Exam 11: Hypothesis Testing Iv: Chi Square
Exam 1: Introduction53 Questions
Exam 2: Basic Descriptive Statistics: Percentages, Ratios and Rates, Tables, Charts and Graphs53 Questions
Exam 3: Measures of Central Tendency51 Questions
Exam 4: Measures of Dispersion52 Questions
Exam 5: The Normal Curve42 Questions
Exam 6: Introduction to Inferential Statistics: Sampling and the Sampling Distribution54 Questions
Exam 7: Estimation Procedures57 Questions
Exam 8: Hypothesis Testing I: the One-Sample Case55 Questions
Exam 9: Hypothesis Testing II: the Two-Sample Case53 Questions
Exam 10: Hypothesis Testing III: the Analysis of Variance53 Questions
Exam 11: Hypothesis Testing Iv: Chi Square51 Questions
Exam 12: Bivariate Association: Introduction and Basic Concepts42 Questions
Exam 13: Association Between Variables Measured at the Nominal Level37 Questions
Exam 14: Association Between Variables Measured at the Ordinal Level52 Questions
Exam 15: Association Between Variables Measured at the Interval-Ratio Level51 Questions
Exam 16: Elaborating Bivariate Tables53 Questions
Exam 17: Partial Correlation and Multiple Regression and Correlation53 Questions
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If the row and column marginals of the expected frequencies are not equal to those of the observed frequencies,
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An obtained chi square of 10.78 has been calculated.Critical Chi square is 3.841.What should be concluded?
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A bivariate table in which one variable had three categories and the other variable had four categories would have
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When the null hypothesis in the chi square test for independence is true,there should be
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One limitation of the Chi square test (and all tests of hypothesis)is that they cannot tell us if relationships between variables are
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Which pattern of cell frequencies in a 2x2 table would indicate that the variables are independent?
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In the Chi square test for independence,the null hypothesis and the research hypothesis
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In a research study conducted to determine if arrests were related to the socioeconomic class of the offender,the chi square critical score was 9.488 and the chi square test statistic was 12.2.We can conclude that
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