Paired t-Test on Pre/Post-Treatment Pollution Concentrations: 18-Env-B7 Environmental Sampling and Analysis, May 2017, Q2

Question 2: Paired t-Test on Pre/Post-Treatment Pollution Concentrations (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — Given. Ten pre-treatment and ten post-treatment concentration readings, both taken at the same location before and after a single treatment event.

Pre/post-treatment pollution concentrations
i	1	2	3	4	5	6	7	8	9	10
Pre	68.3	66.5	62.9	64.8	66.6	69.4	65.2	62.5	70.4	62.2
Post	61.8	60.1	60.2	63.5	68.5	61.8	62.1	61.2	65.1	68.9
$d_i=$pre$-$post	6.5	6.4	2.7	1.3	−1.9	7.6	3.1	1.3	5.3	−6.7

Find. Whether post-treatment concentrations are significantly lower than pre-treatment concentrations at α=0.05 (one-tailed).

Approach — test selection. Both readings were taken at the same location, one before and one after treatment, so the ten pre/post pairs share a common sampling occasion — each post-treatment value is naturally matched to its own pre-treatment value, not to an arbitrary one. This is a paired (matched-pairs) design, not two independent samples, so the correct test is a paired (one-sample-on-the-differences) t-test on $d_i=\text{pre}_i-\text{post}_i$, testing $H_0:\mu_d=0$ against $H_1:\mu_d>0$. Using an independent two-sample t-test here would ignore the location-to-location variability common to both readings and understate the evidence for a treatment effect.

Step 1 — compute the mean and standard deviation of the differences. $$\bar d = \frac{1}{n}\sum d_i = \frac{25.6}{10} = 2.560, \qquad s_d = \sqrt{\frac{\sum(d_i-\bar d)^2}{n-1}} = 4.376$$
Step 2 — compute the test statistic. With $n=10$ pairs, $$t = \frac{\bar d - 0}{s_d/\sqrt{n}} = \frac{2.560}{4.376/\sqrt{10}} = \frac{2.560}{1.384} = 1.850, \qquad df = n-1 = 9$$
Step 3 — compare to the critical value. From the supplied t-table at $\upsilon=9$, one-tailed $\alpha=0.05$: $t_{0.05,9}=1.833$. $$\boxed{t_{calc}=1.850 > t_{0.05,9}=1.833 \;\Rightarrow\; \text{reject } H_0}$$ The test statistic just clears the critical value (corresponding one-tailed $p\approx0.049$), so the reduction is significant at the 5% level, though only marginally.

Final Results — Question 2(a)
Quantity	Value
Test used	Paired (matched-pairs) t-test on $d_i=\text{pre}_i-\text{post}_i$
$\bar d$	2.560
$s_d$	4.376
$t_{calc}$ (df=9)	1.850
$t_{0.05,9}$ (critical, one-tailed)	1.833
Conclusion at α=0.05	Reject $H_0$ — post-treatment values are significantly lower

Part (b) — assumption and graphical check. The main assumption of the paired t-test is that the differences $d_i$ are (approximately) normally distributed — the individual pre- and post-treatment readings themselves need not be normal, only their paired differences. This is checked graphically with a normal probability (quantile-quantile) plot of the ten differences: each $d_i$ is plotted against its corresponding standard-normal quantile (rank-based plotting position $z_{(i)}$ for $P_i=(i-0.5)/n$); if the differences are normal, the points fall close to a straight line.

Normal probability plot of the ten paired differences $d_i$. The points track the fitted reference line reasonably closely, with no strong curvature or isolated outlier — consistent with the normality assumption underlying the paired t-test (confirmed by a Shapiro–Wilk check on the differences, $p\approx0.33$, well above 0.05).

Part (c) — alternative tests. If the normality of the differences were in doubt, the natural nonparametric alternatives for paired data are the Wilcoxon signed-rank test (ranks the absolute differences and tests whether positive and negative ranks are balanced — more powerful when the differences are roughly symmetric) and the simpler sign test (counts only how many differences are positive vs. negative, using the binomial distribution — fewer assumptions but less powerful). If the two sets of readings were instead treated as unrelated (unpaired) samples, an independent two-sample t-test (or its nonparametric analogue, the Mann–Whitney U / Wilcoxon rank-sum test) could be used, though this discards the pairing information and is not the preferred choice here.

Topic: Paired-samples hypothesis test for a location shift (before/after treatment).

Key relations: $t=\bar d/(s_d/\sqrt n)$, $df=n-1$; reject $H_0:\mu_d=0$ in favour of $H_1:\mu_d>0$ if $t_{calc}>t_{\alpha,df}$.

Why this works: Reducing paired data to differences removes any consistent location-to-location variability common to both readings, isolating the treatment effect and giving a more powerful test than treating the two sets as independent when they are not.

Common pitfall: Running an independent two-sample t-test on matched before/after data at the same locations understates the evidence for an effect (it adds unrelated between-location variance into the standard error) and is the single most common error on this question type.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, ch. 10 (paired-sample inference), ch. 16 (nonparametric statistics).

Question 2 of 7: Paired t-Test on Pre/Post-Treatment Pollution Concentrations

Question 2: Paired t-Test on Pre/Post-Treatment Pollution Concentrations (20 marks)