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- 1. Are our results reliable enough to support a conclusion? Geoff Browne Anglo-European School Essex, UK
- 2. Imagine we chose two children at random from two class rooms… D8 C1 … and compare their height …
- 3. D8 C1 … we find that one pupil is taller than the other WHY?
- 4. 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
- 5. REASON 2: By chance, we picked a short pupil from D8 and a tall one from C1 D8 C1 TITCH HAGRID (Year 9) (Year 9)
- 6. How do we decide which reason is most likely? MEASURE MORE STUDENTS!!!
- 7. If there is a significant difference between the two groups… D8 … the average or C1 mean height of the two groups should be very… … DIFFERENT
- 8. If there is no significant difference between the two groups… D8 … the average or C1 mean height of the two groups should be very… … SIMILAR
- 9. Remember: Living things normally show a lot of variation, so…
- 10. It is VERY unlikely that the mean height of our two samples will be exactly the same C1 Sample D8 Sample Average height = 162 cm Average height = 168 cm Is the difference in average height of the samples large enough to be significant?
- 11. 16 We can analyse the 14 C1 Sample spread of the heights of 12 Frequency the students in the 10 8 samples by drawing 6 histograms 4 2 Here, the ranges of the 140- 150- 160- 170- 180- 149 159 169 179 189 two samples have a Height (cm) small overlap, so… 16 14 12 D8 Sample … the difference between Frequency 10 8 the means of the two 6 samples IS probably 4 significant. 2 140- 150- 160- 170- 180- 149 159 169 179 189 Height (cm)
- 12. 16 Here, the ranges of 14 C1 Sample the two samples have 12 Frequency a large overlap, so… 10 8 6 4 … the difference 2 between the two 140- 150- 160- 170- 180- samples may NOT be 149 159 169 179 189 Height (cm) significant. 16 14 12 D8 Sample The difference in Frequency 10 means is possibly due 8 to random sampling 6 4 error 2 140- 150- 160- 170- 180- 149 159 169 179 189 Height (cm)
- 13. 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 You can calculate standard deviation using the formula: Where: (Σx)2 Sx is the standard deviation of sample Sx = Σx2 - Σ stands for ‘sum of’ n x stands for the individual measurements in n-1 the sample n is the number of individuals in the sample
- 14. 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 DT 1 3 DT
- 15. Calculate the mean AC SHIFT S-VAR (2 Button) 1 ( x ) = 24 Calculate the standard deviation AC SHIFT S-VAR 3 (xσn-1) = 10.5357
- 16. 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 ) Where: t= x1 is the mean of sample 1 (s1)2 (s2)2 s1 is the standard deviation of sample 1 + n1 is the number of individuals in sample 1 n1 n2 x2 is the mean of sample 2 s2 is the standard deviation of sample 2 n2 is the number of individuals in sample 2
- 17. 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 145 149 152 153 154 148 153 157 161 162 Height 154 158 160 166 166 162 163 167 172 172 (cm) 166 167 175 177 182 175 177 183 185 187 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
- 18. 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 C1: (s1)2 = 10.862 ‚ 15 = 7.86 n1 D8: (s2)2 = 11.742 ‚ 15 = 9.19 n2
- 19. Step 5: Calculate (s1)2 (s2)2 + n1 n2 (s1)2 (s2)2 = (7.86 + 9.19) = 4.13 + n1 n2 Step 6: Calculate t (Step 2 divided by Step 5) x2 – x1 t= 6.67 (s1)2 (s2)2 = = 1.62 + 4.13 n1 n2
- 20. 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
- 21. 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
- 22. Uploaded (with permission from Geoff Browne) by Stephen Taylor For more IB Biology resources: http://sciencevideos.wordpress.com

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