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Let \(X\) be \(N(0, \theta)\) and, in the notation of this section, let \(\theta^{\prime}=4, \theta^{\prime \prime}=9\), \(\alpha_{a}=0.05\), and \(\beta_{a}=0.10 .\) Show that the sequential probability ratio test can be based upon the statistic \(\sum_{1}^{n} X_{i}^{2} .\) Determine \(c_{0}(n)\) and \(c_{1}(n)\).

Short Answer

Expert verified
For the sequential probability ratio test based on the statistic \(\sum_{1}^{n} X_{i}^{2}\), the decision boundaries are \(c_{1}(n)=9.5\) and \(c_{0}(n)=0.05556\). reject \(H_{0}\) if \(L_n > 9.5\), and do not reject \(H_{0}\) if \(L_n<0.05556\). Continue with the test if the value of \(L_n\) lies between these two boundaries.

Step by step solution

01

SPRT Setup

We want to test the null hypothesis \(H_{0}: \theta = \theta^{\prime} = 4\) against the alternative hypothesis \(H_{1}: \theta = \theta^{\prime \prime} = 9\). The likelihood ratio for \(n\) observations \(X_{1},... ,X_{n}\) under \(H_{0}\) and \(H_{1}\) is given by the ratio of the densities, raised to the power \(n\), and multiplied by the exponential of the statistic. In this case, the likelihood ratio \(L_n\) is: \(L_n = \left( \frac{\theta^{\prime}}{\theta^{\prime \prime}} \right)^{\frac{n}{2}} \mathrm{exp}\left( \frac{1}{2} \left( \frac{1}{\theta^{\prime}} - \frac{1}{\theta^{\prime \prime}} \right) \sum_{1}^{n} X_{i}^{2} \right)\)
02

Decision Boundaries

The decision boundaries \(c_{0}(n)\) and \(c_{1}(n)\) can be computed using the formulae for SPRT: - Reject \(H_{0}\) for \(L_n\) > \(c_{1}(n)\) = \(\frac{1-\alpha_{a}}{\beta_{a}}\). - Do not reject \(H_{0}\) for \(L_n\) < \(c_{0}(n)\) = \(\frac{\alpha_{a}}{1-\beta_{a}}\). Substituting the given values \(\alpha_{a}=0.05, \beta_{a}=0.10\) gives: \(c_{1}(n)=\frac{1-0.05}{0.10}=9.5\), \(c_{0}(n)=\frac{0.05}{1-0.10} = 0.05556\)
03

Final Statement

For each \(n\), calculate \(L_n\). Proceed to the next \(X\) if \(c_{0}(n) < L_n < c_{1}(n)\), otherwise stop and make a decision: If \(L_n < c_{0}(n)\), fail to reject the null hypothesis \(H_{0}\), and if the \(L_n > c_{1}(n)\), reject the null and accept the alternate hypothesis \(H_{1}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis Testing
In the world of statistics, hypothesis testing is a critical method used to make decisions about a population parameter based on sample data.
At its core, hypothesis testing involves two main hypotheses:
  • The null hypothesis (\(H_{0}\)) represents the status quo or a statement that there is no effect or no difference.
  • The alternative hypothesis (\(H_{1}\)) suggests a possible outcome or effect that we test against the null.
For example, if you're testing whether a new drug is more effective than an existing one, \(H_{0}\) might claim that both drugs have equal effectiveness, while \(H_{1}\) could state that the new drug is more effective.
In the provided exercise, the null hypothesis is \(H_{0}: \theta = \theta^{\prime} = 4\), and the alternative hypothesis is \(H_{1}: \theta = \theta^{\prime \prime} = 9\). The goal is to determine which one these is more likely given the sample data.
Likelihood Ratio
The likelihood ratio is a fundamental concept in hypothesis testing, especially within the Sequential Probability Ratio Test (SPRT).
It helps determine how much more likely the data is under one hypothesis compared to another.
In simple terms, the likelihood ratio is the probability of observing the data under the alternative hypothesis divided by the probability under the null hypothesis.In mathematical terms, it is represented as:\[L_n = \left( \frac{\theta^{\prime}}{\theta^{\prime \prime}} \right)^{\frac{n}{2}} \mathrm{exp}\left( \frac{1}{2} \left( \frac{1}{\theta^{\prime}} - \frac{1}{\theta^{\prime \prime}} \right) \sum_{1}^{n} X_{i}^{2} \right)\]This function tells us which hypothesis is favored by the given sample data. In the exercise, this ratio is computed for a series of observations. By comparing it to predetermined thresholds, it helps decide whether to keep collecting data, reject the null hypothesis, or fail to reject it. The key is that if the likelihood under \(H_{1}\) becomes significantly greater, we lean towards rejecting \(H_{0}\).
Decision Boundaries
Decision boundaries are crucial in hypothesis testing as they dictate when and how decisions should be made based on the likelihood ratio.
In the context of the Sequential Probability Ratio Test (SPRT), these boundaries are known as critical thresholds: one for accepting \(H_{1}\) and another for sticking with \(H_{0}\).The boundaries are calculated using:
  • If the likelihood ratio, \(L_n\), is greater than the upper boundary, \(c_{1}(n)\), you reject \(H_{0}\) and accept \(H_{1}\).
  • If the likelihood ratio is less than the lower boundary, \(c_{0}(n)\), you fail to reject \(H_{0}\).
    • This setup ensures minimal errors in decision making. For instance, in the exercise, the values \(\alpha_{a}=0.05\) and \(\beta_{a}=0.10\) were used to set:
      • The upper boundary: \(c_{1}(n) = 9.5\).
      • The lower boundary: \(c_{0}(n) = 0.05556\).
      These boundaries help determine whether to continue with the sample collection or make a decision based on the data already gathered.

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Most popular questions from this chapter

The effect that a certain drug (Drug A) has on increasing blood pressure is a major concern. It is thought that a modification of the drug (Drug B) will lessen the increase in blood pressure. Let \(\mu_{A}\) and \(\mu_{B}\) be the true mean increases in blood pressure due to Drug \(\mathrm{A}\) and \(\mathrm{B}\), respectively. The hypotheses of interest are \(H_{0}: \mu_{A}=\mu_{B}=0\) versus \(H_{1}: \mu_{A}>\mu_{B}=0 .\) The two-sample \(t\) -test statistic discussed in Example \(8.3 .3\) is to be used to conduct the analysis. The nominal level is set at \(\alpha=0.05\) For the experimental design assume that the sample sizes are the same; i.e., \(m=n .\) Also, based on data from Drug \(A, \sigma=30\) seems to be a reasonable selection for the common standard deviation. Determine the common sample size, so that the difference in means \(\mu_{A}-\mu_{B}=12\) has an \(80 \%\) detection rate. Suppose when the experiment is over, due to patients dropping out, the sample sizes for Drugs \(A\) and \(B\) are respectively \(n=72\) and \(m=68 .\) What was the actual power of the experiment to detect the difference of \(12 ?\)

Let \(X\) and \(Y\) have a joint bivariate normal distribution. An observation \((x, y)\) arises from the joint distribution with parameters equal to either $$ \mu_{1}^{\prime}=\mu_{2}^{\prime}=0, \quad\left(\sigma_{1}^{2}\right)^{\prime}=\left(\sigma_{2}^{2}\right)^{\prime}=1, \quad \rho^{\prime}=\frac{1}{2} $$ Ior $$ \mu_{1}^{\prime \prime}=\mu_{2}^{\prime \prime}=1, \quad\left(\sigma_{1}^{2}\right)^{\prime \prime}=4, \quad\left(\sigma_{2}^{2}\right)^{\prime \prime}=9, \quad \rho^{\prime \prime}=\frac{1}{2} \text { . } $$ Show that the classification rule involves a second-degree polynomial in \(x\) and \(y\).

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.

The power function for the one-sample \(t\) -test is discussed. (a) Plot the power function for the following setup: \(X\) has a \(N\left(\mu, \sigma^{2}\right)\) distribution; \(H_{0}: \mu=50\) versus \(H_{1}: \mu \neq 50 ; \alpha=0.05 ; n=25 ;\) and \(\sigma=10\). (b) Overlay the power curve in (a) with that for \(\alpha=0.01\). Comment. (c) Overlay the power curve in (a) with that for \(n=35\). Comment. (d) Determine the smallest value of \(n\) so the power exceeds \(0.80\) to detect \(\mu=53\). Hint: Modify the R function tpowerg.R so it returns the power for a specified alternative.

Let \(X_{1}, X_{2}, \ldots, X_{10}\) be a random sample from a distribution that is \(N\left(\theta_{1}, \theta_{2}\right)\). Find a best test of the simple hypothesis \(H_{0}: \theta_{1}=\theta_{1}^{\prime}=0, \theta_{2}=\theta_{2}^{\prime}=1\) against the alternative simple hypothesis \(H_{1}: \theta_{1}=\theta_{1}^{\prime \prime}=1, \theta_{2}=\theta_{2}^{\prime \prime}=4\).

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