# How to eliminate constant to derive the decision rule in terms of the sufficient statistic $\bar{X}$ for normal distribution means hypothesis test?

Suppose that we have a random sample, of size $$n$$, from a population that is normally-distributed. Both the mean, $$\mu$$, and the standard deviation, $$\sigma$$, of the population are unknown. We want to test whether the mean is equal to a given value, $$\mu_{0}$$. Thus, our null hypothesis is $$H_{0}: \mu=\mu_{0}$$ and our alternative hypothesis is $$H_{1}: \mu \neq \mu_{0}$$. The likelihood function is $$\mathcal{L}(\mu, \sigma \mid x)=\left(2 \pi \sigma^{2}\right)^{-n / 2} \exp \left(-\sum_{i=1}^{n} \frac{\left(x_{i}-\mu\right)^{2}}{2 \sigma^{2}}\right)$$ With some calculation (omitted here), it can then be shown that $$\lambda=\left(1+\frac{t^{2}}{n-1}\right)^{-n / 2}$$ where $$t$$ is the $$t$$-statistic with $$n-1$$ degrees of freedom. Hence we may use the known exact distribution of $$t_{n-1}$$ to draw inferences.

The likelihood-ratio test provides the decision rule as follows: If $$\Lambda>c$$, do not reject $$H_{0}$$; If $$\Lambda, reject $$H_{0}$$ Reject with probability $$1$$ if $$\Lambda=c$$. Reject with probability 1 if $$\Lambda. The values $$c$$ and $$q$$ are usually chosen to obtain a specified significance level $$\alpha$$, via the relation $$\cdot \mathrm{P}\left(\Lambda=c \mid H_{0}\right)+\mathrm{P}\left(\Lambda

To determine the decision rule/region region for this test

$$P( \lambda = c|\mu=\mu_0) + P(\lambda

$$P(t^2\leq(c^{\frac{-2}{n}}-1)(n-1)| \mu=\mu_{0})=\alpha$$

$$P(({\dfrac{\bar{X}-\mu_0}{\dfrac{s}{\sqrt{n}}}})^2 \leq c^{\dfrac{-2}{n}}(n-1))=\alpha$$

How do I eliminate "$$c$$" to make the decision rule or rejection region only depend on the sufficient statistic, $$\mu_0$$, the sample variance $$s^2$$, and sample-size, $$n$$?

• $\lambda=\Lambda$. t is the $n-1$ degrees of freedom t-distribution statistic
– user318514
Commented Jul 31, 2022 at 21:50

The statistic $$t$$ (and hence $$\lambda$$ also) is continuous. The probability of equality is $$0$$.

That equality term is important for test statistics that are not continuous, where that probability is not zero. You can safely focus on the inequality in this case.

Note that you can simplify your inequality by moving things that are not $$t^2$$ across to the other side of the "$$<$$" symbol (take the same power of both sides, multiply both sides by the relevant constant, etc). You'll get some new constant on the right, $$c_1$$, say, in terms of something with a known distribution. Once you know $$c_1$$, you can compute $$c$$ (if you need it at all).

• I may ignore $n-1$ term?
– user318514
Commented Jul 31, 2022 at 23:20
• Not ignore, no. I don't suggest that anywhere in my answer. Use algebraic manipulations; be mindful of when the direction of the inequality changes. The $n-1$ term will be important if you are attempting to find $c$. Commented Jul 31, 2022 at 23:21
• we have to break the $1+t^2...$ out of the probability operator before we can move non-t terms to the R.H.S.
– user318514
Commented Jul 31, 2022 at 23:27
• I am talking about manipulations inside $P(.)$. That is rearrange the event $(1+\dfrac{t^2}{n-1})^{-\dfrac{n}{2}} <c$ so that it has $t^2$ isolated. Still the same event, but in terms of something whose distribution you know. Commented Jul 31, 2022 at 23:47
• reduce to reciprocal of $P(\dfrac{1}{exp(t^2)}<c)$ but this doesn't work well if we assume n is small.
– user318514
Commented Aug 1, 2022 at 0:03