TRACTOR SKIDDING CONFIDENCE INTERVAL Forest engineers are interested in studying the skidding distances of tractors along...

TRACTOR SKIDDING CONFIDENCE INTERVAL Forest engineers are interested in studying the skidding distances of tractors along a new road in a European Forest. The engineers collect data on a random sample of 12 tractors. The collected data (in meters) are: 350, 285, 574, 439, 295, 184, 261, 273, 400, 311, 141, 425. The standard deviation for this random sample is 118.46. (Source: A First Course in Statistics, McClave & Sincich, 10th Edition) Which statement below is the correct interpretation of the confidence interval? Enter 1,2,3, or 4 in the box. I am 95% confident that the average skidding distance, xbar, is significantly greater than suspected. I am 95% confident that the average skidding distance, µ, is significantly greater than suspected. I am 95% confident that the average skidding distance, µ, is between 252.9 ft. and 403.4 ft. I am 95% confident that the average skidding distance, xbar, is between 252.9 ft. and 403.4 ft.

Solutions

Expert Solution

We have for given data,

Sample mean =328.17
Sample standard deviation =118.46
Sample size =12
Level of significance=   1-0.95=   0.05
Degree of freedom =11

t critical value is (by using t table)=   2.201

Confidence interval formula is

=(252.90,403.4)



Lower confidence limit=252.9

Upper confidence limit=403.4

Option 3) is correct.

I am 95% confident that the average skidding distance, µ, is between 252.9 ft. and 403.4 ft.

orchestra answered 1 year ago

Next > < Previous

Related Solutions

TRACTOR SKIDDING HYPOTHESIS TESTING Forest engineers are interested in studying the skidding distances of tractors along...

TRACTOR SKIDDING HYPOTHESIS TESTING Forest engineers are interested in studying the skidding distances of tractors along a new road in a European Forest. The engineers collect data on a random sample of 12 tractors. The collected data (in meters) are: 350, 285, 574, 439, 295, 184, 261, 273, 400, 311, 141, 425. The average skidding distance for the sample of 12 tractors is 328.17 and the standard deviation is 118.46. Local loggers working on the road believe that the mean...

A recent study investigated tractor skidding distances along a road in a forest. The skidding distances...

A recent study investigated tractor skidding distances along a road in a forest. The skidding distances (in meters) were measured at 20 randomly selected road sites. The data are given in the accompanying table. A logger working on the road claims that the mean skidding distance is at least 425 meters. Is there sufficient evidence to refute this claim? Use α=0.10. 488 347 460 205 278 420 424 590 447 534 3813 288 182 259 270 392 316 313 143...

A newsgroup is interested in constructing a 90% confidence interval for the difference in the proportions...

A newsgroup is interested in constructing a 90% confidence interval for the difference in the proportions of Texans and New Yorkers who favor a new Green initiative. Of the 506 randomly selected Texans surveyed, 436 were in favor of the initiative and of the 525 randomly selected New Yorkers surveyed, 459 were in favor of the initiative. a. With 90% confidence the difference in the proportions of Texans and New Yorkers who favor a new Green initiative is between _______...

A newsgroup is interested in constructing a 90% confidence interval for the difference in the proportions...

A newsgroup is interested in constructing a 90% confidence interval for the difference in the proportions of Texans and New Yorkers who favor a new Green initiative. Of the 582 randomly selected Texans surveyed, 428 were in favor of the initiative and of the 569 randomly selected New Yorkers surveyed, 456 were in favor of the initiative. . With 90% confidence the difference in the proportions of Texans and New Yorkers who favor a new Green initiative is between (round...

Hypotheses for a statistical test are given along with a confidence interval for a sample. Use...

Hypotheses for a statistical test are given along with a confidence interval for a sample. Use the confidence interval to state a formal conclusion of the test for that sample and give the significance level used to make the conclusion. Ho: p = 0.5 vs Ha: p ≠ 0.5 95% confidence interval for p: 0.36 to 0.55. 90% confidence interval for p: 0.32 to 0.48. 99% confidence interval for p: 0.18 to 0.65.

CONFIDENCE INTERVAL IN STATS MODE. At a particular location along a highway, the speed limit is...

CONFIDENCE INTERVAL IN STATS MODE. At a particular location along a highway, the speed limit is 65 mph. The speeds of vehicles are Normally distributed with sigma equals 6 m p h, but is the mean speed 65 mph? A state trooper measures the speeds of a random sample of 16 passing vehicles. The average speed for this sample is x with bar on top= 67.8 mph. Compute a 90% confidence interval for the mean speed at this location. lower...

A smart phone manufacturer is interested in constructing a 95% confidence interval for the proportion of...

A smart phone manufacturer is interested in constructing a 95% confidence interval for the proportion of smart phones that break before the warranty expires. 80 of the 1669 randomly selected smart phones broke before the warranty expired. Round answers to 4 decimal places where possible. a. With 95% confidence the proportion of all smart phones that break before the warranty expires is between and . b. If many groups of 1669 randomly selected smart phones are selected, then a different...

You are interested in finding a 95% confidence interval for the mean number of visits for...

You are interested in finding a 95% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 10 randomly selected physical therapy patients. 25 28 22 19 15 20 7 27 23 6 a. To compute the confidence interval use a ? z t distribution. b. With 95% confidence the population mean number of visits per physical therapy patient is between and visits. c. If many groups of 10 randomly selected...

A fitness center is interested in finding a 98% confidence interval for the mean number of...

A fitness center is interested in finding a 98% confidence interval for the mean number of days per week that Americans who are members of a fitness club go to their fitness center. Records of 213 members were looked at and their mean number of visits per week was 3.5 and the standard deviation was 1.8. Round answers to 3 decimal places where possible. a. To compute the confidence interval use a distribution. b. With 98% confidence the population mean...

You are interested in finding a 90% confidence interval for the mean number of visits for...

You are interested in finding a 90% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 15 randomly selected physical therapy patients. Round answers to 3 decimal places where possible. 13 26 16 13 6 27 26 6 13 25 6 11 20 14 16 a. To compute the confidence interval use a ? z t distribution. b. With 90% confidence the population mean number of visits per physical...

Subjects