1. Bivariate Data – Scatter Plots and Correlation Coefficient…… Section 3.1 and 3.2

2. 2 Quantitative Variables……  We represent 2 variables that are quantitative by using a scatter plot.  Scatter Plot – a plot of ordered pairs (x,y) of bivariate data on a coordinate axis system. It is a visual or pictoral way to describe the nature of the relationship between 2 variables.

3. Input and Output Variables……  X: a. Input Variable b. Independent Var c. Controlled Var  Y: a. Output Variable b. Dependent Var c. Results from the Controlled variable

4. Example……  When dealing with height and weight, which variable would you use as the input variable and why?  Answer:  Height would be used as the input variable because weight is often predicted based on a person’s height.

5. Constructing a scatter plot……  Do a scatter plot of the following data: IndependentDependent Variable AgeBlood Pressure 43128 48120 56135 61143 67141 70152

6. What do we look for?......  A. Is it a positive correlation, negative correlation, or no correlation?  B. Is it a strong or weak correlation?  C. What is the shape of the graph?

7. Answer……With TI AgeBlood Pressure 43128 48120 56135 61143 67141 70152

8. Notice……  Notice the following:  A. Strong Positive – as x increases, y also increases. B. Linear - it is a graph of a line.

9. Example 2……By Hand IndependentDependent Variable # of AbsencesFinal Grade 682 286 1543 974 1258 590 878

10. Example 2……With TI IndependentDependent Variable # of AbsencesFinal Grade 682 286 1543 974 1258 590 878

11. Notice……  Notice the following: A. Strong Negative – As x increases, y decreases B. Linear – it’s the graph of a line.

12. Example 3……By Hand IndependentDependent Variable Hrs. of ExerciseAmt of Milk 348 08 232 564 810 532 1056 272 148

13. Example 3……With TI IndependentDependent Variable Hrs. of ExerciseAmt of Milk 348 08 232 564 810 532 1056 272 148

14. Notice……  Notice:  There seems to be no correlation between the hours or exercise a person performs and the amount of milk they drink.

15. Steps to see on Calculator……  Put x’s in L1 and y’s in L2  Click on “2 nd y=“  Set scatter plot to look like the screen to the right.  Press zoom 9 or set your own window and then press graph.

16. Linear Correlation Section 3.2

17. Correlation……  Definition – a statistical method used to determine whether a relationship exists between variables.  3 Types of Correlation: A. Positive B. Negative C. No Correlation

18.  Positive Correlation: as x increases, y increases or as x decreases, y decreases.  Negative Correlation: as x increases, y decreases.  No Correlation: there is no relationship between the variables.

19. Linear Correlation Analysis ……  Primary Purpose: to measure the strength of the relationship between the variables.  *This is a test question!!!!

20. Coefficient of Linear Correlation  The numerical measure of the strength and the direction between 2 variables.  This number is called the correlation coefficient.  The symbol used to represent the correlation coefficient is “r.”

21. The range of “r” values……  The range of the correlation coefficient is -1 to +1.  The closer to 0 you get, the weaker the correlation.

22. Range……  Strong Negative No Linear RelationshipStrong Positive ____________________________________ -1 0 +1

23. Computational Formula using z-scores of x and y……

24. Example 1……  Find the correlation coefficient (r) of the following example.  Use the lists in the calculator. xy 280 5 170 490 260

25. Find mean and st. dev first……  Since you will be using a formula that uses z-scores, you will need to know the mean and standard deviation of the x and y values.  Put x’s in L1  Put y’s in L2  Run stat calc one var stats L1 – Write down mean & st. dev.  Run stat calc one var stats L2 – Write down mean & st. dev.

26.  X values:  Y values:

27. Write down on your paper……You’ll use them later.  X Values: Mean = 2.8 St. Dev = 1.643167673  Y Values: Mean = 76 St. Dev = 11.40175425

28. Calculator Lists…… Set Formula L1L2L3 = (L1-2.8)/1.643167673L4 = (L2-76)/11.40175425L5 = L3 x L4 xyz(of x)z (of y)z (of x) times z(of y) 280-0.48690.35082-0.1708 5801.33890.350820.46971 170-1.095-0.52620.57646 4900.73031.22790.89672 260-0.4869-1.4030.68321 2.455298358

29. Calculate “r”……  From the lists…..  n = 5

30. What does that mean?  Since r = 0.61, the correlation is a moderate correlation.  Do we want to make predictions from this?  It depends on how precise the answer needs to be.

31. Example 2……  Find the correlation coefficient (r) for the following data.  Do you remember what we found from the scatter plot? AgeBlood Pressure 43128 48120 56135 61143 67141 70152

32. Let’s do this one together……  Remember to use your lists in the calculator.  Don’t round numbers until your final answer.  Find the mean and st. dev. for x and y.  Explain what you found.

33.  X Values:  Y Values:

34. List values you should have…… L1L2L3L4L5 43128-1.368-0.74581.0205 48120-0.8965-1.4481.2978 n=656135-0.1415-0.13160.01863 611430.330280.570310.18836 671410.896470.394830.35395 701521.17961.361.6042 4.483364073

35. Compute “r”……

36. Describe it……  Since r = 0.897 Strong Positive Correlation

37. Example 3……  Find the correlation coefficient for the following data.  Do you remember what we found from the scatter plot? # of AbsencesFinal Grade 682 286 1543 974 1258 590 878

38.  X Values:  Y Values:

39. List Values you should have…… L1L2L3L4L5 682-0.48980.53626-0.2626 286-1.4040.7746-1.088 15431.5673-1.788-2.802 n=79740.195910.059580.01167 12580.88158-0.8938-0.7879 590-0.71831.0129-0.7276 878-0.03270.29792-0.0097 -5.66529102

40. Compute “r”……

41. Describe it……  Since r = -0.944 Strong Negative Correlation

42. Example 4……  Find the correlation coefficient of the following data.  Do you remember what we found from the scatter plot? Hrs of ExerciseAmt of Milk 348 08 232 564 810 532 1056 272 148

43.  X Values:  Y Values:

44. List Values you should have…… Hrs of ExerciseAmt of MilkL3L4L5 348-0.30150.30713-0.0926 08-1.206-1.4761.7804 232-0.603-0.40620.24495 5640.301511.02050.30768 n=98101.206-1.387-1.673 5320.30151-0.4062-0.1225 10561.80910.663791.2008 272-0.6031.3771-0.8304 148-0.90450.30713-0.2778 0.537689672

45. Compute “r”……

46. Describe It……  Since r =.067 No Correlation…..No correlation exists

47. What is  It is the coefficient of determination.  It is the percentage of the total variation in y which can be explained by the relationship between x and y.  A way to think of it: The value tells you how much your ability to predict is improved by using the regression line compared with NOT using the regression line.

48. For Example……  If it means that 89% of the variation in y can be explained by the relationship between x and y.  It is a good fit.

49. Assignment……  Worksheet