Question

Let the independent random variables \(Y\) and \(Z\) be \(N\left(\mu_{1}, 1\right)\) and \(N\left(\mu_{2}, 1\right)\), respectively. Let \(\theta=\mu_{1}-\mu_{2} .\) Let us observe independent observations from each distribution, say \(Y_{1}, Y_{2}, \ldots\) and \(Z_{1}, Z_{2}, \ldots .\) To test sequentially the hypothesis \(H_{0}: \theta=0\) against \(H_{1}: \theta=\frac{1}{2}\), use the sequence \(X_{i}=Y_{i}-Z_{i}, i=1,2, \ldots .\) If \(\alpha_{a}=\beta_{a}=0.05\), show that the test can be based upon \(\bar{X}=\bar{Y}-\bar{Z} .\) Find \(c_{0}(n)\) and \(c_{1}(n)\)

Short answer

The test can be based upon \(\bar{X}=\bar{Y}-\bar{Z}\). The values of \(c_{0}(n)\) and \(c_{1}(n)\) can be calculated as \(c_{0}(n)= \mu_{0} + 1.645/\sqrt{n}\) under \(H_{0}\) and \(c_{1}(n)= \mu_{1} - 1.645/\sqrt{n}\) under \(H_{1}\).

Step-by-step solution

Show that the test can be based upon \(\bar{X}=\bar{Y}-\bar{Z}\)

The difference of observations from two independent normal distributions will also be a normal distribution. Hence, \(X_{i}=Y_{i}-Z_{i}\) is also normal. The sample mean (\(\bar{X}\)) for this distribution can be shown to be \(\bar{X}=\bar{Y}-\bar{Z}\) as we are taking the mean of the differences of the variables.

Calculation of \(c_{0}(n)\) and \(c_{1}(n)\)

By definition, under \(H_{0}\), \(\alpha(n) = P(\bar{X} > c_{0}(n)|H_{0})\) and under \(H_{1}\), \(\beta(n) = P(\bar{X} < c_{1}(n)|H_{1})\). Here, \(\alpha_{a}=\beta_{a}=0.05\) is the error rate and supposed to equal \(\alpha(n)\) and \(\beta(n)\) at some \(n\). Since \(\bar{X}\) is normally distributed, we can use the z-score transformation and standard normal distribution table to find the values of \(c_{0}(n)\) and \(c_{1}(n)\). For a significance level of 0.05, the Z score is approximately 1.645 (for \(\alpha_{a}\)) and -1.645 (for \(\beta_{a}\)). So, \(c_{0}(n)= \mu_{0} + 1.645/\sqrt{n}\) under \(H_{0}\) and \(c_{1}(n)= \mu_{1} - 1.645/\sqrt{n}\) under \(H_{1}\), where \(\mu_{0}\) and \(\mu_{1}\) are means under \(H_{0}\) and \(H_{1}\) respectively, and \(n\) is the sample size.