Stat prep

Stat Prep: HW 5 Student’s Name Institution Affiliation Course Name & Code Instructor’s Name Date Stat Prep: HW 5 3) Significance level = 0.05 Chi-square Test for Independence Hypotheses Null hypotheses (H0): The type of restoration and the presence of adverse health conditions are independent. Alternative hypothesis (Ha): The type of restoration and the presence of adverse health conditions are not independent. Calculating Chi-square Statistics (X2) = ∑[(O-E)2/E] O → observed frequency E → expected frequency Calculate expected frequency E = (Column total * Row total)/Overall total Table with column totals, overall totals, and row totals. Amalgam Composite Row Totals Adverse Health Condition Reported 135 145 280 No Adverse Health Condition Reported 132 122 254 Column Totals 267 267 534 (Overall total) Table with observed and excepted frequencies. Outcome O E O-E (O-E)2 (O-E)2/E Adverse Health Condition Reported Amalgam 135 140 -5 25 0.179 Composite 145 140 5 25 0.179 No Adverse Health Condition Reported Amalgam 1...