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- 1. Correlation.
- 2. What is it? <ul><li>Two things correlate when they vary together. </li></ul><ul><li>E.G such as temperature decreasing with altitude or land values falling with distance from the city centre. </li></ul><ul><li>If, as one variable increases in value so does the other this is positive correlation. </li></ul><ul><li>If one goes up as the other goes down this is a negative correlation. </li></ul>
- 3. Positive. Negative.
- 4. Correlation. <ul><li>Correlation is useful for three reasons. </li></ul><ul><li>It is more precise than a graph. While two graphs showing correlations may look similar, the correlation coefficients for the sets of data may well be slightly different. </li></ul><ul><li>If we wanted to compare several pairs of data, such as the relationship between temperature and altitude on twenty slopes, it would be far easier to compare twenty numbers than twenty graphs. </li></ul><ul><li>It is possible to test the correlation to see if it is really significant or whether it could have occurred by chance. </li></ul>
- 5. WARNING!!!!!!. <ul><li>The fact that two things correlate proves nothing. We can never conclude from statistical evidence alone that, because two things correlate, one must be affecting the other. </li></ul><ul><li>All statistical tests must be supplemented with research regardless of the result. </li></ul>
- 6. Spearmans Rank. <ul><li>This technique is among the most reliable methods of calculating a correlation coefficient. </li></ul><ul><li>This is a number which will summarise the strength and direction of any correlation between two variables. </li></ul>
- 7. Method. <ul><li>Stage 1- Tabulate the data- I will show you how to do this with an example. </li></ul><ul><li>Stage 2 Find the difference between the ranks of each of the paired variables (d). Square these differences (d ²) and sum them ( Σ d ²). </li></ul><ul><li>Stage 3 Calculate the coefficient from (r s ) from the formula… </li></ul>
- 8. <ul><li>R s = 1- 6 Σ d ² </li></ul><ul><li> n ³-n </li></ul><ul><li>Where d= The difference in rank of the values of each matched pair. </li></ul><ul><li>N= the number of pairs. </li></ul>
- 9. <ul><li>The result can be interpreted from the scale. </li></ul><ul><li>+1.0 0 -1.0 </li></ul><ul><li>Perfect no Perfect </li></ul><ul><li>postive Correlation negative </li></ul><ul><li>correlation correlation </li></ul>
- 10. Next. <ul><li>Now you determine whether the correlation you have calculated is really significant, or whether it could have occurred by chance. </li></ul><ul><li>Stage 4 Decide on the rejection level ( ). </li></ul><ul><li>This is simply how certain you wish to be that the correlation you have calculated could not just have occurred by chance. Thus, if you wish to be 95 % certain your rejection level is calculated as follows… </li></ul>
- 11. <ul><li> = 100-95 </li></ul><ul><li> 100 </li></ul><ul><li>=0.05. </li></ul>
- 12. Stage 5. <ul><li>Calculate the formula for T. </li></ul><ul><li>T= R s n-2 </li></ul><ul><li> 1- R s ² </li></ul><ul><li>Where R s = spearmans rank correlation coefficient. </li></ul><ul><li>N= number of pairs. </li></ul>
- 13. <ul><li>Calculate the degrees of freedom. </li></ul><ul><li>Df = n-2. </li></ul><ul><li>Where n = the number of pairs. </li></ul>
- 14. Stage 7 <ul><li>Look up the critical value in the t- table using the degrees of freedom and the rejection level. </li></ul><ul><li>If the critical value is less than your t-value then the correlation is significant at the level chosen (95 %). </li></ul>
- 15. But what if my critical value is higher than my t value??? <ul><li>This means that you cannot be certain that the correlation could not have occurred by chance. This may mean one of two things. </li></ul><ul><li>A- The relationship is not a good one and it is thus not really worth pursuing it any further. </li></ul><ul><li>B- The size of the sample you are using is too small to permit you to prove correlation. </li></ul>
- 16. Example. <ul><li>Population size and number of services in each of 12 settlements. </li></ul><ul><li>Draw a graph for the following data set… </li></ul>
- 17. 19 2 362 12 11 1 016 11 35 4 981 10 73 6 781 9 81 9 982 8 72 8 763 7 114 15 739 6 4 220 5 87 10 714 4 43 6 793 3 41 5 632 2 3 350 1 No. of Services. Population Settlement
- 18. Stages 1-2 0 0 1 114 1 15 739 0 0 2 87 2 10 714 4 2 5 72 3 8 763 1 1 3 81 4 7 982 1 1 6 43 5 6 793 4 2 4 73 6 6 781 0 0 7 41 7 5 632 0 0 8 35 8 4 981 0 0 9 19 9 2 362 0 0 10 11 10 1 016 1 1 12 3 11 350 1 1 11 4 12 220 d ² Difference between ranks (d) Rank No. of services. Rank Settlement Population
- 19. You Complete stages 3-7. <ul><li>R s = 1- 6 Σ d ² n ³-n </li></ul><ul><li>= 1- 6x12. </li></ul><ul><li> 12³-12 </li></ul><ul><li>=+0.96 (a strong positive correlation) </li></ul>
- 20. Stage 4 and 5. <ul><li>Stage 4 Rejection level = 95% </li></ul><ul><li> = 0.05. </li></ul><ul><li>Stage 5. T= R s n-2 </li></ul><ul><li> 1- R s ² </li></ul><ul><li>= 0.96 12-2 </li></ul><ul><li> 1-0.96² </li></ul><ul><li>= 10.73 </li></ul>
- 21. Stage 6 and 7 <ul><li>Stage 6 </li></ul><ul><li>Df= (n-2) = (12-2) = 10. </li></ul><ul><li>Stage 7 df = 10 </li></ul><ul><li> Rejection value = 0.05 </li></ul><ul><li> Therefore critical value of t =2.23. </li></ul><ul><li>The critical value is less than our t- value (10.73). We can therefore conclude that there is a significant correlation between settlement size and the number of services offered in each. </li></ul>

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