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Chapter 7

Chapter 7. Calculation of Pearson Coefficient of Correlation, r and testing its significance. SS xx = Σ x 2 – ( Σ x) 2 n. SS xy = Σ xy – ( Σ x) ( Σ y) n. From previous lecture:. b = SS xy and a = y - b x

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Chapter 7

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  1. Chapter 7 Calculation of Pearson Coefficient of Correlation, r and testing its significance

  2. SSxx = Σx2 – (Σx)2 n SSxy = Σxy – (Σx) (Σy) n • From previous lecture: b = SSxy and a = y - b x SSxx Today’s lecture: we are going to calculate the correlation coefficient of the two variables, x and y, called the Pearson Product Moment Correlation Coefficient, r The values of SSxy, SSxx, SSyy can also be obtained by using the following basic formulas: SSxy = Σ(x – x)(y – y) SSxx = Σ (x – x)2 SSyy = Σ (y – y)2 But these formulas take longer to make calculations since you have to calculate The means x and y NOTE: x and y are denoted as means for this course only. The line should appear on top of the letters x and y.

  3. Pearson Product Moment Correlation Coefficient, r • r measures the strength of the relationship between two variables: x and y Examples of different strengths of relationships between variables x and y: Strong positive correlation Weak positive correlation Weak negative correlation Strong negative correlation

  4. What is the correlation coefficient of the scatterplot below? The value of r ranges from -1 to +1.

  5. Pearson Product Moment Coefficient of Correlation, r is given by: SSxy r = SSxx SSyy Example: Calculate the Pearson Product Moment Coefficient of Correlation, r to show the relationship between Maths and Science marks for Form 5A: Maths Science 35 9 49 15 21 7 39 11 15 5 28 8 25 9

  6. STEP 1: Calculate Σx, Σy, Σxy, Σx2 Maths, x Science, y xy x2 y2 35 9 315 1225 81 49 15 735 2401 225 21 7 147 441 49 39 11 429 1521 121 15 5 75 225 25 28 8 224 784 64 25 9 225 625 81 Σx = 212 Σy = 64 Σxy = 2150 Σx2 = 7222 Σy2 = 646 STEP 2: Calcute SSxy, SSxx and SSyy SSxy = Σxy – (Σx) (Σy) n = 2150 – (212)(64) /7 = 211.7143 SSxx = Σx2 – (Σx)2 n = 7222 – (212)2 / 7 = 801.4286 SSyy = Σy2 – (Σy)2 n = 646 – (64)2 / 7 = 60.8571

  7. STEP 3: Substitute inside the r formula: SSxy 211.7143 r = = = .96 SSxx SSyy (801.4286) (60.8571) The linear correlation coefficient is .96 (rounded to 2 decimal places) Interpretation: Maths and Science marks are strongly correlated. The square of the correlation, called the coefficient of determination, r2 = (.96)2 = .96 indicates that Maths marks account for 96% of the variance of the Science marks in this case.

  8. STEP 4: Test the significance of r obtained by stating the null hypothesis that there are no significant relationship between Maths and Science scores. To test the significance of the r value obtained, you will first need to set the level of significance you wish to test, say at 1% or at p < .01. You can test the hypothesis about the population correlation coefficient ρ using the sample correlation coefficient, r. We can use the t distribution to make this test. n - 2 t = r 1 – r2 Where n – 2 are the degrees of freedom.

  9. The null hypothesis is that the linear correlation coefficient between 2 variables is zero, that is ρ = 0. The alternative hypothesis can be: linear correlation coefficient between the 2 variables is less than zero, ρ < 0 linear correlation coefficient between the 2 variables is more than zero, ρ > 0 linear correlation coefficient between the 2 variables is not equal to zero, ρ≠ 0 State the null hypothesis: (ρ is the population correlation coefficient) Ho: ρ = 0 (The linear correlation coefficient is zero in the population) H1 : ρ > 0 (The linear correlation coefficient is positive in the population)  means One-tailed (We test H1: the positive correlation coefficient only when it is impossible for the correlation to be negative) (Otherwise we have to test H1: ρ≠ 0, when we wish to test for correlations both positive or negative  two-tailed test) STEP 5: Select the distribution to use. The population distribution for both variables are normally distributed. Hence, we can use the t distribution to perform this test about the linear correlation coefficient STEP 6: Determine the rejection and nonrejection regions

  10. STEP 6: Determine the rejection and nonrejection regions The significance level you have chosen for this test is 1%. From the alternative hypothesis, we know that the test is right-tailed. Hence Area in the right tail of the t distribution = .01 df = n – 2 = 7 – 2 = 5 From the t distribution table, the critical value of t is 3.365. The rejection and nonrejection regions for this test are as shown below: Do not Reject Ho Reject Ho 3.365 Critical Value of t

  11. STEP 7: Calculate the value of the test statistic, t n - 2 t = r 1 – r2 7 - 2 t = .96 = 7.667 1 – (.96)2 STEP 8: Make a decision The value of the test statistic t = 7.667 is greater than the critical value of t = 3.365 and it falls in the rejection region. Hence, we reject the null hypothesis and conclude that there is a significant, positive linear relationship between Maths and Science marks

  12. Hypothesis • A hypothesis is a specific statement about on aspect of the population e.g. its mean, or its variance. • A null hypothesis is a specific statement that indicates that something has a “no effect” or “no difference” between two situations. Eg. There is no effect of the treatment on students’ motivation • Or There are no gender differences in Mathematics scores.

  13. Alternative Hypothesis • An alternative hypothesis is the opposite of the null hypothesis. Eg. There is a relationship between academic achievement and motivation  a two-tailed hypothesis • A one-tail hypothesis only tests on one direction. Eg, There boys are better in Mathematics than girls

  14. A hypothesis is a statement about the POPULATION and not the sample. You cannot write a hypothesis as: Ho: This is not correct since can be measured accurately. We need an hypothesis to estimate the population mean .

  15. Hypothesis Testing • 1. State the null and alternative hypothesis • 2. Select the distribution to use • 3. Determine the rejection and nonrejection regions • 4. Calculate the value of the test statistic • 5. Make a decision

  16. Hypothesis Testing – Example using Correlation • Step 1. State the null and alternative hypothesis Ho: ρ = 0 (The linear correlation coefficient is zero in the population) H1 : ρ > 0 (The linear correlation coefficient is positive in the population)  means One-tailed Or H1: ρ ≠ 0 (Means two possibilities, ρ > 0 or ρ < 0 => Two tailed test)

  17. STEP 2. Select the distribution to use The population distribution for both variables are normally distributed. Hence, we can use the t distribution to perform this test about the linear correlation coefficient

  18. STEP 3: Determine the rejection and nonrejection regions The significance level you have chosen for this test is 1%. From the alternative hypothesis, we know that the test is right-tailed. Hence Area in the right tail of the t distribution = .01 df = n – 2 = 7 – 2 = 5 From the t distribution table, the critical value of t is 3.365. The rejection and nonrejection regions for this test are as shown below: Do not Reject Ho Reject Ho 3.365 Critical Value of t

  19. STEP 4: Calculate the value of the test statistic, t n - 2 t = r 1 – r2 7 - 2 t = .96 = 7.667 1 – (.96)2 STEP 5: Make a decision The value of the test statistic t = 7.667 is greater than the critical value of t = 3.365 and it falls in the rejection region. Hence, we reject the null hypothesis and conclude that there is a significant, positive linear relationship between Maths and Science marks

  20. Another way of calculating r – using the standard score method

  21. Do not Reject HO Reject HO r critical = -.878 R obtained = -.90 • Decision: Sig at p < .05 • Significantly negative relationship between Class Size and Achievement Test (r = -.90, p < .05)

  22. Another method of calculating r – using the computational formula

  23. Exercise 1 • Explain the following concept. You may use graphs to illustrate each concept • a) Perfect positive linear correlation • b) Perfect negative linear correlation • c) Strong positive linear correlation • d) Strong negative linear correlation • e) Weak positive linear correlation • f) Weak negative linear correlation • g) No linear correlation • 2) For a sample data set, the linear correlation coefficient r has a positive value. • Which of the following is true about the slope b of the regression line estimated • for the same sample data? • a) The value of b will be positive • b) The value of b will be negative • c) The value of b can be positive or negative • 3) A population data set produced the following information. • N = 250, Σx = 9880, Σy = 1456, Σxy = 85,080 • Σx2 = 485,870 and Σy2 = 135,675 • Find the linear correlation coefficient ρ. Ans: 0.25

  24. 4) A sample data set produced the following information. N = 10, Σx = 100, Σy = 220, Σxy = 3680 Σx2 = 1140 and Σy2 = 25,272 a) Find the linear correlation coefficient r. b) Using the 5% significance level, can you conclude the ρ is different from zero? 5) A sample data set produced the following information. N = 12, Σx = 66, Σy = 588, Σxy = 2244 Σx2 = 396 and Σy2 = 58734 a) Find the linear correlation coefficient r. b) Using the 5% significance level, can you conclude the ρ is negative?

  25. 6) The data on ages (in years) and prices (in hundred of dollars for eight • cars of a specific model are shown below: • Age 8 3 6 9 2 5 6 3 • Prices 18 94 50 21 145 42 36 99 • Do you expect the ages and prices of cars to be positively or negatively • related? Explain. • b) Calculate the linear correlation coefficient. • c) Test at the 5% significance level whether ρ is negative. • 7) The following table lists the midterm and final term exam scores for 7 students • in a statistics class. • Midterm score 79 95 81 66 87 94 59 • Final Exam score 85 97 78 76 94 84 67 • Do you expect the midterm and final exam scores to be positively • or negatively correlated? • b) Plot a scatter diagram. By looking at the scatter diagram, do you expect the • correlation coefficient between these 2 variables to be close to zero, 1, or -1. • c) Find the correlation coefficient. Is the value of r consistent with what you expected • in parts a and b? • d) Using the 1% significance level, test whether the linear correlation coefficient is • Positive.

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