Splinting in mountain climbing accidents. The most common injury that occurs among mountain climbers is trauma to the lower extremity (leg). Consequently, rescuers must be proficient in immobilizing and splinting fractures. In High Altitude Medicine & Biology (Vol. 10, 2009), researchers examined the likelihood of mountain climbers needing certain types of splints. A Scottish Mountain Rescue study reported that there was 1 femoral shaft splint needed among 333 live casualties. The researchers will use this study to estimate the proportion of all mountain casualties that require a femoral shaft splint.

a. Is the sample large enough to apply the large-sample estimation method of this section? Show why or why not.

b. Use Wilson’s adjustment to find a 95% confidence interval for the true proportion of all mountain casualties that require a femoral shaft splint. Interpret the result

Step by step solution

01

Given information

The most common injury that occurs among mountain climbers is trauma to the lower extremity, there was 1 femoral shaft splint needed among 333 live casualties

02

Checking whether this sample is large enough to apply the large sample method

a.

Let p be the proportion of all-mountain casualties femoral shaft splint needed.

$\begin{array}{rcl}\hat{\mathrm{p}}& =& \frac{\mathrm{x}}{\mathrm{n}}\\ & =& \frac{1}{333}\\ & =& 0.003\\ \hat{\mathrm{p}}& =& 0.003\end{array}$

Since p is near 0, an extremely larger sample is required to estimate it by the usual large-sample method.

Here, the number of successes 1, is less than 15.

So, the sample is not large enough to estimate the population proportion.

03

95% confidence interval for the true proportion

b.

Using Wilson’s adjustment,

The adjusted sample proportion is,

$\begin{array}{rcl}\mathrm{p}\%& =& \frac{\mathrm{x}+2}{\mathrm{n}+4}\\ & =& \frac{1+2}{333+4}\\ & =& \frac{3}{337}\\ & =& 0.0089\end{array}$

Using Wilson’s adjustment, the 95% confidence interval for true proportion is,

$\begin{array}{rcl}\mathrm{p}\%\pm {\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\sqrt{\frac{\mathrm{p}\%\left(1-\mathrm{p}\%\right)}{\mathrm{n}+4}}& =& 0.0089\pm {\mathrm{z}}_{0.025}\sqrt{\frac{0.0089\left(1-0.0089\right)}{333+4}}\\ & =& 0.0089\pm 1.960\sqrt{\frac{0.0088}{337}}\left[\mathrm{Using}\mathrm{Standard}\mathrm{Normal}\mathrm{Table}\right]\\ & =& 0.0089\pm \left(1.960\times 0.0051\right)\\ & =& 0.0089\pm 0.0099\\ & =& \left(-0.001,0.0188\right)\end{array}$

Here, p cannot be negative.

So, the 95% confidence interval for the true proportion is $\left(0,0.019\right)$.

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