Hypothesis tests, size and power

A single observation from an exponential distribution with mean θ is used to test $H_0:\theta=2$ against $H_1:\theta=5$, rejecting when $X>4$. Since $P(X>k\mid\theta)=e^{-k/\theta}$,

For a test of size exactly 0.05, solve $e^{-k/2}=0.05$: $k=-2\log 0.05\approx5.991$.

For $X_1,\dots,X_n\sim\mathrm{Exp}(\lambda)$ the likelihood is $\lambda^n e^{-\lambda Y}$ with $Y=\sum X_i$, maximised at $\hat\lambda=n/Y$. The likelihood ratio for $H_0:\lambda=\lambda_0$ depends on the data only through Y, and under H₀

where a and b are the α/2 and 1 − α/2 quantiles of that gamma distribution. With n = 50 and λ₀ = 1 the 2020 answer found a ≈ 37.11 and b ≈ 64.78.

The sign test makes almost no assumptions: under $H_0: m = m_0$ each observation is above the median with probability ½, so the number of + signs is $Y\sim\mathrm{Bin}(n,\tfrac12)$. Testing whether the median daily count was below 15, only small values of Y are evidence against H₀:

Fourteen of the 21 days had at least 15 new cases, so there is no evidence that the median was below 15 — although the counts were clearly falling, which the test (treating days as a random sample) ignores.

The Wilcoxon rank-sum test compares two groups through the ranks of the pooled sample. R reports$\;W=\sum_{i\in X}\operatorname{rank}(x_i)-\tfrac{n_X(n_X+1)}{2}\;$, the number of (x, y) pairs with x > y (ties count ½). With ties in the data R switches to a normal approximation with a tie-corrected variance and a continuity correction: