Statistical Tests

Question 1

Purpose of t-test.

Answer 1

• To compare 2 sets of mean
• If they are significantly different or not

Question 2

When is t-test used?

Answer 2

Data are:
- continuous
- normally distributed

Question 3

Formula of t-test.

Answer 3

• t = |Ā - B| / Square root of (SA^2/nA + SB^2/nB)
• degrees of freedom = (nA - 1) + (nB - 1) = nA + nB - 2

Question 4

Steps for t-test.

Answer 4

1. let Ho: Ā = B
2. Calculate s. d. of each sample.
3. Calculate t-value.
4. Find degrees of freedom.
5. Find probability of Ā = B
6. When P <= 0.05, Ā does not = B
   When P > 0.05, Ā = B

Question 5

How to conclude for t-test?

Answer 5

1. critical value (at degrees of freedom = n & p = 0.05) = x
2. calculated t-value is, greater / lesser, than critical (t-) value
3. null hypothesis, accepted / rejected (at p = 0.05)

Question 6

Purpose of chi-squared test.

Answer 6

• To test the goodness of fit between
• O & E (observed & expected values)
• When analysing results from:
• genetics / breeding experiments
• ecological sampling

Question 7

When is chi-squared test used?

Answer 7

Data are:
- discrete
- categoric

Question 8

Formula of chi-squared test.

Answer 8

χ^2= ∑(O-E)^2/E

Question 9

Steps for chi-squared test.

Answer 9

1. Set up null hypothesis, Ho: there is no significant difference between O & E.
2. Calculate chi-squared value.
3. Calculate degrees of freedom = n-1 (n = no. of classes)
4. Use a chi-squared table to find P(χ2)
5. If P(χ2) > 0.05:
   - chances of goodness of fit > 5%
   - accept Ho
   - O = E
6. If P(χ2) <= 0.05:
   - chances of goodness of fit <= 5%
   - reject Ho
   - O P(χ2) > 0.05:
   - chances of goodness of fit > 5%
   - accept Ho
   - O ≠ E

Question 10

How to conclude for chi-squared test?

Answer 10

• Conclusion is made with 95% confidence.
• (Because) the rejected value could be rejected wrongly
• Allow 5% in conclusion

Question 11

Purpose of Pearson’s linear correlation.

Answer 11

Determines whether there is linear correlation between 2 variables

Question 12

When is Pearson’s linear correlation used?

Answer 12

Data must be:
- quantitative
- show normal distribution

Question 13

Formula of Pearson’s linear correlation.

Answer 13

r = [ ∑ xy - nx̄ȳ ] / (n-1) SxSy

• r = correlation coefficient
• x = no. of species A
• y = no. of species B
• n = no. of readings
• Sx = standard deviation of species A
• Sy = standard deviation of species B
• x̄ = mean no. of species A
• ȳ = mean no. of species B

Question 14

Steps for Pearson’s linear correlation.

Answer 14

1. Create a scatter graph of data & identify if a linear correlation exists.
2. State a null hypothesis, Ho = there is no correlation between the abundance of species A & species B.
3. Use the following equation to work out Pearson’s correlation coefficient, r

Question 15

How to conclude for Pearson’s linear correlation?

Answer 15

If the correlation coefficient, r is close to 1 or -1, then it can be stated that there is a strong linear correlation between the 2 variables & the null hypothesis can be rejected.

Question 16

Purpose of Spearman’s Rank Correlation.

Answer 16

Determines whether there is correlation between variables that don’t show a normal distribution.

Question 17

When is Spearman’s Rank Correlation used?

Answer 17

• There is an apparent relationship between 2 variables but
• Data does not show a normal distribution
• Data can be ranked

Question 18

Formula for Spearman’s Rank Correlation.

Answer 18

rs = 1 - (6 x ∑D^2 / n^3 - n)

• rs = spearman’s rank coefficient
• D = difference in rank
• n = number of samples

Question 19

Steps for Spearman’s Rank Correlation.

Answer 19

1. Create a scatter graph & identify possible linear correlation.
2. State a null hypothesis, Ho = there is no correlation between the abundance of species A & species B.
3. Use the Spearman’s Rank Correlation formula to work out the coefficient r.
4. Refer to a table that relates critical values of rs to levels of probability.

Question 20

How to conclude for Spearman’s Rank Correlation?

Answer 20

If the value calculated for Spearman’s rank is greater than the critical value for the number of samples in the data (n) at the 0.05 probability level (p), then the Ho can be rejected, meaning there is a (positive / negative) correlation between 2 variables.

Question 21

Purpose of Simpson’s Index of diversity (D).

Answer 21

To calculate species diversity & evenness of abundance across different species in an area.

Question 22

When is Simpson’s Index used?

Answer 22

When data for species abundance has been calculated

Question 23

Formula for Simpson’s Index of Diversity.

Answer 23

d = 1 - (∑ (n / N)^2)

• n = number of individuals
• N = total number of organisms

Question 24

Steps for calculating Simpson’s Index of Diversity.

Answer 24

1. Calculate (n / N) for each species
2. Square each of these values
3. Add them together & subtract the total from 1.

Question 25

How to conclude for Simpson's Index of Diversity?

Answer 25

- The value of D can fall between 0 and 1. - Value near 1 = high levels of biodiversity - Value near 0 = low levels of biodiversity

Question 26

Purpose of Lincoln's Index.

Answer 26

Calculate an estimate of population size

Question 27

Formula for Lincoln's Index.

Answer 27

N = n1 x n2 / m2 - N = population estimate - n1 = number of marked individuals released from 1st sample - n2 = number of individuals in 2nd sample (marked & unmarked) - m2 = number of marked individuals in the 2nd sample

Question 28

State what standard deviation shows.

Answer 28

- Spread of data from the mean value - Larger value = less reliable results