Tomato Fertilizer Trial: Confounded t-Test and Confidence Interval
In a Biology practical, a student investigates whether a new organic fertilizer increases tomato plant height. Ten seeds are planted in each of two identical trays. Both trays are placed in the same greenhouse, but the treatment tray is placed on the sunny side (south-facing) while the control tray is on the north side. After six weeks, the control group (n=10) has a mean height of 24.3 cm with a standard deviation of 3.1 cm. The treatment group (n=10) has a mean height of 28.7 cm with a standard deviation of 4.2 cm. The student performs an independent two-sample t-test and obtains t = 2.58, p = 0.019. The student concludes that the fertilizer caused a 4.4 cm increase. You are asked to evaluate this conclusion. You may use the pooled standard deviation and a t-critical value of 2.10 for 18 degrees of freedom at α = 0.05.
1. Explain whether the p-value of 0.019 supports the student's causal claim. Identify the main confounding variable and describe how it could produce the observed difference.
Read the worked answer
The p-value of 0.019 means that, if the null hypothesis were true and there were no difference between treatments, we would expect a difference at least as large as 4.4 cm in about 1.9% of repeated experiments. It does not tell us why the difference occurred. The student's causal claim requires that the only systematic difference between groups is the fertilizer. That condition is violated: the treatment tray was on the sunny south side while the control tray was on the north side. Light intensity and temperature are therefore confounded with the fertilizer. Tomato plants on a sunnier side could photosynthesise more and grow taller regardless of fertilizer. The p-value is compatible with a light effect, so it cannot isolate the fertilizer's causal contribution. The correct conclusion is that the data show an association between tray position/fertilizer and height, not proof that the fertilizer caused the increase.
2. Calculate the 95% confidence interval for the difference in means (treatment minus control). Show your formula and steps. Interpret the interval in the context of the confounding variable.
Read the worked answer
Difference in means: 28.7 − 24.3 = 4.4 cm.
Pooled variance: sp² = [(n1−1)s1² + (n2−1)s2²] / (n1+n2−2) = [9(3.1²) + 9(4.2²)] / 18 = [9(9.61) + 9(17.64)] / 18 = (86.49 + 158.76) / 18 = 245.25 / 18 = 13.625. Pooled SD = √13.625 ≈ 3.69 cm.
Standard error: SE = sp × √(1/n1 + 1/n2) = 3.69 × √(0.1 + 0.1) = 3.69 × √0.2 ≈ 3.69 × 0.447 ≈ 1.65 cm.
Margin of error: t(0.05, 18) × SE = 2.10 × 1.65 ≈ 3.47 cm.
95% CI: 4.4 ± 3.47 = [0.93, 7.87] cm.
Interpretation: We estimate that the true mean difference lies between about 0.9 cm and 7.9 cm. However, because tray position is confounded with treatment, this interval describes the combined effect of fertilizer plus light/temperature, not the fertilizer alone. The interval excludes zero, but that does not rescue the causal claim; a design flaw can produce a statistically significant but biased estimate. A follow-up experiment with randomised tray positions is needed before the interval can be interpreted as a fertilizer effect.
3. Propose one specific change to the experimental design that would reduce confounding, and explain how it would affect the validity of the t-test.
Read the worked answer
A specific change is to randomise the position of individual plants within the greenhouse, or to rotate the two trays every three days so that both groups receive similar average light exposure. Better still, use multiple trays per treatment and randomise tray placement, then include tray as a blocking factor.
If positions are randomised, the sunny-side advantage is no longer systematically tied to the fertilizer group. The t-test assumes independent observations and comparable groups apart from treatment; randomisation makes that assumption more plausible. The t-test would then test the fertilizer effect with reduced confounding bias, although it would still not control for unmeasured differences if sample sizes are small. Rotating trays reduces the light confound but may introduce handling effects; randomisation is the stronger design change because it distributes nuisance variables probabilistically rather than assuming they are absent.