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Business 205

Business 205. Review. Correlation MS5. Preview. Chi-Square. Types of Tests. Parametric Tests Assume normal and homogeneity of variance Require parameters Z-test, T-test, 2-Group T, ANOVA Non-parametric Tests Few (if any) assumptions about the population distribution

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Business 205

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  1. Business 205

  2. Review • Correlation • MS5

  3. Preview • Chi-Square

  4. Types of Tests • Parametric Tests • Assume normal and homogeneity of variance • Require parameters • Z-test, T-test, 2-Group T, ANOVA • Non-parametric Tests • Few (if any) assumptions about the population distribution • Rarely state a hypotheses in terms of a specific parameter • Chi-Square

  5. Chi-Square • Tests for “goodness of fit” • Looks at sample data to test hypothesis about the shape or proportions of a population distribution • How well does the sample proportions fit the population proportions?

  6. Regular Hypothesis Specifies preferences Specifies differences in population Null Hypothesis Specifies no preferences Specifies no difference in population Chi-Square Hypotheses

  7. One-Sample Chi-Square χ2 • Compares the goodness of fit of the data to that of the null hypothesis • Compares observed frequencies against expected frequencies • Looks at the difference between 1 IV with multiple levels

  8. Observed Frequencies • Number of individuals from the sample who are classified in a particular category • How many times it occurs

  9. Expected Scores/Frequencies • The frequency value that is predicted from the null hypothesis and the sample size.

  10. One-Sample Chi-Square

  11. 1 Sample Chi-Square Example

  12. Fill out the chart

  13. Chi-Square • Chi-Square = (6.72+.06+0+2.72+.06+3.56+.50+2.72) = 16.34 • Critical Chi-Square value at an alpha of .05 is 3.84. • Accept that your sample is not from the population because 16.34 is larger than 3.84.

  14. Two Sample Chi-Square TestChi-Square Test for Independence • 2 Factors • Many levels • Tests whether or not there is a relationship between 2 variables

  15. Two Sample Chi-Square TestChi-Square Test for Independence df = (number of rows - 1)(number of columns - 1)

  16. Two Sample Chi-Square TestChi-Square Test for Independence • Three different drug treatments are used to control hypertension. At the end of treatment, the investigator classifies patients as having either a favorable or unfavorable response to the medication. Your hypothesis is that there is a different between treatments.

  17. Two Sample Chi-Square TestChi-Square Test for Independence 1.) Total up each column 2.) Total up each row 3.) Find the expected score for each cell

  18. Two Sample Chi-Square TestChi-Square Test for Independence

  19. Two Sample Chi-Square TestChi-Square Test for Independence

  20. Two Sample Chi-Square TestChi-Square Test for Independence • Chi-Square = (1.15+4.15+.00+.01+.48+1.89) = 8.02 • Critical value for an alpha of .05 and at 2 df = 5.99 • Ours falls above the critical region so we can accept the hypothesis and say that the treatments differ.

  21. Reporting findingsChi-Square Test for Independence The chi-square value is in the critical region. Therefore, we can reject the null hypothesis. There is a relationship between drug treatments and responses towards medication (2, n =500) = 8.02, p <.05.

  22. Chi-Square In Class Example H1: There is a difference between hotels.

  23. Excel: Chi-Square • DO NOT USE: • CHIDIST( )  Only returns a probability • For Single Chi-Squares, you must program that in by hand • CHITEST( )  Only returns a probability • For a test of independence (2 sample) you must program that in by hand

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