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:
Col1 Unlogged 2218 222015121 13 13 1913 19 15

Col2 Logged 17 4 181418151510 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 methoo. Interval: ________to__________.

Answer 1

Answer: -8.16 to 15.84

Step-by-step explanation: Confidence Interval is an interval in which we are a percentage sure the true mean is in the interval.

A confidence interval for a difference between two means and since sample 1 and sample 2 are under 30, will be

$x_{1}-x_{2}$ ± $t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }$

where

x₁ and x₂ are sample means

t is t-score

$S_{p}$ is estimate of standard deviation

n₁ and n₂ are the sample numbers

The estimate of standard deviation is calculated as

$S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} }$

where

s₁ and s₂ are sample standard deviation of each sample

Degrees of freedom is:

$df=n_{1}+n_{2}-2$

df = 12 + 9 - 2

df = 19

Checking t-table, with 90% Confidence Interval and df = 19, t = 1.729.

The mean and standard deviation for 12 unlogged forest plots are 17.5 and 3.53, respectively.

The mean and standard deviation for 9 logged plots are 13.66 and 4.5, respectively.

Calculating estimate of standard deviaton:

$S_{p}=\sqrt{\frac{(12-1)(3.53)^{2}+(9-1)(4.5)^{2}}{12+9-2} }$

$S_{p}=\sqrt{\frac{299.07}{19} }$

$S_{p}=$ 15.74

The difference between means is

$x_{1}-x_{2}$ = 17.5 - 13.66 = 3.84

Calculating the interval:

$t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }$ = $1.729.15.74.\sqrt{\frac{1}{12} +\frac{1}{9} }$

$t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }$ = $27.21\sqrt{\frac{21}{108} }$

$t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }$ = $27.21\sqrt{0.194}$

$t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }$ = 12

Then, interval for the difference in mean is 3.84 ± 12, which means the interval is between:

lower limit: 3.84 - 12 = -8.16

upper limit: 3.84 + 12 = 15.84

The interval is from -8.16 to 15.84.