Question

Minimizing tractor skidding distance. Refer to the Journal of Forest Engineering (July 1999) study of minimizing tractor skidding distances along a new road in a European forest, Exercise 6.37 (p. 350). The skidding distances (in meters) were measured at 20 randomly selected road sites. The data are repeated below. Recall that a logger working on the road claims the mean skidding distance is at least 425 meters. Is there sufficient evidence to refute this claim? Use \(\alpha = .10\)

Short answer

There is sufficient evidence to refute this claim

Step-by-step solution

Given Information

The sample size is 20.

The hypothesis are given by

\(\begin{aligned}{l}{H_0}:\mu = 425\\{H_a}:\mu > 425\end{aligned}\)

Compute and standard deviation

The mean is calculated as

\(\begin{aligned}\bar x &= \frac{{\sum\limits_{i = 1}^n {{X_i}} }}{n}\\ &= \frac{{7169}}{{20}}\\ &= 358.45\end{aligned}\)

The standard deviation is calculated as

\(\begin{aligned}sd &= \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({X_i} - \bar X)}^2}} }}{{n - 1}}} \\ &= \sqrt {\frac{\begin{aligned}{l}16783.2 + 71.4025 + 9712.103 + 25424.3 + 5394.903 + 2555.303 + \\5859.903 + 46461.8 + 6488.303 + 35175 + 704.9025 + 4025.903 + \\30432.8 + 9496.503 + 7301.703 + 1726.403 + 2251.503 + 2157.603 + \\47284.5 + 4428.903\end{aligned}}{{19}}} \\ &= \sqrt {\frac{{266292.3}}{{19}}} \\ &= \sqrt {14015.38} \\ &= 118.38\end{aligned}\)

Therefore, the mean and standard deviation are 358.45 and 118.38.

Test statistic

The test statistic is computed as

\(\begin{aligned}t &= \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\\ &= \frac{{358.45 - 425}}{{\frac{{118.38}}{{\sqrt {20} }}}}\\ &= \frac{{ - 66.55}}{{26.47}}\\ &= - 2.514\end{aligned}\)

Therefore, the test statistic is -2.514.

Conclusion

For,\(\alpha = .10\,and\,n - 1 = 19\)

\(\begin{aligned}{t_{\alpha ,n - 1}} &= {t_{0.10,19}}\\ &= 1.327\end{aligned}\)

The calculated value is less than tabulated value.

Therefore, we reject the null hypothesis.