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Quantitative Methods Part 3 Chi - Squared Statistic

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Recap on T-Statistic It used the mean and standard error of a population sample The data is on an “interval” or scale Mean and standard error are the parameters This approach is known as parametric Another approach is non-parametric testing

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Introduction to Chi-Squared It does not use the mean and standard error of a population sample Each respondent can only choose one category (unlike scale in T-Statistic) The expected frequency must be greater than 5 for the test to succeed. If any of the categories have less than 5 for the expected frequency, then you need to increase your sample size

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Example using Chi-Squared “Is there a preference amongst the UW student population for a particular web browser? “ (Dr C Price’s Data) ◦ They could only indicate one choice ◦ These are the observed frequencies responses from the sample FirefoxIExplorerSafariChromeOpera Observed frequencies 306482

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Was it just chance? How confident am I? ◦ Was the sample representative of all UW students? ◦ Was it just chance? Chi-Squared test for significance ◦ Some variations on test ◦ Simplest is Null Hypothesis : The students show “no preference” for a particular browser

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Chi-Squared: “Goodness of fit” (No preference) : The students show no preference for a particular browser This leads to Hypothetical or Expected distribution of frequency ◦ We would expect an equal number of respondents per category ◦ We had 50 respondents and 5 categories FirefoxIExplorerSafariChromeOpera Expected frequencies 10 Expected frequency table

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Stage1: Formulation of Hypothesis : There is no preference in the underlying population for the factor suggested. : There is a preference in the underlying population for the factors suggested. The basis of the chi-squared test is to compare the observed frequencies against the expected frequencies

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Stage 2: Expected Distribution As our “null- hypothesis” is no preference, we need to work out the expected frequency: ◦ You would expect each category to have the same amount of respondents ◦ Show this in “Expected frequency” table ◦ Has to have more than 5 to be valid FirefoxIExplorerSafariChromeOpera Expected frequencies 10

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Stage 3a: Level of confidence Choose the level of confidence (often 0.05) ◦ 0.05 means that there is 5% chance that conclusion is chance ◦ 95% chance that our conclusions are certain Stage 3b: Degree of freedom We need to find the degree of freedom This is calculated with the number of categories ◦ We had 5 categories, df = 5-1 (4)

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Stage 3: Critical value of Chi-Squared In order to compare our calculated chi- square value with the “critical value” in the chi-squared table we need: ◦ Level of confidence (0.05) ◦ Degree of freedom (4) Our critical value from the table = 9.49

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Stage 4: Calculate statistics We compare the observed against the expected for each category We square each one We add all of them up FirefoxIExplorerSafariChromeOpera Observed306482 Expected10 = 52

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Stage 5: Decision Can we reject the That students show no preference for a particular browser? ◦ Our value of 52 is way beyond 9.49. We are 95% confident the value did not occur by chance So yes we can safely reject the null hypothesis Which browser do they prefer? ◦ Firefox as it is way above expected frequency of 10

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Chi-Squared: “No Difference from a Comparison Population”. RQ: Are drivers of high performance cars more likely to be involved in accidents? ◦ Sample n = 50 and Market Research data of proportion of people driving these categories ◦ Once null hypothesis of expected frequency has been done, the analysis is the same as no preference calculation High Performance CompactMidsizeFull size FOFO 201499 MR% 10%40%30%20% FEFE 5 (10% of 50)201510

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Chi-Squared test for “Independence”. What makes computer games fun? Review found the following ◦ Factors (Mastery, Challenge and Fantasy) ◦ Different opinion depending on gender Research sample of 50 males and 50 females MasteryChallengeFantasy Male 10328 Female 24818 Observed frequency table

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What is the research question? A single sample with individuals measured on 2 variables ◦ RQ: ”Is there a relationship between fun factor and gender?” ◦ H O : “There is no such relationship” Two separate samples representing 2 populations (male and female) ◦ RQ: ““Do male and female players have different preferences for fun factors?” ◦ H O : “Male and female players do not have different preferences”

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Chi-Squared analysis for “Independence”. Establish the null hypothesis (previous slide) Determine the critical value of chi-squared dependent on the confidence limit (0.05) and the degrees of freedom. ◦ df = (R – 1)*(C – 1) = 1 * 2 = 2 (R=2, C=3) Look up in chi-squared table ◦ Chi-squared value = 5.99 MasteryChallengeFantasy Male 10328 Female 24818

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Chi-Squared analysis for “Independence”. Calculate the expected frequencies ◦ Add each column and divide by types (in this case 2) ◦ Easier if you have equal number for each gender (if not come and see me) MasteryChallengeFantasyRespondents Male (FO) 1032850 Female (FO) 2481850 Cat total 344026 Male (FE) 172013 Female (FE) 172013

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Chi-Squared analysis for “Independence”. Calculate the statistics using the chi-squared formula ◦ Ensure you include both male and female data MasteryChallengeFantasy Male (FO) 10328 Female (FO) 24818 Male (FE) 172013 Female (FE) 172013

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Stage 5: Decision Can we reject the null hypothesis? ◦ Our value of 24.01 is way beyond 5.99. We are 95% confident the value did not occur by chance Conclusion: We are 95% confident that there is a relationship between gender and fun factor But else can we get from this? ◦ Significant fun factor for males = Challenge ◦ Significant fun factor for females = Mastery and Fantasy MasteryChallengeFantasy Male (FO) 10328 Female (FO) 24818 Male (FE) 172013 Female (FE) 172013

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Workshop Work on Workshop 7 activities Your journal (Homework) Your Literature Review (Complete/update) References Dr C. Price’s notes 2010 Gravetter, F. and Wallnau, L. (2003) Statistics for the Behavioral Sciences, New York: West Publishing Company

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