Presentation is loading. Please wait.

Presentation is loading. Please wait.

S519: Evaluation of Information Systems

Similar presentations


Presentation on theme: "S519: Evaluation of Information Systems"— Presentation transcript:

1 S519: Evaluation of Information Systems
Social Statistics Inferential Statistics Chapter 13: correlation coefficient

2 This week Testing correlation coefficient The interpretation PEARSON
SPEARMAN Using Excel and SPSS to calculate correlation coefficient

3 Which test to use Figure 13.1
The relationship between variables, and not the difference between groups, is being examined. Only two variables are being used The appropriate test statistic to use is the t test for the correlation coefficient

4 Example Quality of Marriage Quality of parent-child relationship 76 43
81 33 78 23 34 31 51 56 98 44 88 45 32 66 28 67 39 65 59 38 87 21 77 27 79 85 46 68 41 48 99 55 54

5 Correlation coefficient
CORREL() and PEARSON() Same value There is no significant difference Spearman’s rank correlation coefficient Kendall's tau

6 T test for the significance of the correlation coefficient
Step1: A statement of the null and research hypotheses Null hypothesis: there is no relationship between the quality of the marriage and the quality of the relationship between parents and children Research hypothesis: (two-tailed, nondirectional) there is a relationship between the two variables

7 T test for the significance of the correlation coefficient
Step2: setting the level of risk (or the level of significance or Type I error) associated with the null hypothesis 0.05 or 0.01 What does it mean? on any test of the null hypothesis, there is a 5% (1%) chance you will reject it when the null is true when there is no group difference at all. Why not ? So rigorous in your rejection of false null hypothesis that you may miss a true one; such stringent Type I error rate allows for little leeway

8 T test for the significance of the correlation coefficient
Step 3 and 4: select the appropriate test statistics The relationship between variables, and not the difference between groups, is being examined. Only two variables are being used The appropriate test statistic to use is the t test for the correlation coefficient

9 T test for the significance of the correlation coefficient
Step5: determination of the value needed for rejection of the null hypothesis using the appropriate table of critical values for the particular statistic. Table B4 compute the correlation coefficient (r=0.393) Compute df=n-2 (df=27) If obtained value>the critical value reject null hypothesis If obtained value<the critical value accept null hypothesis

10 T test for the significance of the correlation coefficient
Step6: compare the obtained value with the critical value obtained value: 0.393 critical value: 0.349

11 T test for the significance of the correlation coefficient
Step 7 and 8: make decisions What could be your decision? And why, how to interpret? obtained value: > critical value: (level of significance: 0.05) Coefficient of determination is 0.154, indicating that 15.4% of the variance is accounted for and 84.6% of the variance is not. There is a 5% chance that the two variables are not related at all

12 A simplified solution Using statistical software packages as SPSS, we do not need to determine level of risk or critical value. Instead, the exact probability will be calculated. Moreover, numbers will be automatically highlighted if they are bigger than corresponding critical values.

13 Causes and associations
Two variables are related to each other One causes another having a great marriage cannot ensure that the parent-child relationship will be of a high quality as well; The two variables maybe correlated because they share some traits that might make a person a good husband or wife and also a good parent; It’s possible that someone can be a good husband or wife but have a terrible relationship with his/her children.

14 A critique a correlation can be taken as evidence for a possible causal relationship, but cannot indicate what the causal relationship, if any, might be. These examples indicate that the correlation coefficient, as a summary statistic, cannot replace the individual examination of the data.

15 Exercises S-P267 1 2 3

16 S-P267-Q1 n degree of freedom correlation coefficient level tail
critical value 20 18 0.567 0.01 one 0.516 80 78 -0.45 0.05 0.183 50 48 0.37 two 0.273

17 S-P267-Q2 Excel SPSS Pearson Spearman Kendal


Download ppt "S519: Evaluation of Information Systems"

Similar presentations


Ads by Google