Chi-Square Sample Problem Using Jamovi.pptx

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1. THE CHI-SQUARE TEST for Homogeneity of Samples

2. TWO INDEPENDENT SAMPLES Example : In a study conducted on the use of seat belts in preventing fatalities, records of the last 100 vehicular accidents were reviewed. These 100 accidents involved 238 persons. Each person was classified as using or not using seat belts when the accident happened and as injured fatally or a survivor.

3. Research Question: Is there a significant difference in the proportion of persons who are fatally injured between those who wear seatbelts and those who do not? Null Hypothesis: There is no significant difference in the proportion of persons who are fatally injured between those who wear seatbelts and those who do not.

4. Alternative Hypothesis (Non-directional): There is a significant difference in the proportion of persons who are fatally injured between those who wear seatbelts and those who do not. Alternative Hypothesis (Directional): The proportion of persons who are fatally injured is higher for those who do not wear seatbelts than those who wear seatbelts.

5. Data: Injured Fatally? Wearing Seat Belt? Total Yes (1) No (0) Yes (1) 9 88 97 No (0) 23 118 141 Total 32 206 238

6. For contingency tables, the degrees of freedom is given by df = (r-1)(c-1) where r is the number of categories of the row variable while c is the number of categories of the column variable. Where o is the “observed frequency” and e is the expected frequency. Test Statistic:

7. The tabled data is actually a summary of the responses of the respondents which can be encoded directly using SPSS Use codes for the responses: 1 – Yes; 2 – No Encode the frequency counts in a column Put the corresponding codes for the two variables So that SPSS will read the data as frequency counts, we have to set on the “weight cases” command

8. Click “Data”; then “Weight Cases” to show the window on top right. 2. Highlight the data “ Freq ”, Click “Weight cases by” then Click the arrow button; Click “ok”

9. Check if “Weight On” already appears at the bottom right corner of the worksheet

10. To Analyze the data: Click “ Analyze ” ”Descriptive Statistics”  “Crosstabs” to obtain the window at the right. Transfer the row variable “Injured Fatally” and the column variable “Wearing Seatbelts” by highlighting them one at a time and clicking the arrow button.

11. Click “Statistics” then check “Chi-square”, then click “Continue”; Click “Cells” then check “Column”; then Click “Continue”; then Click “OK”

12. Output (Descriptive Statistics): Note that 7.1% of those who wore seatbelts were injured fatally while 29.9% of those who did not wear seatbelts were injured fatally. Is 29.9% significantly higher than 7.1%?

13. Output (Inferential Statistics): If there is no cell with expected frequency of less than 5, we report the Pearson Chi-square value. Since  2 = 26.014 and the associated probability is <.001 (which is less than  = .05), the null hypothesis is rejected. If there is a cell with expected frequency of less than 5, we have to report the Fisher’s Exact Test . Note that the Fisher’s Exact Probability is less than .001. Hence the null hypothesis is rejected.