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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 H0 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: Category Less than 12 years 12 years More than 12 years Frequency (observed) 29 20 11

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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 Category Less than 12 years 12 years More than 12 years Frequency (observed) 29 20 11 Expected

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**GOF Hypothesis Test Step 5 Step 6 Reject Ho**

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 Educ 25 to 34 35 to 54 55+ Total No HS 5,325 9,152 16,035 30,512 HS 14,061 24,070 18,320 56,451 C. 1-3 11,659 19,926 9,662 41,247 C. 4+ 10,342 19,878 8,005 38,225 41,388 73,028 52,022 166,438

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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**

HO: Sex and political party affiliation are independent (have no relationship). HA: 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 Democrat Republican Independent Total Male 36 45 24 105 Female 48**

33 16 97 84 78 40 202

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Expected Counts If HO 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 Cell – Male and Independent

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**Two-Way Table of Expected Counts**

Democrat Republican Independent Total obs exp 1 36 43.66 45 40.54 24 20.79 105 2 48 40.34 33 37.46 16 19.21 97 84 78 40 202

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**Expected Counts Expected cell count is close to observed cell count**

Evidence Ho is true Expected cell count is far from observed cell count Evidence Ho 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 H0 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 χ2 = 4.85**

As long as the assumptions are met χ2 will have a χ2 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(χ2 > 4.85)= 0.09**

Decision: Since p-value > α = 0.05, we will Do Not Reject HO. 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 B Mall C Total Will Participate 52 45 36 133 Will not participate 40 47 56 143 92 276

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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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Copyright © 2012 Pearson Education. All rights reserved. 15-1 Copyright © 2012 Pearson Education. All rights reserved. Chapter 15 Inference for Counts:

Copyright © 2012 Pearson Education. All rights reserved. 15-1 Copyright © 2012 Pearson Education. All rights reserved. Chapter 15 Inference for Counts:

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