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 ______.

Answer 1

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!

Answer 2

The confidence interval of the tropical rain forest is 0.80 to  6.86

The confidence interval is calculated using:

$CI = (\bar x_1 - \bar x_2) \pm t_{df} * \sigma_a * \sqrt{1/n_1 + 1/n_2}$

Where:

$df = n_1 + n_2 - 2$ --- the degrees of freedom.

n1 = 12 and n2 = 9

So, we have:

$df = 12 + 9 - 2$

$df = 19$

Also, we have:

$\bar x_1 = \frac{22+ 18+ 22 +20 +15 +21 +13 +13 +19+ 13+ 19 +15}{12} = 17.5$

$\bar x_2 = \frac{17 +4 +18+ 14 +18 +15 +15+ 10+ 12}{9} = 13.67$

Using a calculator, we have:

$\sigma_1 = 3.53$

$\sigma_2 = 4.50$

So, we have:

$CI = (\bar x_1 - \bar x_2) \pm t_{df} * \sigma_a * \sqrt{1/n_1 + 1/n_2}$

$CI = (17.5 - 13.67) \pm t_{19} * \sigma_a * \sqrt{1/12 + 1/9}$

$CI = 3.83 \pm t_{19} * \sigma_a * 0.441$

Next, we have:

$\sigma_a =\sqrt{\frac{(n_1 - 1)* \sigma_1^2 +(n_2- 1)* \sigma_2^2} {df}}$

So, we have:

$\sigma_a =\sqrt{\frac{(12 - 1)* 3.53^2 +(9- 1)* 4.50^2} {19}}$

$\sigma_a =\sqrt{\frac{299.0699}{19}}$

$\sigma_a =3.97$

So, we have:

$CI = 3.83 \pm t_{19} * 3.97* 0.441$

$CI = 3.83 \pm t_{19} * 1.751$

At 90% confidence interval, and a degree of freedom of 19, the t value is:

$t = 1.729$

So, we have:

$CI = 3.83 \pm 1.729 * 1.751$

$CI = 3.83 \pm 3.03$

Split

$CI = (3.83 - 3.03,3.83 + 3.03)$

$CI = (0.80,6.86)$

Hence, the confidence interval is 0.80 to  6.86

Read more about confidence intervals at:

https://brainly.com/question/17097944