Breakfast and Tardiness: Conditional Probability, Significance, and Causation
A school counselor surveys 240 students about breakfast and tardiness. Of the 90 students who report eating breakfast at least four mornings a week, 18 have been late to school at least twice this month. Of the 150 students who report rarely eating breakfast, 45 have been late at least twice. The counselor concludes that skipping breakfast causes students to be late and proposes a mandatory breakfast program. The principal wants to know whether the survey supports a causal claim, whether the difference is large enough to matter, and what a stronger study would look like. The counselor's newsletter draft says, 'Students who skip breakfast are 50% more likely to be late.'
1. Build the two-way table, compute P(late | breakfast) and P(late | no breakfast), and compute the relative risk. Check whether the two variables are independent.
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The table is:
Breakfast, late: 18; breakfast, not late: 72; breakfast total: 90. No breakfast, late: 45; no breakfast, not late: 105; no breakfast total: 150. Late total: 63; not late total: 177; grand total: 240.
Conditional probabilities: P(late | breakfast) = 18/90 = 0.20. P(late | no breakfast) = 45/150 = 0.30.
Relative risk = 0.30/0.20 = 1.5, so the newsletter's '50% more likely' is a relative-risk statement, not a 50-percentage-point difference.
Independence check: P(late) = 63/240 = 0.2625 and P(breakfast) = 90/240 = 0.375. Their product is about 0.0984, but P(late and breakfast) = 18/240 = 0.075. Since these differ, the variables are not independent in this sample.
2. The observed difference is 0.30 - 0.20 = 0.10. Compute the standard error and a z-statistic, then assess whether the difference is statistically significant at the 5% level. Does the survey establish causation?
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For a difference of two proportions, the standard error is
SE = sqrt[p1(1 - p1)/n1 + p2(1 - p2)/n2] = sqrt[(0.20)(0.80)/90 + (0.30)(0.70)/150] = sqrt(0.001778 + 0.001400) = sqrt(0.003178) ≈ 0.0564.
Then z = (0.30 - 0.20)/0.0564 ≈ 1.77. A two-sided p-value is about 0.077, which is greater than 0.05, so the difference is not statistically significant at the 5% level. A 95% confidence interval for the difference is about 0.10 ± 1.96(0.0564), or (-0.011, 0.211), which includes 0.
The data suggest an association, but they do not establish causation. This is an observational survey; students who skip breakfast may differ in many other ways that affect tardiness.
3. Identify a plausible confounding variable and describe a stronger study design that could support a causal conclusion.
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A plausible confounder is morning routine stability: students who skip breakfast may also have later bedtimes, longer commutes, less parental supervision, or more morning responsibilities. Any of these could increase tardiness independently of breakfast.
A stronger design would randomly assign students to a breakfast program or a control condition, then compare tardiness rates. Random assignment tends to balance confounding variables across the groups, so a difference in tardiness could be attributed more credibly to the program. The study should include a control group, a clear definition of tardiness, and ideally have the people recording tardiness unaware of which students were assigned to the program. The newsletter should also distinguish the 50% relative increase from the 10-percentage-point absolute difference.