Question

Consider a distribution having a pmf of the form \(f(x ; \theta)=\theta^{x}(1-\theta)^{1-x}, x=\) 0,1, zero elsewhere. Let \(H_{0}: \theta=\frac{1}{20}\) and \(H_{1}: \theta>\frac{1}{20} .\) Use the Central Limit Theorem to determine the sample size \(n\) of a random sample so that a uniformly most powerful test of \(H_{0}\) against \(H_{1}\) has a power function \(\gamma(\theta)\), with approximately \(\gamma\left(\frac{1}{20}\right)=0.05\) and \(\gamma\left(\frac{1}{10}\right)=0.90\)

Short answer

The main steps are understanding the Central Limit Theorem, understanding power functions, transforming the hypotheses and power probabilities into Z-scores, setting up a system of equations based on these Z-scores, and then solving this system for the sample size. Rounding up to the nearest whole number will ensure an integer sample size. Empirical verification of the hypothesis also required.

Step-by-step solution

Understanding the question

The problem involves hypothesis testing for a probability mass function (pmf) \(f(x ; \theta)=\theta^{x}(1-\theta)^{1-x}, x=\{ 0,1\}\). The central limit theorem (CLT) is used to determine the sample size \(n\) which provides a uniformly most powerful test between the two hypotheses.

Writing down the power function of the test

Start by writing down the power function of the test \(\gamma(\theta)\). There is a relationship between the sample size \(n\) and the power of a test. The power function is given here as \(\gamma\left(\frac{1}{20}\right)=0.05\) and \(\gamma\left(\frac{1}{10}\right)=0.90\).

Apply the Central Limit theorem

In the Central Limit Theorem, the sum of a large number of independent and identically distributed variables will be approximately normally distributed. We can use this to transform our hypotheses concerning \(\theta\) into hypotheses concerning the sample mean \(\bar{X}\). Specifically, if we consider the statistic \(Z= \sqrt{n}(\bar{X}-E(X))/\sqrt{Var(X)}\), as \(n\) becomes large, \(Z\) tends to a standard normal distribution.

Transform the power probabilities into Z-scores

The Z-scores corresponding to the probabilities of \(0.05\) and \(0.90\) can be found from the standard normal distribution table or using a computation tool. The corresponding Z-scores are \(Z_{0.05}= -1.645\) and \(Z_{0.90}= 1.28\) respectively.

Setting up the system of equations

Having the transformed hypotheses and the z-scores, we can formulate the following system of equations: 1) \(\sqrt{n}\left(\frac{1}{20} - E(X)\right) / \sqrt{Var(X)} = -1.645\), 2) \(\sqrt{n}\left(\frac{1}{10} - E(X)\right) / \sqrt{Var(X)} = 1.28\)

Solving system of equations for sample size \(n\)

To find the required sample size \(n\), the system of equations can be solved. Since \(E[X] = \theta\) and \(Var(X)=\theta(1-\theta)\) for a Bernoulli distribution, we can substitute these values into our equations and solve for \(n\). The sample size \(n\) will likely be a real number, but since we can't have a fraction of a sample, always round up to the nearest whole number.

Empirical verification

Once you have determined the sample size, you would then carry out the hypothesis test, determining whether to accept or reject the null hypothesis based on your samples.