Topic 1: Statistical Analysis Assessment Statements 1.1.1 State that error bars are a graphical representation of the variability of data 1.1.2 Calculate the mean and standard deviation of a set of values. 1.1.3 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. 1.1.4 Explain how the standard deviation is useful for comparing the means and the spread of data between 2 or more samples 1.1.5 Deduce the significance of the difference between two sets of data using calculated values for t and the appropriate tables. 1.1.6 Explain that the existence of a correlation does not establish that there is a causal relationship between two variables.

Why use statistics? In science, we are always making observations about the world around us. Many times these observations result in the collection of measurable, quantitative data. Using this data, we can ask questions based upon some of these observations and try to answer those questions.

Essential Statistics Terms Mean: the average of the data points Range: the difference between the largest and smallest observed values in a data set (the measure of the spread of the data) Standard Deviation: a measure of how the individual observations of a data set are dispersed or spread out around the mean Error bars: Graphical representation of the variability of data

Standard Deviation Standard deviation is used to summarize the spread of values around the mean and to compare the means of data between two or more samples. In a normal distribution, about 68% of all values lie within ±1 standard deviation of the mean. 95% will lie within ±2 standard deviations of the mean. This normal distribution will form a bell curve The shape of the curve tells us how close all of the data points are to the mean.

Calculating Standard Deviation We can use the formula to solve for standard deviation… Σ(xi - (Σx)2 Sx = n n - 1 Example data: Ward pg.5 For calculator help, go to http://www.heinemann.co.uk/hotlinks ,enter code 4242p and click weblink 1.4a or 1.4b HOWEVER the easiest way to calculate…find an app online!!!!

Are our results reliable enough to support a conclusion?

Imagine we chose two children at random from two class rooms… … and compare their height …

… we find that one pupil is taller than the other C1 … we find that one pupil is taller than the other WHY?

REASON 1: There is a significant difference between the two groups, so pupils in C1 are taller than pupils in D8 D8 C1 YEAR 7 YEAR 11

REASON 2: By chance, we picked a short pupil from D8 and a tall one from C1 D8 C1 Sammy (Year 9) HAGRID (Year 9)

How do we decide which reason is most likely? MEASURE MORE STUDENTS!!!

… the average or mean height of the two groups should be very… If there is a significant difference between the two groups… D8 C1 … the average or mean height of the two groups should be very… … DIFFERENT

… the average or mean height of the two groups should be very… If there is no significant difference between the two groups… D8 C1 … the average or mean height of the two groups should be very… … SIMILAR

Living things normally show a lot of variation, so… Remember:

It is VERY unlikely that the mean height of our two samples will be exactly the same C1 Sample Average height = 162 cm D8 Sample Average height = 168 cm Is the difference in average height of the samples large enough to be significant?

Here, the ranges of the two samples have a small overlap, so… We can analyse the spread of the heights of the students in the samples by drawing histograms 2 4 6 8 10 12 14 16 Frequency 140-149 150-159 160-169 170-179 180-189 Height (cm) C1 Sample D8 Sample Here, the ranges of the two samples have a small overlap, so… … the difference between the means of the two samples IS probably significant.

Here, the ranges of the two samples have a large overlap, so… 2 4 6 8 10 12 14 16 Frequency 140-149 150-159 160-169 170-179 180-189 Height (cm) C1 Sample Here, the ranges of the two samples have a large overlap, so… … the difference between the two samples may NOT be significant. 2 4 6 8 10 12 14 16 Frequency 140-149 150-159 160-169 170-179 180-189 Height (cm) D8 Sample The difference in means is possibly due to random sampling error

… and the spread of heights in each sample. To decide if there is a significant difference between two samples we must compare the mean height for each sample… … and the spread of heights in each sample. Statisticians calculate the standard deviation of a sample as a measure of the spread of a sample Sx = Σ(xi - (Σxi)2 n n - 1 Where: Sx is the standard deviation of sample Σ stands for ‘sum of’ xi stands for the individual measurements in the sample n is the number of individuals in the sample You can calculate standard deviation using the formula:

It is much easier to use the statistics functions on a scientific calculator! e.g. for data 25, 34, 13 Set calculator on statistics mode MODE 2 (CASIO fx-85MS) Clear statistics memory SHIFT CLR 1 (Scl) = Enter data 2 5 DT (M+ Button) 3 4 1

Calculate the standard deviation Calculate the mean 24 AC SHIFT S-VAR (2 Button) 1 ( x ) = Calculate the standard deviation 10.5357 AC SHIFT S-VAR 3 = (xσn-1)

We start by calculating a number, t Student’s t-test The Student’s t-test compares the averages and standard deviations of two samples to see if there is a significant difference between them. We start by calculating a number, t t can be calculated using the equation: ( x1 – x2 ) (s1)2 n1 (s2)2 n2 + t = Where: x1 is the mean of sample 1 s1 is the standard deviation of sample 1 n1 is the number of individuals in sample 1 x2 is the mean of sample 2 s2 is the standard deviation of sample 2 n2 is the number of individuals in sample 2

Worked Example: Random samples were taken of pupils in C1 and D8 Their recorded heights are shown below… Students in C1 Students in D8 Student Height (cm) 145 149 152 153 154 148 157 161 162 158 160 166 163 167 172 175 177 182 183 185 187 Step 1: Work out the mean height for each sample C1: x1 = 161.60 D8: x2 = 168.27 Step 2: Work out the difference in means x2 – x1 = 168.27 – 161.60 = 6.67

Step 3: Work out the standard deviation for each sample C1: s1 = 10.86 D8: s2 = 11.74 Step 4: Calculate s2/n for each sample (s1)2 n1 = C1: 10.862 ÷ 15 = 7.86 (s2)2 n2 = D8: 11.742 ÷ 15 = 9.19

Step 5: Calculate (s1)2 n1 + (s2)2 n2 (s1)2 n1 + (s2)2 n2 = (7.86 + 9.19) = 4.13 Step 6: Calculate t (Step 2 divided by Step 5) t = (s1)2 n1 + (s2)2 n2 = x2 – x1 6.67 4.13 = 1.62

Step 7: Work out the number of degrees of freedom d.f. = n1 + n2 – 2 = 15 + 15 – 2 = 28 Step 8: Find the critical value of t for the relevant number of degrees of freedom Use the 95% (p=0.05) confidence limit Critical value = 2.048 Our calculated value of t is below the critical value for 28d.f., therefore, there is no significant difference between the height of students in samples from C1 and D8

Follow the instructions CAREFULLY Do not worry if you do not understand how or why the test works Follow the instructions CAREFULLY You will NOT need to remember how to do this for your exam