Interpreting Survey Data
A survey of 200 students asked about their preferred study method: reading, writing, or listening. The results are: 80 prefer reading, 70 prefer writing, and 50 prefer listening. The survey also recorded gender: 90 males and 110 females. Among the males, 40 prefer reading, 30 prefer writing, and 20 prefer listening. The student council wants to know if there is an association between gender and preferred study method. They plan to use a chi-square test for independence at the 5% significance level. However, they are unsure if the conditions for the test are met.
1. Construct a two-way table of observed frequencies. Then calculate the expected frequencies for each cell under the assumption of independence. Show the formula and calculations.
Read the worked answer
Observed table:
| Gender | Reading | Writing | Listening | Total |
|---|---|---|---|---|
| Male | 40 | 30 | 20 | 90 |
| Female | 40 | 40 | 30 | 110 |
| Total | 80 | 70 | 50 | 200 |
| Expected frequency for a cell = (row total * column total) / grand total. | ||||
| For Male-Reading: (90*80)/200 = 7200/200 = 36. | ||||
| Male-Writing: (90*70)/200 = 6300/200 = 31.5. | ||||
| Male-Listening: (90*50)/200 = 4500/200 = 22.5. | ||||
| Female-Reading: (110*80)/200 = 8800/200 = 44. | ||||
| Female-Writing: (110*70)/200 = 7700/200 = 38.5. | ||||
| Female-Listening: (110*50)/200 = 5500/200 = 27.5. |
2. Check whether the conditions for a chi-square test for independence are satisfied. State the conditions and verify them using the expected frequencies. If not satisfied, suggest a possible remedy.
Read the worked answer
Conditions for chi-square test: (1) Random sample or random assignment; (2) Large sample: all expected frequencies should be at least 5 (some texts say at least 1, but no more than 20% less than 5). Here, the smallest expected frequency is 22.5, which is greater than 5. So the large sample condition is met. (3) Independence: each observation contributes to only one cell; assuming the survey was random, this is likely met. However, we must assume the sample is random. If the sample is not random, results may not generalize. Since all expected counts are ≥5, the chi-square test is appropriate. No remedy needed.