Question

Study of aircraft bird-strikes. As worldwide air traffic volume has grown over the years, the problem of airplanes striking birds and other flying wildlife has increased dramatically. The International Journal for Traffic and Transport Engineering (Vol. 3, 2013) reported on a study of aircraft bird strikes at Aminu Kano International Airport in Nigeria. During the survey period, a sample of 44 aircraft bird strikes were analyzed. The researchers found that 36 of the 44 bird strikes at the airport occurred above 100 feet. Suppose an airport air traffic controller estimates that less than 70% of aircraft bird strikes occur above 100 feet. Comment on the accuracy of this estimate. Use a 95% confidence interval to support your inference.

Short answer

At a 5% level of significance, it can be concluded that the estimate is not good.

Step-by-step solution

Given information

The International Journal for Traffic and Transport Engineering reported on a study of aircraft bird strikes at Aminu Kano International Airport in Nigeria, a sample of 44 aircraft bird strikes was analyzed.

The researchers found that 36 of the 44 bird strikes at the airport occurred above 100 feet.

Finding the 95% confidence interval

The sample proportion is the point estimator of the population proportion p.

The sample proportion is,

$\begin{array}{rcl}\hat{\mathrm{p}}& =& \frac{\mathrm{X}}{\mathrm{n}}\\ & =& \frac{36}{44}\\ & =& 0.8182\end{array}$

Then the level of$100\left(1-\mathrm{\alpha }\right)\%$ confidence interval for p (proportion) is,

$\hat{\mathrm{p}}\pm {\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\sqrt{\frac{\hat{\mathrm{p}}\left(1-\hat{\mathrm{p}}\right)}{\mathrm{n}}}$

For a 95% confidence interval, the value of$\frac{\mathrm{\alpha }}{2}$ is,

$\begin{array}{rcl}100\left(1-\mathrm{\alpha }\right)\%& =& 95\%\\ \left(1-\mathrm{\alpha }\right)& =& 0.95\end{array}$

For,$\mathrm{\alpha }=0.05\mathrm{and}\frac{\mathrm{\alpha }}{2}=0.025$

The 95% confidence interval is,

$\begin{array}{rcl}\hat{\mathrm{p}}\pm {\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\sqrt{\frac{\hat{\mathrm{p}}\left(1-\hat{\mathrm{p}}\right)}{\mathrm{n}}}& =& 0.8182\pm 1.960\sqrt{\frac{0.8182\left(1-0.8182\right)}{44}}\left[\mathrm{From}\mathrm{Standard}\mathrm{Normal}\mathrm{Table}\right]\\ & =& \left(0.8182\pm 1.960\sqrt{0.0034}\right)\\ & =& \left(0.8182\pm 0.1143\right)\\ & =& \left(0.7039,0.9325\right)\end{array}$

Here the 95% confidence interval for the aircraft bird strikes occurs is (0.704, 0.933),an airport air traffic controller estimates that less than 70% of aircraft bird strikes occur above 100 feet. So here the estimate is outside of the 95% confidence interval.

So, at a 5% level of significance, it can be concluded that the estimate is not good.