Researchers have observed that rainforest areas next to clear-cuts (less than 100 meters away) have a reduced tree biomass compared to rainforest areas far from clear-cuts. To go further, Laurance et al. (1997) tested whether rainforest areas more distant from the clear-cuts were also affected. They compiled data on the biomass change after clear-cutting (in tons/hectare/year) for 36 rainforest areas between 100 m and several km from clear-cuts. The data are as follows:

-10.8, -4.9, -2.6, -1.6, -3, -6.2, -6.5, -9.2, -3.6, -1.8, -1, 0.2, 0.2, 0.1, -0.3, -1.4, -1.5, -0.8, 0.3, 0.6, 1, 1.2, 2.9, 3.5, 4.3, 4.7, 2.9, 2.8, 2.5, 1.7, 2.7, 1.2, 0.1, 1.3, 2.3, 0.5

Test whether there is a change in biomass of rainforest areas following clear-cutting.

If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that we don't have a significant change at 5% of signficance.

Step-by-step explanation:

Data given and notation

-10.8, -4.9, -2.6, -1.6, -3, -6.2, -6.5, -9.2, -3.6, -1.8, -1, 0.2, 0.2, 0.1, -0.3, -1.4, -1.5, -0.8, 0.3, 0.6, 1, 1.2, 2.9, 3.5, 4.3, 4.7, 2.9, 2.8, 2.5, 1.7, 2.7, 1.2, 0.1, 1.3, 2.3, 0.5

We can calculate the mean and deviation with the following formulas:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

[tex] s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2)}{n-1}}[/tex]

And we obtain:

[tex]\bar X=-0.506[/tex] represent the sample mean

[tex]s=3.557[/tex] represent the sample standard deviation

[tex]n=36[/tex] sample size

[tex]\mu_o =0[/tex] represent the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test. (assumed)

t would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if there is a change in biomass of rainforest areas following clear-cutting, the system of hypothesis are :

Null hypothesis:[tex]\mu = 0[/tex]

Alternative hypothesis:[tex]\mu \neq 0[/tex]

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]t=\frac{-0.506-0}{\frac{3.557}{\sqrt{36}}}=-0.854[/tex]

P-value

First we need to calculate the degrees of freedom given by:

[tex]df=n-1=36-1=35[/tex]

Since is a two tailed test the p value would be:

[tex]p_v =2*P(t_{(35)}<-0.854)=0.399[/tex]

Conclusion

If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that we don't have a significant change at 5% of signficance.