Let $X_1,X_2,…,X_n$ be a random sample from the normal distribution with mean $\mu$ and known variance $\sigma^2$.

Consider testing the simple versus simple hypotheses

\[H_0:\mu=3\\ H_1:\mu=5\]

Definition/Notation:

\[\begin{align} \text{Let}\quad&=P(\text{Type I Error})\\ &=P(\text{Reject $H_0$ when it's true})\\ &=P(\text{Reject $H_0$ when $\mu=3$}) \end{align}\]

$\alpha$ is called the level of significance of the test.

It is also sometimes referred to as the size of the test.

Developing a Test

Step One:

Choose an estimator for $\mu$.

\[\widehat{\mu}=\overline{X}\] Step Two:

The form of the test:

  • We are looking for evidence that $H_1$ is true.
  • The $N(3,/sigma^2)$ takes on values from $-\infty$ to $\infty$.
  • $\overline{X}\sim N(\mu,\sigma^2)\implies\overline{X}$ also takes on values from $-\infty$ to $\infty$.
  • It is entirely possible that $\overline{X}$ is very large even if the mean of its distribution is $3$.
  • However, if $\overline{X}$ is very large, it will start to seem more likely that $\mu$ is larger than $3$.
  • Eventually, a population mean of $5$ will seem more likely than a population mean of $3$.

Give the “form” of test.

Reject $H_0$, in favor of $H_1$, if $\overline{X}\gt c$ for some $c$ to be determined.

Step Three:

Find $c$.

Reject $H_0$, in favor of $H_1$, if $\overline{X}\gt c$.

  • If $c$ is too large, we are making it difficult to reject $H_0$.

    We are more likely to fail to reject when it should be rejected.

    This is Type II Error.

  • If $c$ is too small, we are making it too easy to reject $H_0$.

    We are more like reject when it should not be rejected.

    This is Type I Error.

This is where $\alpha$ comes in.

\[\begin{align} \alpha&=P(\text{Type I Error})\\ &=P(\text{Reject $H_0$ when true})\\ &=P(\overline{X}\gt c\quad\text{when}\quad\mu=3) \end{align}\] Step Four:

Give a conclusion!

Example

\[X_1,X_2,...,X_{10}\stackrel{iid}{\sim}N(\mu,4)\]

Find a hypothesis test for

\[H_0:\mu=5\quad\text{vs}\quad H_1:\mu=3\]

Use level of significance $\alpha=0.05$.

Step One:

Choose an estimator for $\mu$.

\[\widehat{\mu}=\overline{X}\] Step Two:

Give the “form” of the test.

Reject $H_0$, in favor of $H_1$, if $\overline{X}\lt c$ for some $c$ to be determined.

Step Three: \[\begin{align} 0.05&=P(\text{Type I Error})\\ &=P(\text{Reject $H_0$ when true})\\ &=P(\overline{X}\lt c\text{ when }\mu=5)\\ &=P\Bigg(\frac{\overline{X}-\mu_0}{\sigma/\sqrt{n}}\lt\frac{c-5}{2/\sqrt{10}}\text{ when }\mu=5\Bigg) \end{align}\]

We have

\[0.05=P\Bigg(Z\lt\frac{c-5}{2/\sqrt{10}}\Bigg)\\ \text{cut off point = pnorm(0.05)=-1.645}\\ \begin{align} &\implies \frac{c-5}{2/\sqrt{10}} = -1.645\\ &\implies c=3.9596 \end{align}\] Step Four:

Give a conclusion.

Reject $H_0$, in favor of $H_1$, if

\[\overline{X}\lt3.9596\]

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