p}\text{ is the sample proportion}\text{.}\\ z\text{ is the tabulated value for desired confidence interval}\text{.}\\ n\text{ is the sample size}\text{.}\end{array}$

Now, compute the 95% confidence interval as follows:

$\begin{array}{c}\text{Confidence interval}=\left[\widehat{p}\pm z\left(\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\right)\right]\\ =\left[0.52\pm 1.96\left(\sqrt{\frac{0.52\left(1-0.52\right)}{3000}}\right)\right]\\ =\left(0.52\pm 0.018\right)\\ =\left(0.502,0.538\right)\end{array}$

Interpretation: It can be concluded with 95% confidence that proportion of adults who feel that basic medical care in the United States was affordable lies between approximately 50.2% and 53.8%.

(d)

To determine

To explain: The difference in results obtained in part (a), part (b), and part (c).

Answer to Problem 16E

Solution: The width of confidence interval reduces as the sample size increases.

Explanation of Solution

The confidence interval obtained in part (a) with sample size 750 is approximately 48.4% to 55.6%.

The confidence interval obtained in part (b) with sample size 1500 is approximately 49.5% to 54.5%.

The confidence interval obtained in part (c) with sample size 3000 is approximately 50.2% to 53.84%.

So, from the confidence intervals obtained in parts (a), (b), and (c), it can be concluded that as the size of sample increases, the width of confidence interval reduces.

From the results of part (a), part (b), and part (c), it can be concluded that with the increase in sample size, the width of confidence interval reduces.