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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

Expert verified

There is sufficient evidence to refute this claim.

Step by step solution

01

Given Information

The sample size is 20.

The hypothesis are given by

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

02

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

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

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

03

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.

04

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.

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Most popular questions from this chapter

鈥淪treaming鈥 of television programs is trending upward. According to The Harris Poll (August 26, 2013), over one-third of American鈥檚 qualify as 鈥渟ubscription streamers,鈥 i.e., those who watch streamed TV programs through a subscription service such as Netflix, Hulu Plus, or Amazon Prime. The poll included 2,242 adult TV viewers, of which 785 are subscription streamers. On the basis of this result, can you conclude that the true fraction of adult TV viewers who are subscription streamers differs from one-third? Carry out the test using a Type I error rate of =.10. Be sure to give the null and alternative hypotheses tested, test statistic value, rejection region or p-value, and conclusion.

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