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Student's T test/ Chi squared

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Total cards: 14•Content version: v4•Public deck•Updated: 8/1/2026

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#1
Front (Question)

When is chi squared stastical test used?

Back (Answer)

• 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.

#2
Front (Question)

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?

Back (Answer)

• 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.

#3
Front (Question)

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?

Back (Answer)

• 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.

#4
Front (Question)

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

Back (Answer)

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

#5
Front (Question)

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

Back (Answer)

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

#6
Front (Question)

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?

Back (Answer)

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.

#7
Front (Question)

What does a P value tell you?

Back (Answer)

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

#8
Front (Question)

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

Back (Answer)

• 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

#9
Front (Question)

What is the unpaired T test?

Back (Answer)

• Two sets of data collected may not have same no samples.

#10
Front (Question)

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

Back (Answer)

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

#11
Front (Question)

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?

Back (Answer)

• 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.

#12
Front (Question)

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?

Back (Answer)

• 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.

#13
Front (Question)

Worked Example Miss Estruch Data: Values - mean “change” in species richness from 2019 and 2020. Habitat A: -3.22 P value = < 0.05 Habitat B: +2.31 P value = < 0.01 Habitat C: + 0.12 P value = > 0.05 Scientists concluded that communities DID change. Using only the data above, evaluate this conclusion.

Back (Answer)

• 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.

#14
Front (Question)

What is a “null hypothesis?”

Back (Answer)

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