Answer

If the null parameter space is \(\Omega = {\mu: \mu = 3}\), then the alternative parameter space is everything that is in \(\Omega = {\mu: −∞ < \mu < ∞}\) that is not in \(\Omega\). That is, the alternative parameter space is \(\Omega ' = {\mu: \mu ≠ 3}\). And, so the alternative hypothesis is:

\(H_A : \mu \ne 3\)

In this case, we'd be interested in deriving a two-tailed test.

Answer

If the alternative parameter space is (\omega ' = {\mu: \mu > 3}\), then the null parameter space is \(\omega = {\mu: \mu ≤ 3}\). And, so the null hypothesis is:

\(H_0 : \mu \le 3\)

Now, the reality is that some authors do specify the null hypothesis as such, even when they mean \(H_0: \mu = 3\). Ours don't, and so we won't. (That's why I put that "technically" in parentheses up above.) At any rate, in this case, we'd be interested in deriving a one-tailed test.

Answer

Because we are interested in testing the null hypothesis \(H_0: \mu = 10\) against the alternative hypothesis \(H_A: \mu ≠ 10\) for a normal mean, our total parameter space is:

\(\Omega =\left \{\mu : -\infty < \mu < \infty \right \}\)

and our null parameter space is:

\(\omega = \left \{10\right \}\)

Now, to find the likelihood ratio, as defined above, we first need to find \(L(\hat{\omega})\). Well, when the null hypothesis \(H_0: \mu = 10\) is true, the mean \(\mu\) can take on only one value, namely, \(\mu = 10\). Therefore:

\(L(\hat{\omega}) = L(10)\)

We also need to find \(L(\hat{\Omega})\) in order to define the likelihood ratio. To find it, we must find the value of \(\mu\) that maximizes \(L(\mu)\). Well, we did that back when we studied maximum likelihood as a method of estimation. We showed that \(\hat{\mu} = \bar{x}\) is the maximum likelihood estimate of \(\mu\). Therefore:

\(L(\hat{\Omega}) = L(\bar{x})\)

Now, putting it all together to form the likelihood ratio, we get:

which simplifies to:

Now, let's step aside for a minute and focus just on the summation in the numerator. If we "add 0" in a special way to the quantity in parentheses:

we can show that the summation can be written as:

\(\sum_{i=1}^{n}(x_i - 10)^2 = \sum_{i=1}^{n}(x_i - \bar{x})^2 + n(\bar{x} -10)^2 \)

Therefore, the likelihood ratio becomes:

which greatly simplifies to:

\(\lambda = exp \left [-\dfrac{n}{4}(\bar{x}-10)^2 \right ]\)

Now, the likelihood ratio test tells us to reject the null hypothesis when the likelihood ratio \(\lambda\) is small, that is, when:

\(\lambda = exp\left[-\dfrac{n}{4}(\bar{x}-10)^2 \right] \le k\)

where k is chosen to ensure that, in this case, \(\alpha = 0.05\). Well, by taking the natural log of both sides of the inequality, we can show that \(\lambda ≤ k\) is equivalent to:

\( -\dfrac{n}{4}(\bar{x}-10)^2 \le \text{ln} k \)

which, by multiplying through by −4/n, is equivalent to:

\((\bar{x}-10)^2 \ge -\dfrac{4}{n} \text{ln} k \)

which is equivalent to:

\(\dfrac{|\bar{X}-10|}{\sigma / \sqrt{n}} \ge \dfrac{\sqrt{-(4/n)\text{ln} k}}{\sigma / \sqrt{n}} =k* \)

Aha! We should recognize that quantity on the left-side of the inequality! We know that:

\(Z = \dfrac{\bar{X}-10}{\sigma / \sqrt{n}} \)

follows a standard normal distribution when \(H_0: \mu = 10\). Therefore we can determine the appropriate \(k^*\) by using the standard normal table. We have shown that the likelihood ratio test tells us to reject the null hypothesis \(H_0: \mu = 10\) in favor of the alternative hypothesis \(H_A: \mu ≠ 10\) for all sample means for which the following holds:

\(\dfrac{|\bar{X}-10|}{ \sqrt{2} / \sqrt{n}} \ge z_{0.025} = 1.96 \)

Doing so will ensure that our probability of committing a Type I error is set to \(\alpha = 0.05\), as desired.