Editing Likelihood-ratio test (section)

===An example===
The following example is adapted and abridged from {{Harvtxt|Stuart|Ord|Arnold|1999|loc=§22.2}}.

Suppose that we have a random sample, of size {{mvar|n}}, from a population that is normally-distributed. Both the mean, {{mvar|&mu;}}, and the standard deviation, {{mvar|&sigma;}}, of the population are unknown. We want to test whether the mean is equal to a given value, {{math|''&mu;''{{sub|0}} }}.

Thus, our null hypothesis is {{math|''H''{{sub|0}}:&nbsp; ''&mu;'' {{=}} ''&mu;''{{sub|0}}&nbsp;}} and our alternative hypothesis is {{math|''H''{{sub|1}}:&nbsp; ''&mu;'' ≠ ''&mu;''{{sub|0}}&nbsp;}}. The likelihood function is 
:<math>\mathcal{L}(\mu,\sigma \mid x) = \left(2\pi\sigma^2\right)^{-n/2} \exp\left( -\sum_{i=1}^n \frac{(x_i -\mu)^2}{2\sigma^2}\right)\,.</math>

With some calculation (omitted here), it can then be shown that 
:<math>\lambda_{LR} = n \ln\left[ 1 + \frac{t^2}{n-1}\right] </math> 
where {{mvar|t}} is the [[t-statistic|{{mvar|t}}-statistic]] with {{math|''n''&thinsp;&minus;&thinsp;1}} degrees of freedom. Hence we may use the known exact distribution of {{math|''t''{{sub|''n''&minus;1}}}} to draw inferences.