Question

Consider a normal distribution of the form \(N(\theta, 4)\). The simple hypothesis \(H_{0}: \theta=0\) is rejected, and the alternative composite hypothesis \(H_{1}: \theta>0\) is accepted if and only if the observed mean \(\bar{x}\) of a random sample of size 25 is greater than or equal to \(\frac{2}{5}\). Find the power function \(\gamma(\theta), 0 \leq \theta\), of this test.

Short answer

The power Function is given by \( \gamma(\theta) = 1- \Phi(5\theta - 2)\), where \( \Phi(\cdot)\) denotes the cumulative distribution function of a standard normal distribution.

Step-by-step solution

Realize what the standard form of distribution is

The given distribution is \(N(\theta, 4)\), a normal distribution with mean \(\theta\) and variance 4. In usual practice, while computing probabilities involved in a normal distribution, we deal with standard normal distribution, which is \(N(0,1)\). Any normal distribution \(N(\mu, \sigma^2)\) can be standardized by using the transformation \(Z = \frac{X - \mu}{\sigma}\), where \(Z\) follows \(N(0, 1)\) distribution.

Determine the rejection region

Since the null hypothesis is rejected when observed mean \( \bar{x} \) is greater than or equal to \( \frac{2}{5} \), the rejection region \(R\) in the standardized normal distribution is given by \( R: Z \geq \frac{\frac{2}{5} - \theta}{\frac{2}{\sqrt{25}}} = \frac{\frac{2}{5} - \theta}{\frac{2}{5}} \).

Compute the Power function

The power function \(\gamma(\theta)\) is the probability of rejecting the null hypothesis given that the true value of \(\theta\) is \(\theta\). In terms of standardized normal distribution, this is the probability that \(Z\) lies in the rejection region \(R\). This is given by \( \gamma(\theta) = P(Z \geq \frac{\frac{2}{5} - \theta}{\frac{2}{5}}) = 1 - \Phi(\frac{\frac{2}{5} - \theta}{\frac{2}{5}}) \), where \(\Phi(\cdot)\) denotes the cumulative distribution function of a standard normal distribution.

Simplify the Power function

\(\gamma(\theta) = 1 - \Phi(\frac{\frac{2}{5} - \theta}{\frac{2}{5}})\), after simplification, we get \( \gamma(\theta) = 1- \Phi(5\theta - 2)\) where \( \Phi(\cdot)\) denotes the cumulative distribution function of a standard normal distribution. This is the power function \(\gamma(\theta), 0 \leq \theta \) of this test.