/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q37 E Minimizing tractor skidding dist... [FREE SOLUTION] | 91影视

91影视

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?

Short Answer

Expert verified
  1. The true mean skidding distance for the road with a 95% confidence interval is (303.39,413.50).
  2. 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.
  3. The given samples must be from a normal distribution. As the standard deviation of the population is unknown, there should be used t-distribution to get the confidence interval.
  4. The logger working on the road claims that the mean skidding distance is at least 425 meters is correct and agreed.

Step by step solution

01

Given information

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.

02

Estimate the true mean with a 95% interval

a.

Let鈥檚 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).

03

Determine the practical interpretation

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.

04

Required condition for inference

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.
05

Conclusion of the statement

d.

The confidence interval, which is calculated in part 鈥榓鈥, 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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In each of the following instances, determine whether you would use a z- or t-statistic (or neither) to form a 90%confidence interval and then state the appropriate z- ort-statistic value for the confidence interval.

a. Random sample of size n = 32 from a normal distribution with a population mean of 60 and population standard deviation of 4.

b. Random sample of size n = 108 from an unknown population.

c. Random sample of size n = 12 from a normal distribution witha sample mean of 83 and sample standard deviation of 2.

d. Random sample of size n = 24 from a normal distribution withan unknown mean and sample standard deviation of 3.

Crude oil biodegradation. Refer to the Journal of Petroleum Geology (April 2010) study of the environmental factors associated with biodegradation in crude oil reservoirs, Exercise 2.29 (p. 85). One indicator of biodegradation is the level of dioxide in the water. Recall that 16 water specimens were randomly selected from various locations in a reservoir on the floor of a mine and the amount of dioxide (milligrams/liter) as well as presence of oil was determined for each specimen. These data are reproduced in the next table.

a. Estimate the true mean amount of dioxide present in water specimens that contain oil using a 95% confidence interval. Give a practical interpretation of the interval.

b. Repeat part a for water specimens that do not contain oil.

c. Based on the results, parts a and b, make an inference about biodegradation at the mine reservoir.

Lead and copper in drinking water. Periodically, the Hillsborough County (Florida) Water Department tests the drinking water of homeowners for contaminants such as lead and copper. The lead and copper levels in water specimens collected for a sample of 10 residents of the Crystal Lakes Manors subdivision are shown below, followed by a Minitab printout analysing the data.

a. Locate a 90% confidence interval for the mean lead level in water specimens from Crystal Lakes Manors on the printout.

b. Locate a 90% confidence interval on the printout for the mean copper level in water specimens from Crystal Lakes Manors.

c. Interpret the intervals, parts a and b, in the words of the problem.

d. Discuss the meaning of the phrase 鈥90% confident.鈥

Do social robots walk or roll? Refer to the InternationalConference on Social Robotics(Vol. 6414, 2010) studyof the trend in the design of social robots, Exercise 6.48(p. 357). Recall that you used a 99% confidence interval toestimate the proportion of all social robots designed withlegs, but no wheels. How many social robots would need tobe sampled in order to estimate the proportion to within.075 of its true value?

A random sample of 90 observations produced a mean x = 25.9 and a standard deviation s = 2.7.

a. Find an approximate 95% confidence interval for the population mean

b. Find an approximate 90% confidence interval for

c. Find an approximate 99% confidence interval for

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.