Goodness of Fit: Chi-Square Test Examples - Prof. Raoul Lepage, Exams of Data Analysis & Statistical Methods

Download Goodness of Fit: Chi-Square Test Examples - Prof. Raoul Lepage and more Exams Data Analysis & Statistical Methods in PDF only on Docsity! Chapter 26 Chi-Square Tests Goodness of fit: df = #cells of table - | (C-1 for cells arranged in a row). Homogeneity-Independence: df = (R-1)(C-1). Analyzed the same. Homogeneity is when "row counts are sampled separately." (obs — exp)? m Chi-Square Statistic is always calculated > cots of a table of counts =0. exp a Significance level (P-value) = probability of getting chi-square statistic that is at least as large as your data gave if the null hpothesis is correct. m Using a chi-square table: P-value df 0.0145 30 49 34 Prob(chi-sq with df 30 > 49.34) = 0.0145 —_ m Require all expected counts = 5. Not required of observed counts! m Can merge cells to achieve = 5 requirement. m Can add independent chi-square statistics to combine experimental results. Add df to get the applicable df for the combined data. m Remember: If you choose to "reject Hp whenever P < 0.001" then your type I error probability is 0.001. That is, if Ho is true then you will "reject Ho" with probability 0.001 (error of type I). m= Chance of error of type II > 0 with lots of data. That is, if Ho is false you are nearly certain to reject Hy with enough data. Goodness of fit example: Is the coin fair? Suppose we find 63 heads in 100 tosses? Is full moon statistically related to incidence of crime? a. Some expected counts are less than 5. b. Possible “confounding factors.” Merge cells to meet the “minimum of 5” requirement. Seems that we don’t have to worry over confounding factors. Is there a statistical association between tatooing and hepatitis C? But wait! Are all of the expected counts at least 5?