Question

Life Support. In $2005$, the Terri Schiavo case focused national attention on the issue of withdrawal of life support from terminally ill patients or those in a vegetative state. A Harris Poll of $1010$ U.S. adults was conducted by telephone on April $5-10$, $2005$ Of those surveyed, $140$ had experienced the death of at least one family member or close friend within the last $10$ years who died after the removal of life support. Find and interpret a $90\%$ confidence interval for the proportion of all U.S. adults who had experienced the death of at least one family member or close friend within the last $10$ years after life support had been withdrawn.

Short answer

There is a $90\%$ chance that between $12.07\%$ and $15.65\%$ of U.S. adults have lost a family member or close friend in the recent ten years.

Step-by-step solution

Given Information

According to a Harris study, $140$ out of $1010$ persons in the United States have lost a family member or close friend after life support was stopped in the previous $10$ years.

Explanation

The sample size for this study is $n=1010$.

The success rate is $x=140$.

The fraction of the sample would be $\hat{p}=\frac{x}{n}$

$=\frac{140}{1010}$

$=0.1386$

With a $90\%$confidence interval,

$1-\alpha$,

To find $z_{\alpha /2}$using table II.

The level of confidence is at $90\%$. i.e. $\alpha =0.10$.

Table II

$z_{\alpha /2}=z_{0.10/2}$

$=z_{0.05}$

$=1.645$

Explanation

The range of confidence for $p$is :-

$\hat{p}-z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\text{to}\hat{p}+z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$\Rightarrow 0.1386-1.645\sqrt{\frac{0.1386(1-0.1386)}{1010}}\text{to}0.1386+1.645\sqrt{\frac{0.1386(1-0.1386)}{1010}}$$\Rightarrow 0.1386-0.0179\text{to}0.1386+0.0179$

$\Rightarrow 0.1207\text{to}0.1565$

$\Rightarrow 12.07\%\text{to}15.65\%$