Question
A rail operator times the same 25 drivers on matched routes before and after a scheduling change. Define each driver's saving as \(D=\text{before time}-\text{after time}\); a positive D means the new schedule is faster. The observed mean saving is 2 minutes.
The operator knows that across independently sampled drivers the before and after times form a bivariate normal pair, with variance 36 min² before, variance 25 min² after, and covariance 24 min² within each pair. These population variance values are treated as known.
Test \(H_0\): the mean saving is zero against \(H_1\): it is positive, using a one-sided 5% test. Give the correct paired standard error, z-statistic and p-value. What would happen if an analyst incorrectly treated the two sets of 25 times as independent?
Get hint
Hint 1
Find \(\operatorname{Var}(\text{before}-\text{after})\), remembering the covariance term. The 25 differences are independent across drivers.
Check your answer
Solution
Let \(X_i\) and \(Y_i\) be a given driver's before and after durations. Define \(D_i=X_i-Y_i\). For one driver,
\[\operatorname{Var}(D_i)=\operatorname{Var}(X_i)+\operatorname{Var}(Y_i)-2\operatorname{Cov}(X_i,Y_i)=36+25-2\times24=13\] min².
The 25 matched pairs are independent, so \(\operatorname{Var}(\bar D)=\frac{13}{25}\) min² and \(\operatorname{SE}(\bar D)=\frac{\sqrt{13}}5\approx0.72111\) minutes. Under \(H_0\) the normally distributed standardized statistic is
\[Z=\frac{\bar D-0}{\sqrt{13}/5}=\frac{2\times5}{\sqrt{13}}=\frac{10}{\sqrt{13}}\approx2.77350\].
The one-sided p-value is \(1-\Phi\left(\frac{10}{\sqrt{13}}\right)\approx0.002773\), below 0.05. Thus the observed mean saving is statistically significant under the stated normal model and known covariance.
If the analyst falsely discarded pairing, the variance assigned to the difference of two sample means would be \(\frac{36}{25}+\frac{25}{25}=\frac{61}{25}\); the resulting standard error would be \(\frac{\sqrt{61}}5\approx1.56205\) minutes. That gives \(Z_{\mathrm{wrong}}=\frac{10}{\sqrt{61}}\approx1.28037\) and one-sided \(p\approx0.10021\), failing the same 5% test. This does not mean the actual measurements became less precise: the unpaired calculation simply neglects genuine positive covariance.
For an algebraic cross-check, write \(X_i=\mu_{\mathrm{before}}+U_i+e_i\) and \(Y_i=\mu_{\mathrm{after}}+U_i+f_i\) with mutually independent centered random terms having variances \(\operatorname{Var}(U_i)=24\), \(\operatorname{Var}(e_i)=12\), \(\operatorname{Var}(f_i)=1\). Then \(\operatorname{Var}(X_i)=36\), \(\operatorname{Var}(Y_i)=25\) and \(\operatorname{Cov}(X_i,Y_i)=24\), while \(D_i=\mu_{\mathrm{before}}-\mu_{\mathrm{after}}+e_i-f_i\) has variance 12+1=13. The shared driver effect \(U_i\) cancels exactly, as pairing is intended to exploit. This representation is an equivalent covariance model, not additional observed data.
Assumptions
The 25 before-after pairs are independent across drivers, and each pair has the stated joint normal distribution.
The population variance and covariance parameters are known, so a normal z-test rather than an estimated-variance t-test is stipulated.
The hypothesis concerns the mean of before-minus-after savings, not two independently sampled driver populations.
Final Answer
Paired \(\operatorname{SE}=\frac{\sqrt{13}}5\approx0.72111\) min, \(z=\frac{10}{\sqrt{13}}\approx2.77350\), one-sided \(p\approx0.00277\): reject \(H_0\). Incorrect independent \(\operatorname{SE}=\frac{\sqrt{61}}5\approx1.56205\), \(z=\frac{10}{\sqrt{61}}\approx1.28037\), \(p\approx0.10021\): do not reject.