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Published byCarlos Wolfe Modified about 1 year ago

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Chi square

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Non-parametric test that’s useful when your sample violates the assumptions about normality required by other tests ◦ All other tests we’ve discussed require interval or ratio outcomes, but sometimes you have categories (yes/no) ◦ Also chi-square does not require a normal distribution (only present in large samples) ◦ Or equal variance between groups

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Uses sample data to evaluate hypotheses about proportions of relationships that exist within populations Is the number of y equally distributed in the population? Test between what is observed and what is expected by chance

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X 2 = Σ X 2 = chi square O = observed frequency E = expected frequency (O-E) 2 E

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1. State hypotheses ◦ Null hypothesis: there is no difference in the frequency of occurrences of preference for Coke versus Pepsi P Coke = P Pepsi ◦ Research hypothesis: there is a difference in the frequency of occurrences of preference for Coke versus Pepsi P Coke ≠ P Pepsi

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6. Determine whether the statistic exceeds the critical value ◦ 10 > 3.84 & 6.64 ◦ So it does exceed the critical value 7. If over the critical value, reject the null & conclude that there is a difference in the proportion of people that prefer Coke versus Pepsi

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In results ◦ Significantly more participants reported a preference for Coke (75%) versus Pepsi (25%), Χ 2 (1) = 10.00, p <.05 (can say p <.01).

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