1. Topic 1 and Data analysis
REVISION:
Topic 1 and Data analysis
Statistics

2. Topic 1 –guide
STATE that error bars are a graphical representation of the variability of data.
CALCULATE the mean and standard deviation of a set of values.
STATE that the term standard deviation is used to summarize the spread of values around the mean, and that 68% of the values fall within one standard deviation of the mean.
EXPLAIN how the standard deviation is useful for comparing the means and the spread of data between two or more samples.
DEDUCE the significance of the difference between two sets of data using calculated values for t and the appropriate tables.
EXPLAIN that the existence of a correlation does not establish that there is a causal relationship between two variables.

3. Topic 1 –guide
Mean
Standard deviation
t-tests and significant differences
Correlation

4. Topic 1 –guide
Mean
Standard deviation
t-tests and significant differences
Correlation

5. 𝒙 = 𝒔𝒖𝒎 𝒐𝒇 𝒂𝒍𝒍 𝒗𝒂𝒍𝒖𝒆𝒔 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒗𝒂𝒍𝒖𝒆𝒔
Calculating a mean
What is a mean?
Average value in a data set
Why would we need to calculate it?
Gives us a measure of central tendency for the data
𝒙 = 𝒔𝒖𝒎 𝒐𝒇 𝒂𝒍𝒍 𝒗𝒂𝒍𝒖𝒆𝒔 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒗𝒂𝒍𝒖𝒆𝒔

6. Calculating a mean – question time
You and your friends have just measured the heights of your dogs (in millimeters):
The heights (at the shoulders) are: 600mm, 470mm, 170mm, 430mm and 300mm.
Find out the mean height of your dogs.
394mm

7. Topic 1 –guide
Mean
Standard deviation
t tests and significant differences
Correlation

8. Calculating the standard deviation
What is the standard deviation?
a value for how spread out the data is
Why would we need to calculate it?
Gives an idea of the variability & reliability of the data
𝒔= 𝟏 𝑵−𝟏 𝒊=𝟏 𝑵 𝒙 𝒊 − 𝒙 𝟐

9. Calculating the standard deviation – question time
You and your friends have just measured the heights of your dogs (in millimeters):
The heights (at the shoulders) are: 600mm, 470mm, 170mm, 430mm and 300mm.
Find out the standard deviation in the height of your dogs.
147mm

10. Calculating the standard deviation
DID YOU KNOW 
that your little scientific calculators also calculate the mean and standard deviation for you????

11. Interpreting the standard deviation
STATE 
that the term standard deviation is used to summarize the spread of values around the mean, and that 68% of the values fall within one standard deviation of the mean.

12. Interpreting the standard deviation
68.2% of data falls within one S.D. of the mean
95.4% of data falls within two S.D. of the mean
99.6% of data falls within three S.D. of the mean

13. Interpreting the standard deviation
A small S.D. indicates a small range of data
A large S.D. indicates a large range of data

14. Interpreting the standard deviation
S.D. used to compare means and variability of data between two or more samples
If the S.D. is greater than the difference between two means, then the difference is not statistically significant
this will lead to overlapping error bars on a graph
Remember: small samples are unreliable!!!!!!!!!!!!!!

15. Using the standard deviation
STATE 
error bars are a graphical representation of the variability of data.

16. Mean & standard deviation – question time
The lengths of a sample of tiger canines were measured. 68% of the lengths fell within a range between 15 mm and 45 mm. The mean was 30 mm. What is the standard deviation of this sample?
15 mm

17. Mean & SD – question time
Compare the range of variation in beak length of the Yellow-throated Warblers in Midwest to the beak length of the Yellow-throated Warblers in Delmarva.

18. Mean & SD – question time
Yellow-throated Warblers have greater
variation (of beak length) in Delmarva than in Midwest

19. Mean & SD – question time
Compare the gain and loss of ions in the male moths which have drunk from laboratory solutions with the changes in those that have drunk from natural puddles.
Mean & SD – question time

20. Mean & SD – question time
Sodium was retained from lab solutions and natural puddles;
Potassium was lost from lab solutions but uncertain loss/gain from natural puddles;
Slight loss of magnesium from lab solutions and uncertain gain/loss from natural puddles;
Variation in data for calcium;
More conclusive results in lab solutions / conditions more reliable in lab / greater variation in natural puddles;

21. Topic 1 –guide
Mean
Standard deviation
t-tests and significant differences
Correlation

22. Student’s t-test What is a t-test?
a statistical test used to determine the significance of the difference between two means.
Why would we need to calculate it?
If we want to confirm an experimental hypothesis and determine with confidence that the IV has contributed significantly to the change in the DV.

23. Student’s t-test What would you need to be able to do a t-test?
2 sets of data
t-test data table
hypothesis
information about data
null
experimental
Paired/ unpaired data
One/two tailed
IV will not affect DV
IV will affect DV
Paired data = same subjects tested in both groups
Two tailed = looks for either +ve or –ve effect

24. Student’s t-test Example: Temperatures in Miami Vs. Honolulu
In the following data pairs A = Average monthly temperature in Miami B = Average monthly temperature in Honolulu
The data are paired by month.
Reference: U.S. Department of Commerce Environmental Data Service
A = MIAMI
B = HONOLULU
67.5
74.4
68.0
72.6
71.3
73.3
74.9
74.7
78.0
76.2
80.9
82.2
79.1
82.7
79.8
81.6
79.5
77.8
78.4
72.3
76.1
68.5
73.7

25. Student’s t-test Paired, two-tailed Degrees of freedom
= (# in A) + (# in B) – (# of data sets)
= – 2 = 22
Confidence interval
Usually 99.95%
So p value = 0.05
A = MIAMI
B = HONOLULU
67.5
74.4
68.0
72.6
71.3
73.3
74.9
74.7
78.0
76.2
80.9
82.2
79.1
82.7
79.8
81.6
79.5
77.8
78.4
72.3
76.1
68.5
73.7

26. Student’s t-test Calculated t-value t = 0.431
Critical t-value from data table
t = 2.074
Compare t-value to t-test data table:
If t-value exceeds p=0.05 value, then data is significant
So our answer???
Accept the null hypothesis and reject the experimental hypothesis
No significant difference

27. t-test – question time
The difference between the means in two samples is not significant.
Which hypothesis can be tested using the t-test?
A. The difference in variation between two samples is not significant.
B. The difference between observed values and expected values is not significant.
C. The change in one variable is not correlated with a change in another variable.
D. The difference between the means in two samples is not significant.

28. t-test – question time
The levels of potassium in blood samples from 12 males and 11 females with coronary heart disease were compared using the t-test to see if there was a significant difference at the 5% level.
What is the critical value above which the two samples can be considered significantly different?

29. Topic 1 –guide
Mean
Standard deviation
t-tests and significant differences
Correlation

30. Correlation Correlation is not causation!
Correlation means that there is a relationship between two (or more) things.
This DOES NOT MEAN that one has caused the other to occur.
Correlation describes the strength and direction of a linear relationship between two variables
Use the Pearson correlation coefficient (r), ranging from -1 to +1.

31. Correlation

32. Correlation What does the scatter graph show?
Strong positive correlation between these variables.
What does the scatter graph show?

33. Correlation What does the scatter graph show?
Moderate negative correlation between these variables.
What does the scatter graph show?

34. Practice questions Answers: 1.
(a) as the diameter of the molecule increases the permeability / relative ability to move decreases (accept converse); the relationship is logarithmic / non-linear / negative; for molecules above 0.6 (± 0.1) nm relative ability to move changes little / for molecules below 0.6 (± 0.1) nm relative ability to move changes rapidly; max
(b) (i) 10 mmol cm–3 cells hr–1 (accept values within ±5); 1
(ii) 370 mmol cm–3 cells hr–1 (accept values within ± 10); 1
(c) (i) glucose uptake in facilitated diffusion levels out whereas uptake in simple diffusion does not level out / continues to rise; glucose uptake increases in both; glucose uptake is higher in facilitated diffusion (than in simple diffusion); glucose uptake in simple diffusion is constant / linear whereas in facilitated diffusion uptake increases rapidly at the beginning / increase is not constant; max
(ii) little / no change in glucose uptake; most / all (protein) channels in use;

35. Practice questions Answers: 2.
(a) standard deviation summarizes the spread of values around the mean / 68% of all values fall within one standard deviation of the mean / gives a measure of variability of the data / OWTTE max
(b) November had 113 (+2) ciliates ml-1 sediment (units required) 1
(c) production by treated and untreated samples is almost the same;
production by untreated samples is usually slightly higher than treated samples;
except November, January when the treated samples have a slightly higher methane production; max
(d) endosymbionts do not seem to be responsible for methane production;
methane production is almost the same whether the ciliates are alive (untreated samples) or killed (treated samples);
no apparent correlation between methane production and number of ciliates;
months when the population of ciliates is highest are not the months when the methane production is highest / ciliate numbers high in November when methane production is low / methane production highest in July and August when ciliate numbers are not high; max

36. Practice questions Answers: 2.
(e) greenhouses gases collect in atmosphere;
layer of gases allows incoming short-wave radiation (from sun) to pass through to earth’s surface where it is converted to longer-wave radiation;
long wave radiation cannot all pass through layer of gases but some reflected back to earth causing earth’s surface to become warmer; 2 max
(f) first name / Nacella refers to the genus and the second name / concinna refers to the species 1
(g) negative correlation / inversely proportional / as temperature increases the percentage righting in N. concinna decreases 1
(h) percentage of N. concinna able to right themselves decrease by 50% / decreases from 95% to less than 50% / less than half able to right themselves 1
(i) model suggests two degree rise in temperature which would mean summer temperatures of 3°C;
at this temperature less than 50% of organisms able to carry out basic behaviour;
decreased survival of species / decreased ability to avoid predation; 2 max