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.

Answer 1

Answer:

Step-by-step explanation:

Hello!

You need to estimate by 90% CI the difference between the average number of tree species of unlogged forest plots and the average number of tree species of plots that were logged 8 years ago in Borneo.

You have two samples:

Sample 1: unlogged plots

n₁= 12 plots

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

X[bar]₁= 17.50

S₁= 3.53

S₁²= 12.4609

Sample 2: logged plots

n₂= 9 plots

17 4 18 14 18 15 15 10 12

X[bar]₂= 13.67

S₂= 4.50

S₂²= 16.2409

Considering that both populations are normal and their population variances are unknown and different, the statistic to use to estimate the difference between the average species in both populations is a Welch's t-test.

To calculate the degrees of freedom for the Welch t-statistic you have to use the following formula:

$Df_w= \frac{(\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} )^2}{\frac{(\frac{S_1^2}{n_1} )^2}{n_1-1} +\frac{(\frac{S_2^2}{n_2} )^2}{n_2-1} }$

$Df_w= \frac{(\frac{12.4609}{12} +\frac{16.2409}{9})^2 }{\frac{(\frac{12.4609}{12} )^2}{11} +\frac{(\frac{16.2409}{9} )^2}{8} } = 16.002$

$t_{16;0.95}= 1.746$

(X[bar]₁-X[bar]₂)±$t_{Df_w;1-\alpha /2}$*$\sqrt{\frac{S^2_1}{n_1} +\frac{S^2_2}{n_2} }$

(17.50-13.67)±1.746*$\sqrt{\frac{12.4609}{12} +\frac{16.2409}{9} }$

[0.89;6.77]

Using a confidence level of 90% you'd expect that the interval [0.89;6.77] will contain the difference between the average number of tree species in unlogged plots and the average number of tree species in plots logged 8 years ago.

I hope this helps!

Answer 2

Using the t-distribution, it is found that the 90% confidence interval for the difference in mean number of species between unlogged and logged plots is (0.5, 7.1).

The first step is find the mean and the standard deviation for both samples, using a calculator, hence:

$\mu_{U} = 17.5, s_U = 3.53$

$\mu_{L} = 13.7, s_L = 4.5$

The standard errors are given by:

$s_U = \frac{3.53}{\sqrt{12}} = 1.019$

$s_L = \frac{4.5}{\sqrt{9}} = 1.5$

The distribution of the difference is given by:

$\overline{x} = \mu_U - \mu_L = 17.5 - 13.7 = 3.8$

$s = \sqrt{s_U^2 + s_L^2} = \sqrt{1.019^2 + 1.5^2} = 1.8$

The interval is:

$\overline{x} \pm ts$

• By the conservative method, the amount of degrees of freedom is one less than the smaller sample size, hence:

The critical value, using a t-distribution calculator, for a two-tailed 90% confidence interval, with 9 - 1 = 8 df, is t = 1.8595.

Then:

$\overline{x} - ts = 3.8 - 1.8595(1.8) = 0.5$

$\overline{x} + ts = 3.8 + 1.8595(1.8) = 7.1$

The 90% confidence interval for the difference in mean number of species between unlogged and logged plots is (0.5, 7.1).

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