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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

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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 Observed 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 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 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 Female 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 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) Female (FO) Cat total Male (FE) Female (FE)

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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) Female (FO) Male (FE) Female (FE)

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Stage 5: Decision Can we reject the null hypothesis? ◦ Our value of is way beyond 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) Female (FO) Male (FE) Female (FE)

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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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