Minimizing tractor skidding distance. When planning for a new forest road to be used for tree harvesting, planners must select the location to minimize tractor skidding distance. In the Journal of Forest Engineering(July 1999), researchers wanted to estimate the true mean skidding distance along a new road in a European forest. The skidding distances (in meters) were measured at 20 randomly selected road sites. These values are given in the

accompanying table.

1. Estimate the true mean skidding distance for the road with a 95% confidence interval.
2. Give a practical interpretation of the interval, part a.
3. What conditions are required for the inference, part b, to be valid? Are these conditions reasonably satisfied?
4. A logger working on the road claims the mean skidding distance is at least 425 meters. Do you agree?

There is a journal, in which some researchers wanted to estimate the true mean of skidding distance along a new road in a European forest. There measured 20 randomly selected roads distance.

a.

Let’s consider the mean of the sample is\(\bar X\).

So, \(\bar X = \frac{1}{n}\sum\limits_{i = 1}^{20} {{X_i} = \frac{{7169}}{{20}} = 358.45} \)

The sample standard deviation is

\(\begin{aligned}\sigma &= \sqrt {\frac{{\sum\limits_{i = 1}^{20} {{{\left( {{X_i} - \bar X} \right)}^2}} }}{{n - 1}}} \\ &= \sqrt {\frac{{{{\left( {488 - 358.45} \right)}^2} + \cdots + {{\left( {425 - 358.45} \right)}^2}}}{{19}}} \\ &= \sqrt {\frac{{263737}}{{19}}} \\ &= \sqrt {13880.89} \\ &= 117.81\end{aligned}\)

Now the confidence interval is 95%.

Therefore, the level of significance \(\alpha = 1 - 0.95 = 0.05\).

So, from the t-table, the statistic\({t_{\alpha ,\left( {n - 1} \right)}} = {t_{0.05,19}} = 2.09\).

Therefore, the confidence interval is\(\left( {\bar x \pm {t_{\alpha ,\left( {n - 1} \right)}} \times \frac{\sigma }{{\sqrt n }}} \right)\).

Thus,

\(\begin{aligned}\left( {\bar x \pm {t_{\alpha ,\left( {n - 1} \right)}} \times \frac{\sigma }{{\sqrt n }}} \right) &= \left( {358.45 \pm {t_{0.05,19}} \times \frac{{117.81}}{{\sqrt {20} }}} \right)\\ & \left( {\left( {358.45 - 2.09 \times 26.34} \right),\left( {358.45 + 2.09 \times 26.34} \right)} \right)\\ &= \left( {303.39,413.50} \right)\end{aligned}\)

Therefore, the confidence interval is (303.39,413.50).

b.

The population means \(\bar x = 358.45\). This is between the 95% confidence interval. So, there can be concluded that the population means will be within this interval.

c.

For this case study, there should require two assumptions or conditions to validate the intervals. That is,

1. The given samples must be from a normal distribution.
2. As the standard deviation of the population is unknown, there should be used t-distribution to get the confidence interval.

d.

The confidence interval, which is calculated in part ‘a’, does not contain the given mean skidding distance of 425 meters, So, the null hypothesis will be rejected.

Therefore, there can be concluded that the logger working on the road claims that the mean skidding distance is at least 425 meters is correct and agreed.