Experiment 1

• Investigating the effect of temperature on the rate of respiration of Crustae larvae.

• Basic expt: Measure the respiration rate of larvae at 3oC then every 3 degrees up to 18oC. See results on next page.

Effect of temperature on the respiration rate of Crustacea Larvae

•State null hypothesis•Which stats test? Give a reason.•Calculate test statistic and explain what this shows.

Temp ° C Respiration. rate.μg O2 h-1

3 0.12

6 0.17

9 0.15

12 0.21

15 0.22

18 0.35

Check your answers:

• Null hypothesis:• There is no significant correlation between

a change in temperature and the rate of respiration.

• Or• The temperature does not effect the rate

of respiration.

Which stats test with reason:

Reason: You are looking for a correlation between a change in

temperature and the rate of respiration

Calculating and interpreting test statistic.

Temperature/ ° C

Rank (R1) Respiration. rate.μg O2

h-1 2

Rank (R2)

D (R1-R2) D2

3 1 0.12 1 0 0

6 2 0.17 3 -1 1

9 3 0.15 2 1 1

12 4 0.21 4 0 0

15 5 0.22 5 0 0

18 6 0.35 6 0 0

Σ

Rs = 1 - 6 x (0+1+1+0+0+0) = 1 - 12 = 1- 0.0571 = 0.943 (6x6x6) – 6 210

Interpreting the test statistic:• State what it shows:• My value of 0.943 is greater than 0.89

which is the critical value for 6 pairs of measurements P < 0.05

• Explain the probability of your results being significant:There is less than 5% probability the correlation between temperature and respiration rate is due to chance

• Accept or reject your null hypothesis• I reject my null hypothesis

Experiment 2

• Investigating the growth of Pleurococcus on the north and south facing sides of tree trunks.

• Basic expt: Measure the coverage of Pleurococcus on the north and south facingtrunks of a tree. Repeat the experiment 14 more times.

Pleurococcus coverage on north facing tree trunk / cm2

Pleurococcus coverage on south facing tree trunk / cm2

28 6

36 21

27 35

34 25

43 34

73 63

16 8

12 7

28 26

45 18

67 32

45 27

58 13

39 19

62 5

• State null hypothesis

• Which stats test? Give a reason.

• Calculate test statistic and explain what this shows.

Check your answers:

• Null hypothesis:• There is no significant difference in the

coverage/amount of Pleurococcus on the north and south facing sides of tree trunks.

• Or• The direction the trunk faces does not

effect the amount of Pleurococcus.

Which stats test with reason:

Reason: You are looking for differences in the mean coverage of Pleurcoccus on the north and south facing sides of the trunk

Calculating and interpreting test statistic.

Pleurococcus on North side Pleurococcus on South side

Mean 613 = 40.87

15

Sd = 18.01

SE = 18.01

√15

SE = 18.01

3.873

SE = 4.65

Mean 348 = 22.6

15

Sd = 15.14

SE = 15.14

√15

SE = 15.29

3.873

SE = 3.909

Pleurococcus on North side Pleurococcus on South side

SE = 4.652 X SE = 9.395% confidence

limits =Mean + 2 SE =

40.87+9.3 = 50.17Mean – 2 SE =40.87 – 9.3 = 31.57

ie 31.57 to 55.04Draw a graph of the meanand show the 95%

confidence limits on to it

SE = 3.9092 X SE = 7.81895% confidence

limits =Mean + 2 SE =

22.6+7.818 = 30.418

Mean – 2 SE =22.6 – 7.818 =

14.787

ie 14.787 to 30.418Draw a graph of the meanand show the 95%

confidence limits on to it

Use of Standard Error in showing 95% Confidence Limits on a bar chart of Mean Values

Show 2 SE value up from the mean and show 2 SE

value downwards from the mean for each bar. This illustrates the range for

the 95% confidence Limits for each mean value.

Side of tree

North facing South facing

Mean Coverage of Pleurococcus / cm2

+ 2 S.E

- 2 S.E

Interpreting the test statistic:

• State what it shows:• The error bars do not overlap.• Explain the probability of your results

being significant:• There is less than 5% probability the

difference in the mean coverage on the north and south sides is due to chance.

• Accept or reject your null hypothesis• I reject my null hypothesis

Experiment 3

• Investigating the distribution of banded snails in the environment

• Basic Expt.: Count the number of banded snails in 1m2 of hedgerow then repeat in 1m2 of grassland.

Number of snails in 1m2 hedgerow

Number of snails in 1m2 grassland

24 12

• State null hypothesis

• Which stats test? Give a reason.

• Calculate test statistic and explain what this shows.

Check your answers:

• Null hypothesis:• There is no significant difference between

the number of snails in a hedgerow and the number of snails in the grassland.

• Or

The type of area has no effect on the number of snails.

Which stats test with reason:

Reason: You are looking for a difference in the number of snails

in particular categories.

Calculating and interpreting test statistic.

Observed frequency (O)

Expected frequency (E)

(O - E)

(O - E)2

(O - E)2 / E

Short plants

Long plants

Hedgerow

grassland

24

12

18

18

6

-6

36

36

2

2

Calculating and interpreting test statistic.

= 2 + 2 = 4

Interpreting the test statistic:• State what it shows:• For 1 degree of freedom the critical value

is 3.84. My value of 4 is greater than the critical value so P < 0.05

• Explain the probability of your results being significant: significant:There is less than 5% probability the difference in number of snails in each area is due to chance

• Accept or reject your null hypothesis• I reject my null hypothesis