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Chi-Square Test for Association/Independence

Chi-Square Test for Association/Independence. Ch. 14.

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Chi-Square Test for Association/Independence

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  1. Chi-Square Test for Association/Independence Ch. 14

  2. Anger rating on the Spielberger scale is a categorical variable that takes three possible values: low, medium, and high. Whether or not someone gets heart disease is also a categorical variable. The following two-way table shows the relationship between anger rating and heart disease for a random sample of 8474 people. Do these data provide convincing evidence of an association between the variables in the larger population?

  3. Data The study followed a random sample of 8474 people with normal blood pressure for about four years. All individuals were free of heart disease at the beginning of the study. Each person took the Spielberger Trait Anger Scale test, which measures how prone a person is to sudden anger. Researchers also recorded whether each individual developed coronary heart disease (CHD)

  4. The sample data are easy to investigate: turn them into percents and look for a relationship between the variables. Is the association in the sample evidence of an association between these variables in the entire population? Or could the sample association easily arise just from the luck of random sampling? This is the question for a significance test.

  5. Hypotheses Ho: There is no association between two categorical variables in the population of interest. Ha: There is an association between two categorical variables in the population of interest. OR Ho: Two categorical variables are independent in the population of interest Ha: Two categorical variables are not independent in the population of interest

  6. Chi-Square Test Suppose the Random, Large Sample Size and Independent Conditions are met. You can use the chi-square test for association/independence to test the appropriate hypotheses. Calculate the chi-square statistic 2 = where the sum is over all the cells in the two way table. If Ho is true, the 2 statistic has approximately a chi-square distribution with df = (r-1)(c-1). The P-value is the area to the right of 2 under the corresponding chi-square density curve.

  7. Expected Counts expected count = Calculate the expected counts for the anger level and heart disease two-way table

  8. Do Angry People Have More Heart Disease? Do the data provide convincing evidence of an association between anger level and heart disease in the population of interest? Carry out an appropriate test to help answer this question.

  9. Using Chi-Square Tests Wisely • A chi-square test for homogeneity tests whether the distribution of a categorical variable is the same for each of several populations or treatments. • The chi-square test for association/independence test whether two categorical variable are associated in some population of interest. • Focus on how the data were produced. If the data come from two or more independent random samples or treatment groups in a randomized experiment, do the chi-square test for homogeneity • If the data come from a single random sample, with the individuals classified according to two variables, use a chi-square test for association/independence.

  10. Scary Movies and Fear Are men and women equally likely to suffer lingering fear from watching scary movies as children? Researchers asked a sample of 117 college students to write narrative accounts of their exposure to scary movies before the age of 13. More than one-fourth of the students said that some of the fright symptoms are still present when they are awake. The following table breaks down these results by gender.

  11. Data Explain why a chi-square test for association/independence and not a chi-square test for homogeneity should be used in this setting. Researchers decided to use the null hypothesis Ho: Gender and ongoing fright symptoms are independent in the population of interest. State the correct alternative hypothesis. Perform the appropriate hypothesis test with  = 0.01. Which cell contributes most to the chi-statistic? In what way does this cell differ from what the null hypothesis suggests?

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