Question
A study tests whether a treatment raises a normally distributed measurement above its known baseline. Under the null hypothesis, the four planned patient measurements \(X_1,\dots,X_4\) are independent \(N(0,1)\); under the alternative they have positive common mean and the same known variance.
Investigators are allowed to look once after two patients and, if they have not rejected, again after four. They use the cumulative statistics
\[Z_1=(X_1+X_2)/\sqrt2,\qquad Z_2=(X_1+X_2+X_3+X_4)/2.\]
At either look they reject for \(Z_i>c\) and stop testing. A single-look one-sided 5% threshold is \(c\approx1.64485\).
If that unadjusted threshold is used at both looks, what is the overall false-positive probability? Find a common threshold \(c\) that makes the overall type-I error 5%, accounting for the fact that the final statistic includes the first two observations. A numerical answer with a justified integral is sufficient.
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Hint 1
The rejection event is a union: first-look rejection or final-look rejection. The two events overlap, because the final statistic contains the interim data.
Check your answer
Solution
Under the null, both \(Z_1\) and \(Z_2\) are standard normal. Their overlap in the first two observations gives
\[ \operatorname{Cov}(Z_1,Z_2)=\frac{\operatorname{Var}(X_1+X_2)}{2\sqrt2}=\frac1{\sqrt2}=:\rho. \]
Thus the statistics are jointly normal with correlation \(\rho=1/\sqrt2\), not independent. The trial rejects if \(Z_1>c\) or \(Z_2>c\); stopping at the first rejection does not change this union event. By inclusion–exclusion, under the null
\[ \begin{aligned} \alpha(c)&=P(Z_1>c\text{ or }Z_2>c)\\ &=2[1-\Phi(c)]-P(Z_1>c,Z_2>c). \end{aligned} \]
The conditional normal distribution is \(Z_2\mid Z_1=z\sim N(\rho z,1-\rho^2)\), where \(1-\rho^2=1/2\). Hence an explicit one-dimensional integral for the joint tail is
\[ P(Z_1>c,Z_2>c)=\int_c^\infty\phi(z)\,\Phi\!\left(\frac{\rho z-c}{\sqrt{1-\rho^2}}\right)dz =\int_c^\infty \phi(z)\Phi(z-\sqrt2c)\,dz. \]
Here \(\phi\) and \(\Phi\) are the standard normal density and CDF. At the unadjusted one-sided 5% cutoff \(c=\Phi^{-1}(0.95)\approx1.6448536\), each marginal rejection probability is 0.05, and numerical quadrature gives a joint upper-tail probability about 0.0199245. Thus \(\alpha(c)=0.10-0.0199245=\boxed{0.0800755}\), approximately 8.01%.
For a common boundary with total error exactly 0.05, solve the displayed continuous, strictly decreasing union-tail function \(\alpha(c)=0.05\). Numerical integration and one-dimensional root finding give \(\boxed{c\approx1.8754233}\). The resulting false-positive probability is 0.050000 to the shown precision. As a conservative independent check, using \(\Phi^{-1}(0.975)\approx1.959964\) at both looks bounds the union by 0.05 via Bonferroni; the actual correlated-test error there is about 0.041559. No independent-look approximation is used.
Assumptions
The same four observations are planned before the trial and are iid \(N(0,1)\) under \(H_0\).
The null variance is known, the alternative mean is strictly positive, and rejection is one-sided.
The test stops after interim rejection and otherwise uses all four observations at its final look.
The same common critical value is used at both looks; exact familywise type-I error is measured under \(H_0\).
Final Answer
The two statistics have null correlation \(1/\sqrt2\). Reusing the one-look cutoff \(1.64485\) at both looks gives false-positive probability approximately \(0.08008\) (8.01%). An exactly calibrated common boundary is \(c\approx1.87542\), giving overall 5% type-I error.