Introduction to Business Analytics Course Part 9
Download as PPTX, PDF1 like395 views
The document provides an overview of statistical concepts and tests like chi-squared tests and ANOVA for analyzing data in business contexts. It includes case studies demonstrating the application of these tests to determine associations between variables, such as exposure to radiation and health outcomes, and differences in academic performance among students from different schools. The document serves as a guide for understanding the significance of data analysis in making informed business decisions.
Introduction to Business Analytics Course Part 9
- 2. STATISTICAL CONCEPTS AND THEIR APPLICATIONS IN BUSINESS
- 3. TESTS OF SIGNIFICANCE One sample z-test Two sample z-test One sample t-test Two sample t-test Paired t-test Chi – Squaredtest F test - Analysis of Variance (ANOVA) F test - Regression
- 4. CHI- SQUARED TESTS • Compare the observed result against an expected result based on a hypothesis • Steps: • State the null hypothesis • Prepare the contingency table for the variable • Determine the expected results • Calculate the chi-squared value • Calculate the degrees of freedom • Based on the above, calculate the p-value • If p-value < 0.05, reject the null hypothesis. • Test of independence: • Verify if two variables are independent • Same steps as above.
- 5. CASE STUDY—CHI-SQUARED TEST • A city has a newly opened nuclear plant, and there are families staying dangerously close to the plant. A health safety officer wants to take this case up to provide relocation for the families that live in the surrounding area. To make a strong case, he wants to prove with numbers that an exposure to radiation levels is leading to an increase in diseased population. He formulates a contingency table of exposure and disease. • Does the data suggest an association between the disease and exposure? Disease Total Exposure Yes No Yes 37 13 50 No 17 53 70 Total 54 66 120
- 6. Steps: • Calculate the number of individuals of exposed and unexposed groups expected in each disease category (yes and no) if the probabilities were the same. • If there were no effect of exposure, the probabilities should be same and the chi-squared statistic would have a very low value. Proportion of population exposed = (50/120) = 0.42 Proportion of population not exposed = (70/120) = 0.58 Thus, expected values: Population with disease = 54 Exposure Yes : 54 * 0.42 = 22.5 Exposure No : 54 * 0.58 = 31.5 Population without disease = 66 Exposure Yes : 66 * 0.42 = 27.5 Exposure No : 66 * 0.58 =38.5 CASE STUDY—CHI-SQUARED TEST (CONTD.)
- 7. • Calculate the Chi-squared statistic χ2=Σ = 29.1 • Calculate the degrees of freedom : (Number of rows – 1) X (Number of columns – 1) df = (2 – 1) X (2 – 1) =1 • Calculate the p-value from the chi-squared table For chi-squared value 29.1 and degrees of freedom = 1, from the table, p-value is < 0.001 • Interpretation: There is 0.001 chance of obtaining such discrepancies between expected and observed values if there is no association • Conclusion : There is an association between the exposure and disease. CASE STUDY—CHI-SQUARED TEST (CONTD.)
- 8. ANOVA • Analysis of Variance – used to compare more than two groups • Extension of the independent t-tests • Factor variable – variable defining the groups • Response variable – variable being compared • One wayANOVA • Groups of a single variable • E.g. : Is there a difference in student’s scores based on the row he is seated – front/middle/back? • Two wayANOVA • Two independent variables • E.g. : Does the race and gender affect a person’s yearly income?
- 9. • Marks obtained in the same subject by 3 students belonging to three different schools are given below. • Does the data suggest any association between schools and marks? CASE STUDY—ONE WAY ANOVA Basic Idea : Partition the total variation in the data into the variance between groups and variance within groups. School A B C Marks 82 83 38 83 78 59 97 68 55
- 10. Steps: • Calculate the means • School A : mean(82,83,97) =87.3 • School B : mean(83,78,68) =76.3 • School C : mean(38,59,55) =50.6 • Calculate the grand mean • Grand mean = mean(82,83,97,83,78,68,39,59,55) = 71.4 • Calculating the variations • Sum of Squared Deviations about the grand mean, across all observed values : SSTotal = 2630.2 • Sum of Squared Deviations of group mean about the grand mean – three group means against the grand mean : SSBetween = 2124.2 • Sum of Squared Deviations of observations within a group about their group mean; added across all groups : SSWithin = 506 CASE STUDY—ONE WAY ANOVA (CONTD.)
- 11. • Calculate the degrees of freedom for every variance • dfTotal = Number of observations – 1 = 9 -1 =8 • dfBetween = Number of groups -1 = 3-1 = 2 • dfWithin = Number of observations – number of groups = 9-3 = 6 • Calculate the Mean Squared Variances • Mean Squared variance between groups : MSBetween= SSBetween /dfBetween = 2124.2/2 = 1062.1 • Mean Squared variance within groups : MSWithin= SSWithin /dfWithin = 506/6 = 84.3 • Calculate the f-statistic • F-value : MSBetween /MSWithin= 1062.1/84.3 = 12.59 • Calculate the p-value from the F-table • p-value for given f-value 12.59 and degrees of freedom 2 and 6 is 0.007 • Conclusion : Since the p-value is less than alpha, we can conclude by rejecting the null hypothesis, that there is a difference in the marks obtained by students belonging to different groups. CASE STUDY—ONE WAY ANOVA (CONTD.)
- 12. Thank You If you are looking for business analytics training in Bangalore / Bengaluru then visit: http://beamsync.com/business-analytics-training-bangalore/ Next Part We will Publish Soon.