Presentation on theme: "The Chi-Square Test for Association"— Presentation transcript:
1 The Chi-Square Test for Association Math 137Fall‘13L. Burger
2 -The Chi Square TestA statistical method used to determine goodness of fitGoodness of fit refers to how close the observed data are to those predicted from a hypothesisNote:The chi square test does not prove that a hypothesis is correctIt evaluates to what extent the data and the hypothesis have a good fit
3 -Determine The Hypothesis: Whether There is an Association or Not Ho : The two variables are independent (not matching up well!)Ha : The two variables are associated (variables matching good --- called a ‘good fit!’)
4 Understanding the Chi-Square Distribution You see that there is a range here: if the results were perfect you get a chi-square value of 0 (because obs = exp). This rarely happens: most experiments give a small chi-square value (the hump in the graph).Note that all the values are greater than 0: that's because we squared the (obs - exp) term: squaring always gives a non-negative number.Sometimes you get really wild results, with obs very different from exp: the long tail on the graph. Really odd things occasionally do happen by chance alone (for instance, you might win the lottery).
5 Critical Chi-SquareCritical values for chi-square are found on tables, sorted by degrees of freedom and probability levels. Be sure to use p = 0.05.If your calculated chi-square value is greater than the critical value from the table, you “reject the null hypothesis”.If your chi-square value is less than the critical value, you “fail to reject” the null hypothesis (that is, you accept that your theory about the expected ratio is correct).
6 -Calculating Test Statistics Contrasts observed frequencies in each cell of a contingency table with expected frequencies.The expected frequencies represent the number of cases that would be found in each cell if the null hypothesis were true ( i.e. the nominal variables are unrelated).Expected frequency of two unrelated events is product of the row and column frequency divided by number of cases.Fe= Fr Fc / NMean difference between pairs of values
7 -Calculating Test Statistics Mean difference between pairs of values
8 4. Calculating Test Statistics ObservedfrequenciesExpected frequencyMean difference between pairs of valuesExpected frequency
9 -Determine Degrees of Freedom df = (R-1)(C-1)Number of levels in column variableNumber of levels in row variable
10 -Compare computed test statistic against a tabled/critical value The computed value of the Pearson chi- square statistic is compared with the critical value to determine if the computed value is improbableThe critical tabled values are based on sampling distributions of the Pearson chi-square statisticIf calculated 2 is greater than 2 table value, reject Ho
11 Example 1: One-dimensional Suppose we want to know how people in a particular area will vote in general and go around asking them.How will we go about seeing what’s really going on?RepublicanDemocratOther203010
12 -Determine Hypotheses Ho : There is no difference between what is observed and what is expected.𝐻 𝑎 : There is an association between the observed and expected frequencies.Solution: chi-square analysis to determine if our outcome is different from what would be expected if there was no preference
13 Plug in to formula 20 30 10 Observed Expected Republican Democrat OtherObserved203010Expected
14 Reject H0The district will probably vote democratic, there is association between what is observed and what is expected.However…
15 Conclusion of Example 1Note that all we really can conclude is that our data is different from the expected outcome given a situationAlthough it would appear that the district will vote democratic, really we can only conclude they were not responding by chanceRegardless of the position of the frequencies we’d have come up with the same resultIn other words, it is a non-directional test regardless of the prediction
16 Example 2-Two Dimensional Suppose a researcher is interested in voting preferences on gun control issues.A questionnaire was developed and sent to a random sample of 90 voters.The researcher also collects information about the political party membership of the sample of 90 respondents.
17 Bivariate Frequency Table or Contingency Table FavorNeutralOpposef rowDemocrat103050Republican1540f column25n = 90
18 Bivariate Frequency Table or Contingency Table FavorNeutralOpposef rowDemocrat103050Republican1540f column25n = 90Observedfrequencies
19 Bivariate Frequency Table or Contingency Table Row frequencyBivariate Frequency Table or Contingency TableFavorNeutralOpposef rowDemocrat103050Republican1540f column25n = 90
20 Bivariate Frequency Table or Contingency Table FavorNeutralOpposef rowDemocrat103050Republican1540f column25n = 90Column frequency
21 -Determine The Hypothesis Ho : There is no difference between D & R in their opinion on gun control issue.Ha : There is an association between responses to the gun control survey and the party membership in the population.
27 -Compare computed test statistic against a tabled/critical value α = 0.05df = 2Critical tabled value = 5.991Test statistic, 11.03, exceeds critical valueNull hypothesis is rejected
28 -Conclusion of Example 2 Ho : There is no difference between D & R in their opinion on gun control issue.Ha : There is an association between responses to the gun control survey and the party membership in the population.-Democrats & Republicans differ significantly in their opinions on gun control issues