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Chapter 7: Q59 E (page 418)

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

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

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

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

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

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

An analyst tested the null hypothesis that 渭 鈮� 20against the alternative hypothesis that 渭 <20. The analyst reported a p-value of .06. What is the smallest value of伪for which the null hypothesis would be rejected?

7.83 A random sample of n observations is selected from a normal population to test the null hypothesis that 蟽2=25 . Specify the rejection region for each of the following combinations of Ha,伪and n.

a.Ha:蟽2鈮�25;伪=0.5;n=16

b. Ha:蟽2>25;伪=.01;n=23

c. Ha:蟽2>25;伪=.10;n=15

d. Ha:蟽2<25;伪=.01;n=13

e. Ha:蟽2鈮�25;伪=.10;n=7

f.Ha:蟽2<25;伪=.05;n=25

Trading skills of institutional investors. Refer to The Journal of Finance (April 2011) analysis of trading skills of institutional investors, Exercise 7.36 (p. 410). Recall that the study focused on 鈥渞ound-trip鈥� trades, i.e., trades in which the same stock was both bought and sold in the same quarter. In a random sample of 200 round-trip trades made by institutional investors, the sample standard deviation of the rates of return was 8.82%. One property of a consistent performance of institutional investors is a small variance in the rates of return of round-trip trades, say, a standard deviation of less than 10%.

a. Specify the null and alternative hypotheses for determining whether the population of institutional investors performs consistently.

b. Find the rejection region for the test using伪=.05

c. Interpret the value of in the words of the problem.

d. A Minitab printout of the analysis is shown (next column). Locate the test statistic andp-value on the printout.

e. Give the appropriate conclusion in the words of the problem.

f. What assumptions about the data are required for the inference to be valid?

In a test of the hypothesis \({H_0}:\mu = 10\) versus \({H_a}:\mu \ne 10\), a sample of n = 50 observations possessed mean \(\bar x = 10.7\) and standard deviation s = 3.1. Find and interpret the p-value for this test.

If you select a very small value for 伪when conducting a hypothesis test, will 尾 tend to be big or small? Explain.
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