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

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.

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

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

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

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

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

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

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

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

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.

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

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

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!!!!!!!!!!!!!!

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

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

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.

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

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

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;

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

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.

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

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

Student’s t-test Paired, two-tailed Degrees of freedom = (# in A) + (# in B) – (# of data sets) = 12 + 12 – 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

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

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.

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?

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

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.

Correlation

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

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

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; 2 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; 3 max (ii) little / no change in glucose uptake; most / all (protein) channels in use;

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 1 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; 2 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; 2 max

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