Felt weirdly simple at first and that made me second-guess myself.
Acknowledge that with a single observation, traditional distribution tests are impossible, so you must reframe the problem as a hypothesis test about the observed value's extremeness under the null. Propose using the probability integral transform or the likelihood under the null to compute a p-value, and discuss the limitations and decision criteria.
Pro tip: Emphasize that with n=1, any test has low power, so the focus should be on quantifying evidence (e.g., p-value) rather than making a definitive decision; also mention that the test is essentially checking if the value is an outlier under the null.
State the null hypothesis H0: the observation is drawn from a standard normal distribution (mean 0, variance 1). The alternative H1: it is drawn from some other distribution (e.g., normal with different mean/variance, or non-normal).
Since there is only one observation, the natural test statistic is the observed value itself (or its absolute value). Under H0, the distribution of this statistic is known: standard normal.
Calculate the two-sided p-value as 2 * (1 - Φ(|x|)), where Φ is the standard normal CDF. This represents the probability of observing a value as extreme or more extreme than x under H0.
Compare the p-value to a pre-specified significance level α (e.g., 0.05). If p-value < α, reject H0; otherwise, fail to reject H0. Note that with n=1, failing to reject does not mean H0 is true.
Acknowledge that this test has very low power and cannot detect many departures from normality. Mention that with only one observation, it's impossible to assess distributional shape; consider Bayesian approaches or additional data if possible.
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