Simple versus Composite

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

\[H_0:\mu=\mu_0\qquad H_1:\mu\lt\mu_0\]

where $\mu_0$ is fixed and known.

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 $\overline{X}\lt c$, where $c$ is to be determined.

Step Three:

Find $c$.

\[\begin{align} \alpha&=\underset{\mu=\mu_0}{\text{max}}P(\text{Type I Error})\\ &=\underset{\mu=\mu_0}{\text{max}}P(\text{Reject }H_0;\mu)\\ &=P(\text{Reject }H_0;\mu_0)\\ &=P(\overline{X}\lt c;\mu_0) \end{align}\]

We know that $\overline{X}\sim N(\mu_0,\sigma^2/n)$, we transform $overline{X}$ to Standard Normal distribution:

\[\begin{align} \alpha&=P\Bigg(\frac{\overline{X}-\mu_0}{\sigma/\sqrt{n}}\lt\frac{c-\mu_0}{\sigma/\sqrt{n}};\mu_0\Bigg)\\ &=P\Bigg(Z\lt\frac{c-\mu_0}{\sigma/\sqrt{n}}\Bigg) \end{align}\]

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\[\implies \frac{c-\mu_0}{\sigma/\sqrt{n}} = z_{1-\alpha}\\ \implies c=\mu_0 + z_{1-\alpha}\frac{\sigma}{\sqrt{n}}\] Step Four:

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

$$ \overline{X}\lt\mu_0+z_{1-\alpha}\frac{\sigma}{\sqrt{n}} $$

NOTE:

\[\begin{align} \beta&=\underset{\mu\lt\mu0}{\text{max}}P(\text{Type II error)}\\ &=\underset{\mu\in H_1}{\text{max}}P(\text{Fail to Reject H_0};\mu)\\ &=\underset{\mu\lt\mu0}{\text{max}}P\Bigg(\overline{X}\ge\mu_0+z_{1-\alpha}\frac{\sigma}{\sqrt{n}};\mu\Bigg)\\ &=\underset{\mu\lt\mu0}{\text{max}}P\Bigg(Z\ge\frac{\mu_0+z_{1-\alpha}\frac{\sigma}{\sqrt{n}}-\mu}{\sigma/\sqrt{n}}\Bigg)\\ &=\underset{\mu\lt\mu0}{\text{max}}\Bigg[1-\phi\Bigg(\frac{\mu_0+z_{1-\alpha}\frac{\sigma}{\sqrt{n}}-\mu}{\sigma/\sqrt{n}}\Bigg)\Bigg] \quad\text{(1)} \end{align}\]

$(1)$ get increasing in $\mu$, so in order to maximize $(1)$, we maximize the largest possible in $\mu$. We are going to maximize overall $\mu\lt\mu_0$, which means we get to plug in $\mu_0$, we get:

\[\begin{align} \beta&=1-\phi\Bigg(\frac{\mu_0+z_{1-\alpha}\frac{\sigma}{\sqrt{n}}-\mu_0)}{\sigma/\sqrt{n}}\Bigg)\\ &=1 - \phi(z_{1-\alpha})\\ &=1-\alpha \end{align}\]

Composite versus Composite

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

\[H_0:\mu\ge\mu_0\qquad H_1:\mu\lt\mu_0\]

where $\mu_0$ is fixed and known.

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 $\overline{X}\lt c$, where $c$ is to be determined.

Step Three:

Find $c$.

\[\begin{align} \alpha&=\underset{\mu\ge\mu_0}{\text{max}}P(\text{Type I Error})\\ &=\underset{\mu\ge\mu_0}{\text{max}}P(\text{Reject }H_0;\mu)\\ &=\underset{\mu\ge\mu_0}{\text{max}}P(\overline{X}\lt c;\mu_0)\\ &=\underset{\mu\ge\mu_0}{\text{max}}P\Bigg(Z\lt\frac{c-\mu}{\sigma/\sqrt{n}}\Bigg)\\ &=\underset{\mu\ge\mu_0}{\text{max}}\quad\phi\Bigg(\underbrace{\frac{c-\mu}{\sigma/\sqrt{n}}}_{\text{decreasing in }\mu}\Bigg)\\ \end{align}\]

This is a decreasing function of $\mu$. It’s going down as $\mu$ gets larger. If we want to be largest (max), we need to take the smallest $\mu$, so $\mu=\mu_0$. We have

\[\begin{align} &\alpha=\phi\Bigg(\frac{c-\mu_0}{\sigma/\sqrt{n}}\Bigg)\\ &\implies \frac{c-\mu_0}{\sigma/\sqrt{n}}=z_{1-\alpha}\\ &\implies c=\mu_0+z_{1-\alpha}\frac{\sigma}{\sqrt{n}} \end{align}\] Step Four:

Conclusion:

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

$$ \overline{X}\lt\mu_0+z_{1-\alpha}\frac{\sigma}{\sqrt{n}} $$

Example

In 2019, the average health care annual premium for a family of 4 in the United States, was reported to be $6,015.

In a more recent survey, 100 randomly sampled families of 4 reported an average annual health care premium of $6,537.

Can we say that the true average is currently greater that $6,015 for all families of 4?

Assume that annual health care premiums are normally distributed with a standard deviation of $814.

Let $\mu$ be the true average for all families of 4.

Step Zero:

Set up the Hypotheses.

\[H_0:\mu=6015\qquad H_1:\mu\gt6015\]

Decide on a level of significance.

\[\alpha=0.10\] 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

\[\overline{X}\gt c\]

for some $c$ to be determined.

Step Three:

Find $c$.

\[\begin{align} \alpha&=\underset{\mu=\mu_0}{\text{max}}P(\text{Type I Error};\mu)\\ &=P(\text{Type I Error};\mu_0)\\ &=P(\overline{X}\gt c;\mu_0)\\ &=P\Bigg(\frac{\overline{X}-\mu_0}{\sigma/\sqrt{n}}\gt\frac{c-6015}{814/\sqrt{100}};\mu_0\Bigg)\\ &=P\Bigg(Z\gt\frac{c-6015}{814/\sqrt{100}}\Bigg) \end{align}\]

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We can use qnorm(0.90)=1.28 to calculate the critical value.

\[\begin{align} &\implies \frac{c-6015}{814/\sqrt{100}}=1.28\\ &\implies c=6119.19 \end{align}\] Step Four:

Conclusion.

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

\[\overline{X}\gt6119.19\]

From the data, where $\overline{x}=6537$, we reject $H_0$ in favor of $H_1$.

The data suggests that the true mean annual health care premium is greater than $6015.