Paired T-Test for Two CO₂ Measurement Methods: 18-Env-B7 Environmental Sampling and Analysis, May 2016, Q2

Question 2: Paired T-Test for Two CO₂ Measurement Methods (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.

(a) Which Test, and the Conclusion

Given.

MINITAB two-sample summary and the paired differences $d_i=A_i-B_i$
Location	Method A	Method B	$d_i=A_i-B_i$
1	66.3	71.3	−5.0
2	63.5	60.4	3.1
3	64.9	64.6	0.3
4	61.8	63.9	−2.1
5	64.3	68.8	−4.5
6	64.7	70.1	−5.4
7	65.1	64.8	0.3
8	64.5	68.9	−4.4
9	68.4	65.8	2.6
10	63.2	66.2	−3.0
11	67.4	69.2	−1.8
Mean / StDev / SE	64.918 / 1.885 / 0.568	66.727 / 3.238 / 0.976	−1.809 / 3.016 / 0.909

Find. The appropriate t-procedure, its test statistic, and whether Method A is significantly different from Method B at $\alpha=0.10$.

Approach. Determine whether the two columns are independent or paired, then apply the matching t-procedure to the 11 differences and compare the statistic to the supplied t-table.

Identify the design. Method A and Method B were both applied at the same 11 locations – each pair $(A_i,B_i)$ shares a common sampling occasion, so the two columns are naturally correlated rather than independent. The paired t-test is therefore the statistically appropriate procedure: it works directly on the 11 differences $d_i=A_i-B_i$, so location-to-location variability common to both methods cancels out of the comparison, leaving only the genuine systematic difference between the methods. An independent two-sample test would ignore that correlation, inflate the estimated standard error of the difference, and understate the true significance of any real difference – exactly the pitfall the "partial results" table is built to test for.
Hypotheses. $H_0:\mu_d=0$ (the methods are not significantly different) vs. $H_a:\mu_d\ne0$ (two-sided, matching the "not significantly different" wording, which is a statement of equality).
Test statistic. $t=\dfrac{\bar d}{SE_{\bar d}}=\dfrac{-1.809}{0.909}$ $$t=\boxed{-1.99}\qquad(df=n-1=10)$$
Critical value. For a two-tailed test at $\alpha=0.10$, each tail carries 0.05; from the supplied t-table at $df=10$, $t_{0.05,10}=1.812$.
Decision. $|t|=1.99>1.812=t_{crit}$, so we reject $H_0$. (Equivalently, the two-tailed p-value is $\approx0.075$, which is less than $\alpha=0.10$.)

Conclusion. At the 10% significance level, Method A and Method B produce statistically significantly different $CO_2$ readings (Method A averages about 1.8 ppm lower). This is the opposite of what the exam scenario hoped to demonstrate – running the correct paired test, rather than an independent two-sample test, is exactly what exposes the real difference that a naive independent-samples analysis on the same 11 numbers would tend to miss.

(b) Main Assumption and Graphical Check

The paired t-test's main assumption is that the population of paired differences $d_i=A_i-B_i$ is (at least approximately) normally distributed. With only $n=11$ pairs, the Central Limit Theorem cannot be relied on to normalize a non-normal difference distribution on its own, so approximate normality of the $d_i$ themselves is required for the t reference distribution used in part (a) to be valid.

A simple graphical check is a normal probability (quantile-quantile) plot of the 11 sorted differences against their theoretical normal quantiles: if the points fall close to a straight line, normality is a reasonable working assumption; systematic curvature or a lone far-outlying point would instead signal a problem.

Fig. 1 – Normal probability plot of the 11 paired differences $d_i=A_i-B_i$ against their theoretical normal quantiles, with a fitted reference line through the sample mean and standard deviation. The points track the line reasonably closely with no severe curvature or isolated outlier, supporting the normality assumption used for the paired t-test in part (a).

Final results – Question 2
Item	Result
(a) Appropriate test	Paired t-test – the 11 locations are the same for both methods (correlated, not independent)
(a) Test statistic	$t=-1.99$ ($df=10$)
(a) Critical value ($\alpha=0.10$, two-tailed)	$t_{0.05,10}=1.812$
(a) Decision / Conclusion	Reject $H_0$ – Method A and Method B are significantly different at 10%
(b) Main assumption	Paired differences $d_i$ are approximately normally distributed
(b) Graphical check	Normal probability plot of $d_i$ – points track the reference line, no severe departure

Concept & Theory

Topic. Paired-sample $t$-test for comparing two measurement methods applied at common locations.

Key relations. $t=\bar d/(s_d/\sqrt n)$, $df=n-1$; reject $H_0$ when $|t|>t_{\alpha/2,\,df}$ (two-tailed) or when the corresponding p-value is less than $\alpha$.

Why this works. Pairing removes a nuisance source of variability – location-to-location differences common to both methods – from the comparison, so a paired analysis is strictly more powerful than an independent-samples analysis whenever the pairing is real; here the correlation between $A_i$ and $B_i$ is exactly why the paired standard error (0.909) is much smaller than the individual method standard errors would suggest for an independent comparison.

Common pitfall. Running an independent two-sample test on data that were actually collected in pairs; the independent-samples test is not "more conservative," it is simply the wrong model for correlated data, and it can mask a real, significant difference exactly as it would here.

Source. Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, ch. 10 (paired vs. two-sample inference on means); ch. 8 (normal probability plots).

Question 2 of 6: Paired T-Test for Two CO₂ Measurement Methods

Question 2: Paired T-Test for Two CO₂ Measurement Methods (20 marks)

(a) Which Test, and the Conclusion

(b) Main Assumption and Graphical Check