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Four exercises on using the chi-square test to analyze contingency tables and determine if there is independence between two categorical variables. The exercises involve testing hypotheses, finding test statistics, rejection regions, and interpreting p-values for various scenarios.
Typology: Exams
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Contingency table
Estimated cell count: Eˆij =
ricj n
, where ri: the total for row i and cj : the total for column
j.
Test statistic: X^2 = Σ
(Oij − Eˆij )^2 E^ ˆij
, Oij : observed cell count
df = (r − 1)(c − 1), where r is the number of rows and c is the number of columns
Columns
Rows 1 2 3 Total
1 37 34 93 164
2 66 57 113 236
Total 103 91 206 400
(a) If you wish to test the null hypothesis of ”independence” - that the probability that a response falls in any one row is independent of column it falls in - and you plan to use a chi-square test, how many degrees of freedom will be associated with the χ^2 statistic?
(b) Find the value of the test statistic.
(c) Find the rejection region for α =. 01
(d) Conduct the test and state your conclusions.
(e) Find the approximate p− value for the test and interpret its value.
Treated Untreated
Improved 117 74
Not Improved 83 126
that the serum was effective in improving the condition of arthritic patient. Use the chi-square test of homogeneity to compare the proportions improved in the populations of treated and untreated subjects. Test at the 5% level of significance.
Number of Relationships
Three or Fewer Four or Five Six or More
Cold 49 43 34 No Cold 31 47 62
Total 80 90 96
by the number of relationships you have? Test at the 5% significance level.