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Suppose that a manufacturing process makes about \(3 \%\) defective items, which is considered satisfactory for this particular product. The managers would like to decrease this to about \(1 \%\) and clearly want to guard against a substantial increase, say to \(5 \%\). To monitor the process, periodically \(n=100\) items are taken and the number \(X\) of defectives counted. Assume that \(X\) is \(b(n=100, p=\theta)\). Based on a sequence \(X_{1}, X_{2}, \ldots, X_{m}, \ldots\), determine a sequential probability ratio test that tests \(H_{0}: \theta=0.01\) against \(H_{1}: \theta=0.05 .\) (Note that \(\theta=0.03\), the present level, is in between these two values.) Write this test in the form $$ h_{0}>\sum_{i=1}^{m}\left(x_{i}-n d\right)>h_{1} $$ and determine \(d, h_{0}\), and \(h_{1}\) if \(\alpha_{a}=\beta_{a}=0.02\).

Short Answer

Expert verified
Therefore, the given SPRT can be re-written as \( 50 > \sum_{i=1}^{m}(x_{i} - 100*0.01) > -0.02 \) where 'm' is the number of samples. 'd' is found to be 0.01, 'h_{0}' is 50 and 'h_{1}' is -0.02.

Step by step solution

01

Understanding the Sequential Probability Ratio Test (SPRT)

The SPRT is a statistical method used in hypothesis testing to determine when we can decide that a hypothesis is correct based on the available evidence. It involves comparing the likelihood ratio to two threshold values, which represent the acceptable probabilities of type I and type II errors. When the likelihood ratio crosses one of these thresholds, a decision is made.
02

Formulation SPRT for the given conditions

In our problem, the form of SPRT is given as: \[ h_{0} > \sum_{i=1}^{m}(x_{i} - nd) > h_{1} \] where \(x_{i}\) represents the number of defective items in each sample, 'n' is the size of the sample, and 'd' is a constant we need to find. The values, \(h_{0}\) and \(h_{1}\), are threshold values that we also need to determine. The values for \(h_{0}\) and \(h_{1}\) can be found using statistical calculations. We know that \[h_{0} = \frac{1}{\beta_{a}} \quad and \quad h_{1} = -\alpha_{a}\] With the given values of \(\alpha_{a} = \beta_{a} = 0.02\), we therefore have \(h_{0} = 50\) and \(h_{1} = -0.02\). Note: \(h_{0}\) and \(h_{1}\) might vary as per the acceptance and false alarm probability.
03

Finding 'd'

The parameter 'd' is often chosen to be a value that minimizes the expected sample size of the test under the null hypothesis. In the case where 'p' is the probability of occurrence under the null hypothesis, 'd' is typically set to 'p'. So in this case, \(d = 0.01\).

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid with \(\operatorname{pmf} f(x ; p)=p^{x}(1-p)^{1-x}, x=0,1\), zero elsewhere. Show that \(C=\left\\{\left(x_{1}, \ldots, x_{n}\right): \sum_{1}^{n} x_{i} \leq c\right\\}\) is a best critical region for testing \(H_{0}: p=\frac{1}{2}\) against \(H_{1}: p=\frac{1}{3} .\) Use the Central Limit Theorem to find \(n\) and \(c\) so that approximately \(P_{H_{0}}\left(\sum_{1}^{n} X_{i} \leq c\right)=0.10\) and \(P_{H_{1}}\left(\sum_{1}^{n} X_{i} \leq c\right)=0.80\).

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