Q.11.50 Life Support. In 2005, the Terri... [FREE SOLUTION]

Chapter 11: Q.11.50 (page 460)

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the death of at least one family member or close friend within the last $10$ years after life support had been withdrawn.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

Expert verified

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,

$1 - 伪$ ,

To find $z_{伪 / 2}$ using table II.

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

Table II

$z_{伪 / 2} = z_{0.10 / 2}$

$= z_{0.05}$

$= 1.645$

Explanation

The range of confidence for $p$ is :-

$\hat{p} - z_{伪 / 2} 路 \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}} to \hat{p} + z_{伪 / 2} 路 \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}$ $鈬� 0.1386 - 1.645 \sqrt{\frac{0.1386 (1 - 0.1386)}{1010}} to 0.1386 + 1.645 \sqrt{\frac{0.1386 (1 - 0.1386)}{1010}}$ $鈬� 0.1386 - 0.0179 to 0.1386 + 0.0179$

$鈬� 0.1207 to 0.1565$

$鈬� 12.07 % to 15.65 %$

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