Student's T test/ Chi squared

Question 1

When is chi squared stastical test used?

Answer 1

• Used when data is related to frequency ie. no of plants.
• Tests the significance of difference between expected and observed values.
• Investigating differences between frequencies.

Question 2

Observed no of ivy leaves in dark vs light area:
DARK: 366
LIGHT: 634.
How would you work out EXPECTED no in this case AND what would be your null hypothesis?

Answer 2

• EXPECTED:
• DARK: 500. LIGHT: 500
• 50-50 split.
• Null hypothesis: there will be no significant difference between observed no of ivy plants and expected no of ivy plants in dark vs light areas.

Question 3

Once you have calculated your x² for no of ivy plants in dark vs light areas (as an example) what would you do given critical values table for P = 0.05?

Answer 3

• Work out degree of freedom: number of categories - 1.
• See what critical value this degree of freedom corresponds to.
• If your x² value is greater than your critical value, this shows that there is less than 5% probability that the difference between observed and expected value is due to chance.
• The null hypothesis can be rejected, significant difference between no of ivy plants in dark vs light areas.

Question 4

What does a critical values table for P= 0.05 show for x2 stastical test show?
in chi squared

Answer 4

• Shows probability that difference between expected and observed data is due to chance.

Question 5

When would you use the Student’s t-test (stastical test?)

Answer 5

• When you have continuous data.
• And you are investigating the difference between two mean values in that continuous data.

Question 6

If you are given many critical values in a table that correspond to different p (probability) values, which p value would you pick and why?

Answer 6

P = 0.05.
≤ 0.05 is the cut off in saying whether the difference in values was significant or not.
If it’s too much higher (ie. P ( 0.1)) this will give too high of a probability that the difference in values was significant/ not.

Question 7

What does a P value tell you?

Answer 7

• P value tells you probability that a result/ difference between results is due to chance.

Question 8

What is the paired T test? How is degree of freedom calculated with paired T test?

Answer 8

• Two SETS of data (ie. one with one factor being changed/ the other set with another factor being changed.)
• Collected from same individual.
• Both groups must have same sample size.
• number of categories - 1

Question 9

What is the unpaired T test? How is degree of freedom calculated with unpaired T test?

Answer 9

• Two sets of data collected may not have same no samples/ results.
• (n1 + n2) -2
• n1 = No of samples in data set 1
• n2 = No of samples in data set 2

Question 10

What should you include in your conclusion using a student’s t-test?

Answer 10

• Accept/ reject the null hypothesis.
• Is there more/ less than 5% probability that the difference between the means was due to chance.

Question 11

A student calculated a student’s t-test of 4.82 with 2 degrees of freedom.
In critical values table for P = 0.05, the critical value was 4.31. What can therefore be concluded?

Answer 11

• Reject null hypothesis because t-test CALCULATED value is higher than critical value.
• So, there is less than 5% probability that the difference in means is due to chance.

Question 12

A student calculated a student’s t-test of 2.01 with 5 degrees of freedom.
In critical values table for P = 0.05, the critical value was 2.57. What can therefore be concluded?

Answer 12

• Accept null hypothesis because t-test calculated is less than critical value.
• There is more than 5% probability that the difference in means is due to chance.

Question 13

Worked Example Miss Estruch

Data:
Values - mean “change” in species richness from 2019 and 2020.
() - P values
Habitat A: -3.22
( greater than sign) 0.05
Habitat B: +2.31
(greater than sign) 0.01
Habitat C: + 0.12
(less than sign) 0.05
Scientists concluded that communities DID change. Using only the data above, evaluate this conclusion.

Answer 13

• P value shows probability that differences in mean values are due to chance.
• Habitat C didn’t show a significant change.
• More than 5% probability that increase was due to chance.
• Significant increase in mean species richness for habitat A/ B.
• Less than 1% probability that the increase was due to chance in habitat B.
• Less than 5% probability that decrease was due to chance in habitat A.

Question 14

What is a “null hypothesis?”

Answer 14

• Hypothesis that says difference between observed/ expected data OR difference between mean values ISN’T SIGNIFICANT.