Conservationists have despaired over destruction of tropical rain forest by logging, clearing, and burning.” These words begin a report on a

Question

Conservationists have despaired over destruction of tropical rain forest by logging, clearing, and burning.” These words begin a report on a statistical study of the effects of logging in Borneo. Here are data on the number of tree species in 12 unlogged forest plots and 9 similar plots logged 8 years earlier: Unlogged 22 18 22 20 15 21 13 13 19 13 19 15 Logged 17 4 18 14 18 15 15 10 12 Use the data to give a 90% confidence interval for the difference in mean number of species between unlogged and logged plots. Compute degrees of freedom using the conservative method.Interval: _____ to ______.

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Parker 6 days 2021-10-14T02:30:30+00:00 1 Answer 0

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    2021-10-14T02:32:21+00:00

    Answer:

    [0.841;6.879]

    Step-by-step explanation:

    Hello!

    You have two independent samples:

    Sample 1

    X₁: tree species in an unlogged plot.

    n₁= 12 plots

    Unlogged 22 18 22 20 15 21 13 13 19 13 19 15

    Sample 2

    X₂: tree species in a plot logged 8 years earlier

    n₂= 9

    Logged 17 4 18 14 18 15 15 10 12

    The objective of this experiment is to estimate the difference between the means of the number of species on the unlogged plots and the logged plots. Symbolically: μ₁-μ₂

    To be able to estimate the difference between the two population means, I’ll assume that both variables have a normal distribution and use a pooled t-statistic for the Confidence interval. (I did a quick F-test for variance homogeneity, population variances are unknown but equal)

    The formula is:

    (X₁[bar]-X₂[bar])±t_{(n1+n2-2);1-\alpha /2}*(Sₐ√(1/n₁+1/n₂))

    Where

    sample mean sample 1 X₁[bar]= 17.50

    sample mean sample 2 X₂[bar]= 13.64

    S₁= 3.53

    S₂= 4.50

    t_{n1 + n2 - 2;1-\alpha/2} = t_{19;0.95} = 1.729

    Sₐ²= (n₁ – 1)S₁² + (n₂ – 1)S₂² = 11*(3.53)² + 8*(4.50)² = 15.74

                       n₁+n₂-2                            19

    Sₐ= 3.96

    Now replace the values in the formula:

    (X₁[bar]-X₂[bar])±t_{(n1+n2-2);1-\alpha /2}*(Sₐ√1/n₁+1/n₂)

    [(17.50-13.64) ± 1.729*(3.96*√(1/12+1/9))]

    [3.86 ± 3.019]

    [0.841;6.879]

    With a confidence level of 90%, you’d expect that the interval [0.841;6.879] contains the difference between the means of the number of species on the unlogged plots and the logged plots.

    I hope it helps!

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