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 we see that 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:

And we obtain:

represent the sample mean    

represent the sample standard deviation    

sample size    

represent the value that we want to test    

represent the significance level for the hypothesis test. (assumed)    

t would represent the statistic (variable of interest)    

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:    

Alternative hypothesis:    

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:    

(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:    

P-value    

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

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

Conclusion    

If we compare the p value and the significance level assumed we see that 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.  

c

step-by-step explanation:

2 & 7

step-by-step explanation:

he climbs 300 ft/hr at a constant rate. the peak is at 1,500 ft.

when going up, he reaches 600 ft in 2 hours. remember, that he is going in a constant rate. divide 600 with 300:

600/300 = 2

assuming that he takes the same route back (meaning that it's a 1500 ft route back, you can solve.

note also that you are solving when he gets back to 600 ft after he reaches the peak.

first, find the amount of hours it takes to get up the peak. divide the total amount (1500), with the ft/her (300) to get the amount of hours (x):

1500/300 = x

x = 5

it takes jamie 5 hours to get up. now, he is returning down, and reaches 600 (his current destination). remember that he climbs at a constant 300 ft/hr. divide 600 with 300:

600/300 = x

x = 2

now add the two numbers together.

x = 5 + 2 = 7

jamie takes 7 hours to go up the mountain and reach his current position when descending.

(total amount/constant rate) = (time).

remember that it is a constant rate, so in total, when he gets back down the mountain, it will take the same amount of time as when he goes up. the total amount for the round trip is: 5 + 5 = 10.

~