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Chapter 11 Other Chi-Squared Tests

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Chi-Square Statistic Measures how far the observed values are from the expected values Take sum over all cells in table When is large, there is evidence that H 0 is false.

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Goodness of Fit (GOF) Tests We can use a chi-squared test to see if a frequency distribution fits a pattern. The hypotheses to these tests are written a little different than we have seen in the past because they are usually written in words.

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GOF Hypothesis Test A researcher wishes to see of the number of adults who do not have health insurance is equally distributed among three categories: CategoryLess than 12 years 12 yearsMore than 12 years Frequency (observed) 292011

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GOF Hypothesis Test Step 1 –Ho: The number of people who do not have health insurance is equally distributed over the three categories. –Ha: The number of people who do not have health insurance is not equally distributed over the three categories.

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GOF Hypothesis Test Step 2 –α = 0.05 Step 3

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GOF Hypothesis Test Step 4 Test Statistic = 8.1 –Put Observed in L1 and Expected in L2 –L3 = (L1-L2) 2 / L2 –Sum L3 χ2 cdf (test stat, E99, df) = p-value = 0.017 CategoryLess than 12 years 12 yearsMore than 12 years Frequency (observed) 292011 Expected20

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GOF Hypothesis Test Step 5 –Reject Ho Step 6 –There is enough evidence to suggest that the number of people who do not have health insurance is not equally distributed over the three categories.

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Two-Way Tables Summarizes the relationship between two categorical variables Row = values for one categorical variable Column = values for other categorical variable Table entries = number in row by column class

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Example of Two-Way Table Age Group Educ25 to 3435 to 5455+Total No HS 5,325 9,15216,03530,512 HS14,06124,07018,32056,451 C. 1-311,65919,926 9,66241,247 C. 4+10,34219,878 8,00538,225 Total41,38873,02852,022166,438

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Independence of Categorical Variables. Is there a relationship between two categorical variables. Are two variables related to each other? Unrelated = independent Related = dependent

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Example A 1992 poll conducted by the University of Montana classified respondents by sex and political party. Sex: Male and Female Party: Democrat, Republican, Independent Is there evidence of an association between gender and party affiliation?

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Hypothesis Test for Independence H O : Sex and political party affiliation are independent (have no relationship). H A : Sex and political party affiliation are dependent (are related to each other.)

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Two-Way Tables Describe table with # of rows (r) and # of columns (c) –r x c table Each number in table is called a cell –r times c cells in table We will use the two-way table to test our hypotheses.

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Data DemocratRepublicanIndependentTotal Male364524105 Female48331697 Total847840202

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Expected Counts If H O is true, we would expect to get a certain number of counts in each cell. Expected cell count = row total * column total table total

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Example of Expected Counts Cell – Male and Democrat –Expected Count = Cell – Male and Republican –Expected Count = Cell – Male and Independent –Expected Count =

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Two-Way Table of Expected Counts DemocratRepublicanIndependentTotal obsexpobsexpobsexp 1 3643.664540.542420.79 105 2 4840.343337.461619.21 97 Total847840202

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Expected Counts Expected cell count is close to observed cell count –Evidence H o is true Expected cell count is far from observed cell count –Evidence H o is false

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Chi-Squared Test for Independence To test these hypotheses, we will use a Chi- Squared Test for Independence if the assumptions hold. Assumptions: –Expected Cell Counts are all > 5

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Chi-Square Statistic Measures how far the observed values are from the expected values Take sum over all cells in table When is large, there is evidence that H 0 is false. Your calculator will do this for you.

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Chi-Square Statistic Cell – Male and Democrat –Observed Count = 36 –Expected Count = 43.66 Cell – Male and Republican –Observed Count = 45 –Expected Count = 40.54

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Chi-Square Statistic Repeat this process for all 6 cells χ 2 = 4.85 As long as the assumptions are met –χ 2 will have a χ 2 distribution with d.f. (r-1)(c-1) Sex has 2 categories (so r = 2); party has 3 categories (so c = 3) We have (2-1)(3-1) = 2 degrees of freedom

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Hypothesis Test P-value = P(χ 2 > 4.85)= 0.09 Decision: Since p-value > α = 0.05, we will Do Not Reject H O. Conclusion: There is no evidence of a relationship between sex and political party affiliation.

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Homogeneity of Proportions Samples are selected from different populations and a researcher wants to determine whether the proportion of elements are common for each population. The hypotheses are: –Ho: p1 = p2 = p3 = p4 –Ha: At least one is different

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Homogeneity of Proportions An advertising firm has decided to ask 92 customers at each of three local shopping malls if they are willing to take part in a market research survey. According to previous studies, 38% of Americans refuse to take part in such surveys. At α = 0.01, test the claim that the proportions are equal.

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Homogeneity of Proportions Step 1 –Ho: p1 = p2 = p3 –Ha: At least one is different Step 2 –α = 0.01 Step 3 Mall A Mall B Mall C Total Will Partici pate 524536133 Will not partici pate 404756143 Total92 276

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Homogeneity of Proportions Step 4 –Put into your calculator Observed in matrix A Expected in matrix B –Test statistic = 5.602 –P-value = 0.06

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Homogeneity of Proportions Step 5 –Do Not Reject Ho Step 6 –There is not sufficient evidence to suggest that at least one is different.

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