Discussion
If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from a normally distributed population with mean \(\mu\) and variance \(\sigma^{2}\), then the following follows a standard normal distribution :
Answer: D
The standard normal statistic is \(\dfrac{\bar{X} - u}{\sigma/\sqrt{n}}\) (the paper writes \(u\) for the mean \(\mu\)).
For a random sample from \(N(\mu, \sigma^{2})\), the sample mean is \(N(\mu, \sigma^{2}/n)\). Subtracting its mean and dividing by its standard deviation \(\sigma/\sqrt{n}\) gives a standard normal variable.
Dividing by \(\sigma^{2}\) instead of the standard error gives the wrong scale. The squared forms are non-negative, so they cannot be normal; \((\bar{X} - \mu)^{2}/(\sigma^{2}/n)\) would be chi-square with one degree of freedom, and the reciprocal form has no standard distribution here.
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