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])±
*(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

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])±
*(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!