Multiple-Regression-Chi-Square-Test.pptx

About Chi-square test formulad

1. MULTIPLE REGRESSION The multiple regressions are used in predictions. The dependent variable can be predicted given several independent variables.

2. The statistical formula to be used is y = + + …+   Where: y = the dependent variable to be predicted = the known independent variables that may influence y = the numerical constants which must be determined from observe data  

3. For instance, there are two independent variables x 1 and x 2 and we want to fit the equation, it will become y = + we must solve the three normal equations:   ∑y = n + ∑ ∑ y =∑x 1 ∑ + ∑ ∑ y = ∑ + ∑  

4. The following are the data on the ages and incomes of a random sample of 5 executives working for in RVA Corporation and their academic achievements while in college. Income Age Academic (In thousand pesos) Achievement 81.7 38 1.50 73.3 30 2.00 89.5 46 1.75 79.0 40 1.75 69.9 32 2.50 A. Fit an equation of the form y = to the equation of the given data B. Use the equation obtained in (a) to estimate the average income of a 35-year-old executive with 1.25 academic achievements. Income Age Academic (In thousand pesos) Achievement

5. Complete the table. Y ∑y = ∑ = ∑ = ∑ = ∑ = ∑ = ∑ = ∑ Y ∑y =

6. Substitute the values of all the variables based on the table above in the equation… that is ∑y =n + ∑ ∑ y = ∑x 1 ∑ + ∑ ∑ y = ∑ + ∑  

7. The result is … 393.4 = 5 + 186 + 9.5 14817.4 = 186 + 7084 + 347.5 739.58 = 9.5 347.5 + 18.62 Solve for the value of using the 3 equations…  

8. Step 1: Eliminate using equation 1 and 2. Look at their numerical coefficient. Divide 186 by 5 and the quotient is 37.2. if you multiply 37.2 by 5 the product is 186.so you can eliminate by subtraction. That is …  

9. Step 2. Multiply equation 1 by 1.9. Subtract the equation 1 and equation 3. The result is equation 5.

10. Step 3. Eliminate b 1 using equation 4 and equation 5 by subtraction…multiply equation 5 by the numerical coefficient of b 1 in equation 4. Likewise multiply equation 4 by the numerical coefficient of b 1 in equation 5. It will become…

11. Solve for the value of b1 used either equation 4 or 5, used equation 4 which is…substitute b 2 by its value -3.71, the result is…

12. Step 4. Compute for using either equation 1, 2 or 3… Using equation 1: Replace using its value…the result is…  

13. CHI-SQUARE TEST THIS IS A TEST OF DIFFERENCE BETWEEN THE OBSERVED AND EXPECTED FREQUENCIES. THE CHI-SQUARE TEST IS CONSIDERED A UNIQUE TEST DUE TO ITS 3 FUNCTIONS WHICH ARE AS FOLLOWS: THE TEST OF GOODNESS-OF-FIT THE TEST OF HOMOGENEITY THE TEST OF INDEPENDENCE

14. TEST OF GOODNESS-OF-FIT THIS IS A TEST OF DIFFERENCE BETWEEN THE OBSERVED AND EXPECTED FREQUENCIES. FORMULA Where: = the chi-square test O= observed frequencies E = expected frequencies  

15. Example 1: The theory of Mendel regarding crossing of peas is in the ratio 9:3:3:1, meaning 9 peas are smooth yellow, 3 parts smooth green, 3 parts wrinkled yellow and 1 part wrinkled green. The researcher conducted an experiment and the result was that out of 560 peas, 310 were smooth yellow, 100 were wrinkled yellow, 110 smooth green, and 40 wrinkled green. Is there a significant difference between the observed and the expected?

16. Solve using stepwise method Problem: Is there a significant difference between the observed (actual experiment) and the expected (theory) frequencies? II. Hypotheses: Ho: There is no significant difference between the observed and the expected frequencies. H1: There is a significant difference between the observed and the expected frequencies.

17. Solve using stepwise method III. Level of Significance: α = 0.05 df = h-1 = 4-1 = 3 = 7.815 (TABULAR VALUE)  

18. Solve using stepwise method IV. Statistics: Chi-square Goodness-of-fit Computation: Add the ratio 9:3:3:1 = 16 ATTRIBUTES RATIO ACTUAL RESULT (OBSERVED) THEORY (EXPECTED) SMOOTH YELLOW 9 310 315 WRINKLED YELLOW 3 100 105 SMOOTH GREEN 3 110 105 WRINKLED GREEN 1 40 35 TOTAL 16 560 560

19. Solve using stepwise method IV. Statistics: Chi-square Goodness-of-fit Computation: Add the ratio 9:3:3:1 = 16 For the expected frequencies Total/ total ratio = 560 /16 = 35 35 * 9 = 315 35 * 3 = 105 35 * 3 = 105 35 * 1 = 35

20. Solve using stepwise method IV. Statistics: Chi-square Goodness-of-fit = + + + = 0.079 + 0.238 +0.238 + 0.714 = 1.269 (computed value)  

21. Solve using stepwise method V. Decision Rule: If the chi-square computed value is greater than the chi-square tabular value, reject Ho. VI. Conclusion: The computed value of 1.269 is lesser than the tabular value of 7.815 at 0.05 level of significance with 3 degrees of freedom, so the null hypothesis is accepted. This means that there is no significant difference between the observed and expected frequencies.

22. Activity: A certain machine is supposed to mix peanuts, hazelnuts, cashews, and pecans in the ratio 4:3:2:1. A can containing 500 of these mixed nuts was found to have 270 peanuts, 110 hazelnuts, 80 cashews, and 40 pecans. At 0.05 level f significance, test the hypothesis that the machine is mixing the nuts at the ratio of 4:3:2:1.

23. Chi-Square test of Homogeneity This test is concerned with two or more samples. With only one criterion variable. This test is used to determine if two or more populations are homogeneous. Its data distribution are similar with respect to a particular criterion variable.

24. Chi-Square test of Homogeneity

25. Chi-Square test of Homogeneity

26. Chi-Square test of Homogeneity

27. Chi-Square test of Homogeneity II. Hypotheses: Ho: There is no significant difference between the attitudes of the two political parties on the issue of peace and order in Mindanao. H1: There is a significant difference between the attitudes of the two political parties on the issue of peace and order in Mindanao.

28. Chi-Square test of Homogeneity

29. Chi-Square test of Homogeneity

30. Chi-Square test of Homogeneity