We looked at the distribution of longleaf pine trees in the Wade Tract. One way

Question

We looked at the distribution of longleaf pine trees in the Wade Tract. One way to formulate hypotheses about whether or not the trees are randomly distributed in the tract is to examine the average location in the north– south direction. The values range from 0 to 200, so if the trees are uniformly distributed in this direction, any difference from the middle value (100) should be due to chance variation. The sample mean for the 601601 trees in the tract is 99.0299.02 . A theoretical calculation based on the assumption that the trees are uniformly distributed gives a standard deviation of 5858 . Carefully state the null and alternative hypotheses in terms of this variable. Note that this requires that you translate the research question about the random distribution of the trees into specific statements about the mean of a probability distribution. Test your hypotheses.

H0:=100H0:=100 versus Ha:100Ha:100

H0:=100H0:=100 versus Ha:>100Ha:>100

H0:=100H0:=100 versus Ha:<100Ha:<100

zz (±± 0.001) =

PP (±± 0.001) =

Conclusion:

Reject the null hypotheses

Not reject the null hypotheses

5.

We looked at the distribution of longleaf pine trees in the Wade Tract. One way to formulate hypotheses about whether or not the trees are randomly distributed in the tract is to examine the average location in the north– south direction. The values range from 0 to 200, so if the trees are uniformly distributed in this direction, any difference from the middle value (100) should be due to chance variation. The sample mean for the 601601 trees in the tract is 99.0299.02 . A theoretical calculation based on the assumption that the trees are uniformly distributed gives a standard deviation of 5858 . Carefully state the null and alternative hypotheses in terms of this variable. Note that this requires that you translate the research question about the random distribution of the trees into specific statements about the mean of a probability distribution. Test your hypotheses.

H0:=100H0:=100 versus Ha:100Ha:100

H0:=100H0:=100 versus Ha:>100Ha:>100

H0:=100H0:=100 versus Ha:<100Ha:<100

zz (±± 0.001) =

PP (±± 0.001) =

Conclusion:

Reject the null hypotheses

Not reject the null hypotheses

Equation Editor

Explanation / Answer

The statistical software output for this problem is:

One sample Z summary hypothesis test:
: Mean of population
H0 : = 100
HA : 100
Standard deviation = 58

Hypothesis test results:

Hence,

Hypotheses: Option A

z = -0.414

P value = 0.679

Not reject the null hypothesis.

Mean n Sample Mean Std. Err. Z-Stat P-value 601 99.02 2.3658693 -0.41422406 0.6787

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