Bivariate Analysis. Comparisons of proportions: Chi Square Test (X 2 test) Variable 1. Variable 2 2 LEVELS >2 LEVELS CONTINUOUS

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1 Bivariate Analysis Variable 1 2 LEVELS >2 LEVELS CONTINUOUS Variable 2 2 LEVELS X 2 chi square test >2 LEVELS X 2 chi square test CONTINUOUS t-test X 2 chi square test X 2 chi square test ANOVA (F-test) t-test ANOVA (F-test) -Correlation -Simple linear Regression Comparisons of proportions: Example 1: Research question: Is there a relationship between regular physical activity (PA) and? H o : Proportion of PA among s = Proportion of PA among fes H a : Proportion of PA among s Proportion of PA among fes Statistical test: Chi square test Example 1: Chi square test (SPSS output) * Crosstabulation fe No Yes b Continuity Correction a Fisher's Exact Test a. Computed only for a 2x2 table b. 0 cells (.0%) have expected count less than 5. The minimum expected count is Exact Sig. Exact Sig. (1-sided)

2 This is the p-value for the Chi square test. If the p-value is >0.05 then accept null hypothesis & conclude that the 2 proportions are equal. If the p-value is <0.05 then reject null hypothesis (accept the alternative) & conclude that the 2 proportions are different. Exact Sig. Exact Sig. (1-sided) b Continuity Correction a Fisher's Exact Test a. Computed only for a 2x2 table b. 0 cells (.0%) have expected count less than 5. The minimum expected count is Exact Sig. Exact Sig. (1-sided) b Continuity Correction a Fisher's Exact Test a. Computed only for a 2x2 table b. 0 cells (.0%) have expected count less than 5. The minimum expected count is of statistical test: P-value: Example 1: Research question: Is there a relationship between regular physical activity and? H o : Proportion of PA among s = proportion of PA among fes H a : Proportion of PA among s proportion of PA among fes Statistical test: Chi square test= P=0.000 Conclusion: At significance level of 0.05, we reject null hypothesis and conclude that in the population there is significant difference in the proportions of PA among s and fes.

3 Example 1: Percent (%) * Crosstabulation fe No Yes 23.7% 17.7% 41.4% 41.0% 17.6% 58.6% 64.8% 35.2% 100.0% Example 1: Row Percent (%) * Crosstabulation fe No Yes 57.3% 42.7% 100.0% 70.0% 30.0% 100.0% 64.8% 35.2% 100.0% Example 1: Column Percent (%) * Crosstabulation fe No Yes 36.7% 50.1% 41.4% 63.3% 49.9% 58.6% 100.0% 100.0% 100.0%

4 Example 1: All Percentages * Crosstabulation fe No Yes 57.3% 42.7% 100.0% 36.7% 50.1% 41.4% 23.7% 17.7% 41.4% 70.0% 30.0% 100.0% 63.3% 49.9% 58.6% 41.0% 17.6% 58.6% 64.8% 35.2% 100.0% 100.0% 100.0% 100.0% 64.8% 35.2% 100.0% major * Crosstabulation major health science engineering law business arts No Yes % 36.1% 100.0% 9.5% 9.9% 9.7% 6.2% 3.5% 9.7% % 31.9% 100.0% 26.8% 23.2% 25.5% 17.4% 8.1% 25.5% % 38.2% 100.0% 11.5% 13.1% 12.0% 7.4% 4.6% 12.0% % 37.8% 100.0% 13.3% 15.0% 13.9% 8.6% 5.3% 13.9% % 38.2% 100.0% 17.3% 19.7% 18.2% 11.2% 6.9% 18.2% % 32.3% 100.0% 21.6% 19.0% 20.7% 14.0% 6.7% 20.7% % 35.1% 100.0% 100.0% 100.0% 100.0% 64.9% 35.1% 100.0% a a. 0 cells (.0%) have expected count less than 5. The minimum expected count is Research question: Is there a relationship between regular physical activity and students majors?

5 7.215 a a. 0 cells (.0%) have expected count less than 5. The minimum expected count is H o : Proportions of PA among all majors are equal H a : At least two proportions are different a a. 0 cells (.0%) have expected count less than 5. The minimum expected count is of statistical test: P-value: a a. 0 cells (.0%) have expected count less than 5. The minimum expected count is Conclusion: At significance level of 0.05, we accept the null hypothesis and conclude that in the population there is no significant difference in the proportions of PA among the different students majors.

6 SPSS commands for X 2 test Example 1 Analyze Descriptive statistics Crosstabs select major for row select for column Go to Statistics-chose chi square Go to cells-select row, column, and total from percentages Relation between hypothesis testing & Confidence Interval (CI) Example 1: A cohort study Smoking Lung cancer Yes No No Lung cancer Relative risk (RR): (32/200) / (15/200)= 2.13 Results: Chi square test X 2 = 6.97, p-value < %Confidence Interval around RR: Example 1: A cohort study H o : RR=1 H a : RR 1 Conclusion based on results of Chi square test (X 2 = 6.97, p-value < 0.01): Because p-value of chi square test < 0.05, reject H o. Therefore, there is a significant association between smoking and lung cancer.

7 Example 1: A cohort study H o : RR=1 H a : RR 1 Conclusion based on results of 95% confidence interval around RR: ( ): We are 95% confident that the RR between smoking and lung cancer in the population is between Because the 95% Confidence Interval does not include the value of the null hypothesis (1) then reject the null hypothesis and accept the alternative. Therefore, there is a significant association between smoking & lung cancer. Example 2: A case control study Breast cancer No Breast cancer Artificial sweetener Yes No Odds Ratio (OR): (69 x 134) / (22 x 66)= 6.37 Results: Chi square test (X 2 ) = 46.1, p-value < % Confidence Interval around OR: ( ) Example 2: A case control study H o : OR = 1 H a : OR 1 Results: Chi square test (X 2 ) = 46.1, p-value < % Confidence Interval around OR: ( ) What is the conclusion based on results of X 2 test???? What is the conclusion based on results of 95% CI????

8 Example 2: A case control study H o : OR = 1 H a : OR 1 Conclusion based on Chi square test (X 2 = 46.1, p-value <0.0001): Because p-value of chi square test < 0.05, reject H o. Therefore, there is a significant association between artificial sweeteners and breast cancer. Example 2: A case control study H o : OR = 1 H a : OR 1 Conclusion based on 95% Confidence Interval around OR ( ): We are 95% confident that the OR between artificial sweetener & breast cancer in the population is between Because the 95% Confidence Interval does not include the value of the null hypothesis (1) then reject the null hypothesis & accept the alternative. Therefore, there is a significant association between artificial sweetener & breast cancer. T (True) or F (False): Results of a case control study to assess the association between physical activity and depression were as follows: OR = 4.2 and 95% Confidence Interval ( ). Based on these results, we would conclude that in the population there is no relation between PA and depression because the confidence Interval includes 4.2

9 T (True) or F (False): In studying whether there is an association between engaging in PA and hours spent at work (per week), the investigator found out that the difference in the average weekly hours spent at work between those who engage in PA and those who do not is 10 with a 95% Confidence Interval of (0.9-23). Based on the results; we would reject the null hypothesis and conclude that there is a relation between engaging in PA and hours spent at work.

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