Solution. Hypotheses. \(H_0 : F = F_0\) where \(F_0\) is the Uniform\([0,1]\) cdf, against \(H_1 : F \neq F_0\). Note \(F_0\) is fully specified in advance, with no parameter estimated from the data — which is what makes the tabulated critical values applicable.

The theoretical cdf. For the uniform distribution on \([0,1]\), \(F_0(x) = x\) on that interval. The arithmetic is therefore unusually transparent: the theoretical value at each point is the observation.

Order the sample and tabulate.

The statistics. \[D^{+} = \max _i\left \{\dfrac {i}{n}-F_0\left (x_{(i)}\right )\right \} = 0.290 \quad (\text {at } i = 10),\] \[D^{-} = \max _i\left \{F_0\left (x_{(i)}\right )-\dfrac {i-1}{n}\right \} = 0.229 \quad (\text {at } i = 2),\] \[D = \max \left (D^{+}, D^{-}\right ) = 0.290 .\]

Critical value and decision. For \(n = 10\) at \(\alpha = 0.05\) the two-sided critical value is \(D_{0.05,10} = 0.409\). Since \[0.290 < 0.409,\] we do not reject \(H_0\): the sample is consistent with Uniform\([0,1]\).

Why both \(D^{+}\) and \(D^{-}\) are needed. The empirical distribution function is a step function, so at each observation it jumps from \(\frac {i-1}{n}\) to \(\frac {i}{n}\). The largest gap between \(\widehat {F}_n\) and \(F_0\) may occur immediately after a jump, which \(D^{+}\) measures, or immediately before one, which \(D^{-}\) measures. Taking only one of them would understate the distance — here \(D^{+}\) happens to win, but the maximum deviation of the two occurs at opposite ends of the sample, \(i = 10\) and \(i = 2\).