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

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) The Appropriate Test and Its Conclusion

Given. Eleven locations, each measured by both methods (paired data); printed summary for the differences $d_i=A_i-B_i$: $\bar d=-1.809$, $s_d=3.016$, $SE_{\bar d}=0.909$, $n=11$.

Find. The appropriate t-test, its test statistic, and the conclusion at $\alpha=0.05$.

Approach. Decide independence vs. pairing first, then apply the matching t-procedure and compare the test statistic to the one-tailed critical value from the supplied t-table.

Method A and Method B were both applied at the same 11 locations – each pair $(A_i,B_i)$ shares a sampling occasion, so the two columns are naturally correlated, not independent. The paired t-test is therefore the statistically appropriate procedure: it works directly on the 11 differences $d_i$, so any location-to-location variation common to both methods cancels out of the comparison, leaving only the true systematic difference between methods. An (incorrect) independent two-sample test would throw away that cancellation and inflate the estimated standard error of the difference.

1. Hypotheses. $H_0:\mu_d=0$ vs. $H_a:\mu_d\lt 0$ (one-tailed, since we want to show Method A reads lower than Method B).
2. Test statistic. $t=\dfrac{\bar d-0}{SE_{\bar d}}=\dfrac{-1.809}{0.909}$ $$t=\boxed{-1.99}\qquad df=n-1=10$$
3. Critical value and decision. From the supplied t-table at $df=10$, one-tailed $\alpha=0.05$: $t_{0.05,10}=1.812$. Since $|t|=1.99\gt 1.812$ (equivalently $t=-1.99$ falls left of $-1.812$), we reject $H_0$. Cross-checking against the table: $1.99$ lies between $t_{0.05}=1.812$ and $t_{0.025}=2.228$ at $df=10$, so the one-tailed P-value is bracketed between 0.025 and 0.05 (computed exactly, $P\approx0.037$), confirming the rejection.

Conclusion. At the 5% significance level there is a statistically significant difference: Method A reads, on average, about 1.8 ppm lower than Method B ($P\approx0.037\lt 0.05$). The data support the claim that Method A gives lower pollution readings than Method B.

(b) 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, 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 plot of the 11 differences: each sorted $d_{(i)}$ is plotted against its normal score $z_i=\Phi^{-1}\!\left(\frac{i-0.5}{n}\right)$; if the points fall close to a straight line, normality is a reasonable working assumption.