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Business / Statistics for Business and Economics 12 / Chapter 6 / Problem 35E

Statistics for Business and Economics | 12th Edition | ISBN: 9780321826237 | Authors: James T. McClave, P. George Benson, Terry T Sincich

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QUESTION:

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.

a. Estimate the true mean skidding distance for the road with a 95% confidence interval.

b. Give a practical interpretation of the interval, part a.

c. What conditions are required for the inference, part b, to be valid? Are these conditions reasonably satisfied?

d. A logger working on the road claims the mean skidding distance is at least 425 meters. Do you agree?

Source: Based on Tujek, J., & Pacola, E. “Algorithms for skidding distance modeling on a raster digital terrain model” Journal of Forest Engineering, Vol. 10, No. 1, July 1999 (Table 1).

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QUESTION:

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.

a. Estimate the true mean skidding distance for the road with a 95% confidence interval.

b. Give a practical interpretation of the interval, part a.

c. What conditions are required for the inference, part b, to be valid? Are these conditions reasonably satisfied?

d. A logger working on the road claims the mean skidding distance is at least 425 meters. Do you agree?

Source: Based on Tujek, J., & Pacola, E. “Algorithms for skidding distance modeling on a raster digital terrain model” Journal of Forest Engineering, Vol. 10, No. 1, July 1999 (Table 1).

ANSWER:

Solution :

Step 1 of 4:

Given the skidding distances were measured at 20 randomly selected road sites.

Then the table is given below.

488	350	457	199	285	409	435	574	439	546
385	295	184	261	273	400	311	312	141	425

a). Now we have to estimate the true mean skidding distance for the road with a 95% confidence interval.

We know that the variable N=20.

Now we are using excel to find the distribution.

We use the function =Average() to find the mean.

Then =Average() select all the data, then ok.

We get the mean value.

Then standard deviation =std() select all the data values

Then we get standard deviation.etc.

Using excel we get,

Variable	Skid
N	20
Mean	358.5
Standard deviation	117.8
Minimum	141
Quartile 1	276
Median	367.5
Quartile 3	438
Maximum	574

We know that level of significance $\alpha{}=0.05.$

Then,

$\frac{\alpha{}}{2}=\frac{0.05}{2}$

$\frac{\alpha{}}{2}=0.025$

Now the degrees of freedom is n-1.

n-1 = 20-1

n-1 = 19

Then,

${t}_{\alpha{}/2}={t}_{0.025}$

${t}_{\alpha{}/2}=2.093$

So the confidence interval is

$\overline{x}\pm{}{t}_{\alpha{}/2}$ $\frac{s}{\sqrt{n\ \ }}$

We know that $\overline{x}$ , ${t}_{\alpha{}/2}$ , n, and s.

$\overline{x}\pm{}{t}_{\alpha{}/2}$ $\frac{s}{\sqrt{n\ \ }}=$ $358.5\pm{}2.093$ $\frac{117.8}{\sqrt{20\ \ }}$

$\overline{x}\pm{}{t}_{\alpha{}/2}$ $\frac{s}{\sqrt{n\ \ }}=$ $358.5\pm{}2.093$ $\frac{117.8}{4.4721}$

$\ =$ $(358.5\pm{}55.13)$

$\ =$ $(303.37,\ 413.63)$

Therefore, the 95% confidence interval is between 303.37 and 413.63 meters.

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