Question:medium
A study was done to assess malnutrition among young children. 100 children were selected each from rural and urban areas. Out of these, 30 among rural and 20 among urban were found to be malnourished. Which of the following statistical tests is used to compare the data sets?
A study was done to assess malnutrition among young children. 100 children were selected each from rural and urban areas. Out of these, 30 among rural and 20 among urban were found to be malnourished. Which of the following statistical tests is used to compare the data sets?
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The outcome is a proportion of a categorical variable, not a mean.
Updated On: Jun 23, 2026
- Paired t-test
- Chi-square test
- Standard error of mean
- ANOVA
Show Solution
The Correct Option is B
Solution and Explanation
Mnemonic anchor: t-test and ANOVA are for means, chi-square is for counts. Decide which one your data are.
Here the result is a head count split into two boxes, malnourished or not, in two groups. That is categorical data summarised as proportions (30 percent rural, 20 percent urban). Comparing two proportions for significance is exactly what the chi-square test does, by checking how far the observed cell counts sit from the counts expected if there were no difference.
Knock out the rest: a paired t-test needs two measurements on the same individuals (before and after); ANOVA compares the means of three or more continuous groups; the standard error of the mean is a spread statistic, not a hypothesis test. None handles two proportions. So chi-square wins. Ref: Park PSM, 24e.
Here the result is a head count split into two boxes, malnourished or not, in two groups. That is categorical data summarised as proportions (30 percent rural, 20 percent urban). Comparing two proportions for significance is exactly what the chi-square test does, by checking how far the observed cell counts sit from the counts expected if there were no difference.
Knock out the rest: a paired t-test needs two measurements on the same individuals (before and after); ANOVA compares the means of three or more continuous groups; the standard error of the mean is a spread statistic, not a hypothesis test. None handles two proportions. So chi-square wins. Ref: Park PSM, 24e.
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