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velikii [3]
1 year ago
8

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 ______.
Mathematics
1 answer:
Anna [14]1 year ago
4 0

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= <u>(n₁ - 1)S₁² + (n₂ - 1)S₂² </u>= <u>11*(3.53)² + 8*(4.50)² </u>= 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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