of various trees within ob, you are required to check the health Suppose you are
ID: 3047695 • Letter: O
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of various trees within ob, you are required to check the health Suppose you are a forest ranger, and as part of your j are the diameter at breast height (dbh) of five Douglas Fir trees in part of a forest. Treat this data as the population of Douglas Fir trees. On all questions, round to two decimal points (see my "for example" below). When Iculating means and standard deviations use N in the denominator Tree: Diameter (dbh): #1 12.4 cm. #2 14.5 cm. #3 15.2 cm. #4 17.8 cm. #5 19.1 cm. 1, what is the average diameter of a Douglas Fir tree in the forest? what is the standard deviation of the Douglas Fir trees' diameters? = The next problem requires you to create a sampling distribution for this population, for a given sample size, and ther characterize the sampling distribution, just as we did with the turtles example in lecture 8. 2. Suppose you have a deadline to meet, and Douglas Fir trees are difficult to find within the forest, so you only have time to locate three of the five trees. Identify all the feasible samples of size n = 3 trees and calculate their sample means Sample: Sample mean: 14.03 cm. For example: (1,2,3) cm cm. cm. cm. cm. cm cm cm cm 3. What is the mean of the sampling distribution of sample means for n 3 What is the standard deviation of the sampling distribution of sample means for n-3?Explanation / Answer
1) The mean = sum (dimeter) / n
where n = 5
Standard deviation =sqrt{ [ sum (dimeter^2) / n ] - (mean)^2 }
The mean dimeter of Douglas Fir Tree = 15.8
The Standared deviation dimeter of Douglas Fir Tree = 2.66927
2) After taking the three sample tree amoung the five tree the following results becomes :
The mean of sample mean is = 14.5
The standared deviation of sample mean is = 1.126002
( use n = 10 in above formula )
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Sample Sample mean 1,2,3 14.03 1,4,5 12.2 2,3,4 14.5 3,4,5 15.2 1,2,5 12.4 2,3,5 14.5 1,3,5 12.4 1,4,5 12.4 1,2,4 12.4 2,4,5 14.5Related Questions
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