Presentation on theme: "Chapter 11 Other Chi-Squared Tests. Chi-Square Statistic Measures how far the observed values are from the expected values Take sum over all cells in."— Presentation transcript:
Chapter 11 Other Chi-Squared Tests
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.
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.
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)
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.
GOF Hypothesis Test Step 2 –α = 0.05 Step 3
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 = CategoryLess than 12 years 12 yearsMore than 12 years Frequency (observed) Expected20
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.
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
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 ,65919,926 9,66241,247 C. 4+10,34219,878 8,00538,225 Total41,38873,02852,022166,438
Independence of Categorical Variables. Is there a relationship between two categorical variables. Are two variables related to each other? Unrelated = independent Related = dependent
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?
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.)
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.
Data DemocratRepublicanIndependentTotal Male Female Total
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
Example of Expected Counts Cell – Male and Democrat –Expected Count = Cell – Male and Republican –Expected Count = Cell – Male and Independent –Expected Count =
Two-Way Table of Expected Counts DemocratRepublicanIndependentTotal obsexpobsexpobsexp Total
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
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
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.
Chi-Square Statistic Cell – Male and Democrat –Observed Count = 36 –Expected Count = Cell – Male and Republican –Observed Count = 45 –Expected Count = 40.54
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
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.
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
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.
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 Will not partici pate Total92 276
Homogeneity of Proportions Step 4 –Put into your calculator Observed in matrix A Expected in matrix B –Test statistic = –P-value = 0.06
Homogeneity of Proportions Step 5 –Do Not Reject Ho Step 6 –There is not sufficient evidence to suggest that at least one is different.