[MUSIC]

Hi guys, welcome to the 26th lecture of the course, biological diversity,

theories, measures and data sampling techniques.

Today we will talk about the third part of studies [INAUDIBLE] to

study biological diversity.

I will explain to you how to use the chi-square test to check the study's

significance of the differences in your biodiversity samples.

The chi-square test is a useful test to understand

how the observed frequency of collected data are close to the expected values.

This test is frequently used although it is simple in the formulation it is not

easy in calculation.

This formula is the following in just the sum of the square

of several minus expected values divided by the expected values.

This test is a kind of confrontation of the new hypothesis H zero.

Because it provides that even in the for 2, all the expected frequencies would

be equal if the observant values were the result of of something else.

For the practice of the calculation of the expected frequencies,

just divide the sample of all frequencies detected in the samples, for example,

the number of spiders caught in traps a, b and

c, to the total number of samples in this example, r3.

Then the observer frequencies correspond exactly

to the number of units in the example.

In this case, number of spiders in A, B and C traps.

We proceed to calculate the square value of the seven minus expected

divided by expected values for each samples A, B, and C and

then the shareable that is updated from the sample, she's then calculated.

Then we need to calculate the doubles of freedom.

That is the number of categories of the samples, minus one.

Now our example is 3- 1, so 2.

Then we take the value obtained from the calculation of the chi-square and

the degree of freedom and

we look at the test table that's presented in any statistical book or online.

In the first column of this table there are degree of freedom and

we can select and find those corresponding values to our.

In our example, there are two.

If the chi square value we obtain is higher then the two values on the right

column of the test table.

Which contains the three short values for the probability 0.05 or 0.01.

It can be concluded that the observed association

between the frequencies and the related samples is highly significant.

In the example,

the trap with the higher number of spiders captured is significantly more

effective in capturing these animals, if compared to the other two tested.

If the chi squared value were only higher than the table value of P 0.05,

and lower than P 0.01 this is association between frequency and

sample would be only significant.

This is because of the lack of association between the trap and

frequency that is the null hypothesis H0 In this case, would be valued

less than five times in hundred but at least more than one time out of hundred.

It is important to remember that this test is effective

only if the expected frequency exceeds the value of five.

You can bypass this issue by aggregating two or more categories of samples.

When there is only one degree of freedom, comparison between two samples, for

instance, it is necessary to apply to the chi-squared formula

the Yates correction for continuity,

which simply consists in subtracting 0.5 to the numerator of the formula.

So today I'll show you how to calculate the key square test and to show how

your samples in biological diversity, they are different in a statistical way.

See you in the next section.