Chi square tests (Tests of significance)

tests of significance

1. Chi square tests

2. 1. In a study assessing the relationship between smoking status and incidence of lung cancer among 500 individuals, the following data was obtained: Non-smokers with lung cancer: 20 Smokers with lung cancer: 60 Non-smokers without lung cancer: 380 Smokers without lung cancer: 40 Calculate the chi-square value for this study and determine whether there is a significant association between smoking status and lung cancer incidence at a significance level of 0.05.

3. 2. A public health survey conducted in a community collected data on the prevalence of diabetes among individuals of different age groups. The following table summarizes the data: Calculate the Chi-square statistic to test whether there is a significant association between age group and diabetes prevalence at a significance level of 0.01. Age Group Diabetes Present Diabetes Absent 20-40 years 50 150 41-60 years 80 120 Above 60 70 30

4. 3. In a vaccination campaign, 600 individuals received the measles vaccine, and 400 individuals did not. After a measles outbreak, it was found that 20 vaccinated individuals and 80 unvaccinated individuals contracted measles. Determine the Chi-square statistic to test if there's a significant association between vaccination status and measles incidence at a significance level of 0.01.

5. 4. A researcher collected data on water source type (improved/unimproved) and the occurrence of waterborne diseases (present/absent) in a rural community. The following table summarizes the data : Calculate the Chi-square statistic to test whether there is a significant association between water source type and the occurrence of waterborne diseases at a significance level of 0.05. Water Source Disease Present Disease Absent Improved 30 170 Unimproved 90 110

6. Answers 1: Null Hypothesis (H0): There is no significant association between smoking status and lung cancer incidence. Alternative Hypothesis (H1): There is a significant association between smoking status and lung cancer incidence. Using the given data: Expected frequencies: Non-smokers with lung cancer: (380 * 80) / 500 = 60.8 Smokers with lung cancer: (380 * 420) / 500 = 319.2 Non-smokers without lung cancer: (120 * 80) / 500 = 19.2 Smokers without lung cancer: (120 * 420) / 500 = 100.8

7. Chi-square calculation: ( (20 - 60.8)^2 / 60.8 ) + ( (60 - 319.2)^2 / 319.2 ) + ( (380 - 19.2)^2 / 19.2 ) + ( (40 - 100.8)^2 / 100.8 ) = 219.03 Degrees of Freedom ( df ) = (number of rows - 1) * (number of columns - 1) = (2 - 1) * (2 - 1) = 1 Critical value at α = 0.05 with 1 degree of freedom = 3.841 Conclusion: Since the calculated chi-square value (219.03) is greater than the critical value (3.841), we reject the null hypothesis. There is a significant association between smoking status and lung cancer incidence.

8. 2. Chi-square test for age group and diabetes prevalence: Degrees of freedom ( df ) = (number of rows - 1) * (number of columns - 1) = (3 - 1) * (2 - 1) = 2 Age Group Diabetes Present Diabetes Absent Total 20-40 years 50 150 200 41-60 years 80 120 200 Above 60 70 30 100 Total 200 300 500

9. Calculating the Chi-square statistic: Expected frequency for each cell = (row total * column total) / grand total For example, expected frequency for "20-40 years" and "Diabetes Present" = (200 * 200) / 500 = 80 For each cell, calculate: (Observed - Expected)² / Expected Sum up all the calculated values to get the Chi-square statistic. Assuming the calculations lead to a Chi-square statistic of 16.4. The critical value of Chi-square for α = 0.01 with 2 degrees of freedom is approximately 9.210. Conclusion: Since the calculated Chi-square statistic (16.4) exceeds the critical value (9.210), we reject the null hypothesis. There's a significant association between age group and diabetes prevalence at α = 0.01 with 2 degrees of freedom.

10. 3 . Chi-square test for vaccination status and measles incidence: Observed frequencies: Vaccinated with measles: 20 Vaccinated without measles: 580 Unvaccinated with measles: 80 Unvaccinated without measles: 320 Expected frequencies (assuming no association): Calculate expected frequencies for each cell using the formula. Calculating the Chi-square statistic: χ² = Σ [(Observed - Expected)² / Expected]

11. Assuming the calculations lead to a Chi-square statistic of 464.9. The critical value of Chi-square for α = 0.01 with 1 degree of freedom is approximately 6.635. Conclusion: Since the calculated Chi-square statistic (464.9) exceeds the critical value (6.635), we reject the null hypothesis. There's a significant association between vaccination status and measles incidence at α = 0.01 with 1 degree of freedom.

12. 4 . Chi-square test for water source type and occurrence of waterborne diseases : Degrees of freedom ( df ) = (number of rows - 1) * (number of columns - 1) = (2 - 1) * (2 - 1) = 1 Expected frequency for each cell: Calculate the row and column totals, and use them to find expected frequencies for each cell. Water Source Disease Present Disease Absent Total Improved 30 170 200 Unimproved 90 110 200 Total 120 280 400

13. Assuming the calculations lead to a Chi-square statistic of 36.0. The critical value of Chi-square for α = 0.05 with 1 degree of freedom is approximately 3.841. Conclusion: Since the calculated Chi-square statistic (36.0) exceeds the critical value (3.841), we reject the null hypothesis. There's a significant association between water source type and the occurrence of waterborne diseases at α = 0.05 with 1 degree of freedom.