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Let \(X\) have a Poisson distribution with mean \(\theta\). Find the sequential probability ratio test for testing \(H_{0}: \theta=0.02\) against. \(H_{1}: \theta=0.07\). Show that this test can be based upon the statistic \(\sum_{1}^{n} X_{i}\). If \(\alpha_{a}=0.20\) and \(\beta_{a}=0.10\), find \(c_{0}(n)\) and \(c_{1}(n)\)

Short Answer

Expert verified
Therefore, the constants for this sequential probability ratio test are \(c_{0}(n)= 4.5\) and \(c_{1}(n)= 0.125\).

Step by step solution

01

Understanding Sequential Probability Ratio Test

The sequential probability ratio test (SPRT) is a statistical test that allows for testing while data are still being collected. In SPRT, hypotheses are tested based on the likelihood ratio. The SPRT stops collecting data once sufficient evidence is available to accept or reject a given hypothesis.
02

Calculation Likelihood Ratio

The likelihood ratio for each observation under a Poisson distribution is given by \(L_{i}=\frac{e^{-0.07}(0.07)^{X_i}}{e^{-0.02}(0.02)^{X_i}}\). To find the sequential probability ratio through \(n\) observations, the likelihood ratios need to be multiplied: \(SPR_{n}=\prod_{i=1}^{n}L_{i}\). As we need to find the test based on the statistic \(\sum_{i=1}^{n}X_i\), one would then plug \(X_i\) values sequentially into the equation, multiply the results, and continue until there is enough data to accept or reject the hypothesis.
03

Define the Acceptance and Rejection Constants

In the SPRT, two thresholds are set, \(c_{0}(n)\) and \(c_{1}(n)\). If the \(SPR_{n}\) exceed \(c_{1}(n)\), then the null hypothesis is rejected. If it falls below \(c_{0}(n)\), the null hypothesis is accepted. Formally these are defined as - when error probabilities of type I and type II are mentioned - \(c_{0}(n)=(1-\beta_a) \ /\ \alpha_a\) and \(c_{1}(n)=\beta_a \ /\ (1-\alpha_a)\) using the given \(\alpha_a = 0.20\) and \(\beta_a = 0.10\).
04

Calculation of Constants

Calculating the constants using the relations defined in step 3 and plugging in the values: \(c_{0}(n)=(1-0.10) \ /\ 0.20=4.5\) and \(c_{1}(n)=0.10 \ /\ (1-0.20)=0.125\)

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Key Concepts

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

Poisson Distribution
The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event. It is denoted by the parameter \(\theta\), which is the average number of occurrences in the interval.

The Poisson distribution is especially useful for modeling the number of events in fixed intervals of time with low probability and is applicable in various fields such as biology, finance, and engineering. Its probability mass function is given by:
\[P(X=k)=\frac{e^{-\theta}\theta^k}{k!}\]
where \(k\) is the actual number of events (often the number of arrivals or counts), \(\theta\) is the rate at which events happen, \(e\) is the base of the natural logarithm, and \(k!\) is the factorial of \(k\). It is this distribution that is used in the Sequential Probability Ratio Test for the exercise provided.
Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions using data. It involves making assumptions called hypotheses, which are then tested to determine whether there is enough statistical evidence to reject them.

In the context of the exercise, one begins with two hypotheses: the null hypothesis \(H_0\) and the alternative hypothesis \(H_1\). The null hypothesis represents a default position that there is no effect or no difference, while the alternative hypothesis represents what we are attempting to provide evidence for. In the exercise, \(H_0: \theta=0.02\) is tested against \(H_1: \theta=0.07\).

Steps in Hypothesis Testing

  • State the null and alternative hypotheses.
  • Choose a significance level (\(\alpha\)), which is the probability of rejecting the null hypothesis when it is actually true (Type I error).
  • Calculate the test statistic based on the sample data.
  • Decide to reject or fail to reject the null hypothesis based on the comparison of the test statistic and the critical value or p-value.
Likelihood Ratio
The likelihood ratio in statistical hypothesis testing is a measure of the strength of the evidence. It is calculated as the probability of the observed results under the alternative hypothesis relative to the probability under the null hypothesis.

For the exercise we're discussing, the likelihood ratio for the sequential observations is expressed as:
\[L_i=\frac{e^{-0.07}(0.07)^{X_i}}{e^{-0.02}(0.02)^{X_i}}\]
The idea here is to compare how likely the observed data is under each hypothesis, effectively measuring if the data better supports \(H_0\) or \(H_1\). The likelihood ratio becomes the driving force in the Sequential Probability Ratio Test, directing when enough evidence has been obtained to make a decision regarding the hypotheses.

Using Likelihood Ratio

  • Calculate the likelihood ratio for each piece of data.
  • Multiply the individual likelihood ratios for a cumulative measure (\(SPR_n\)).
  • Use thresholds to make the final decision based on \(SPR_n\).
Statistical Significance
Statistical significance is a term used to determine whether the outcome of a test or experiment is likely to have occurred by chance or if it reflects a true effect. It enables researchers to decide if their hypotheses should be accepted or rejected.

In the sequential probability ratio test (SPRT) from the exercise, statistical significance is expressed through error probabilities \(\alpha_a\) (Type I error) and \(\beta_a\) (Type II error). The constants \(c_0(n)\) and \(c_1(n)\) represent the decision thresholds based on the chosen significance levels:
\[c_0(n)=\frac{1-\beta_a}{\alpha_a}\]
\[c_1(n)=\frac{\beta_a}{1-\alpha_a}\]
These calculated constants determine when to reject or accept the null hypothesis \(H_0\) based on the cumulative likelihood ratio. If the SPRT statistic exceeds \(c_1(n)\), \(H_0\) is rejected; if it falls below \(c_0(n)\), \(H_0\) is accepted, reflecting the significance of the observed data in supporting or refuting the initial assumption namely, the null hypothesis.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{10}\) denote a random sample of size 10 from a Poisson distribution with mean \(\theta .\) Show that the critical region \(C\) defined by \(\sum_{1}^{10} x_{i} \geq 3\) is a best critical region for testing \(H_{0}: \theta=0.1\) against \(H_{1}: \theta=0.5 .\) Determine, for this test, the significance level \(\alpha\) and the power at \(\theta=0.5 .\) Use the \(\mathrm{R}\). function Ppois.

Let \(Y_{1}

Verzani (2014) discusses a data set on healthy individuals, including their temperatures by gender. The data are in the file tempbygender. rda and the variables of interest are maletemp and femaletemp. Download this file from the site listed in the Preface. (a) Obtain comparison boxplots. Comment on the plots. Which, if any, gender seems to have lower temperatures? Based on the width of the boxplots, comment on the assumption of equal variances. (b) As discussed in Example 8.3.3, compute the two-sample, two-sided \(t\) -test that there is no difference in the true mean temperatures between genders. Obtain the \(p\) -value of the test and conclude in terms of the problem at the nominal \alpha-level of \(0.05 .\) (c) Obtain a \(95 \%\) confidence interval for the difference in means. What does it mean in terms of the problem?

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the normal distribution \(N(\theta, 1)\). Show that the likelihood ratio principle for testing \(H_{0}: \theta=\theta^{\prime}\), where \(\theta^{\prime}\) is specified, against \(H_{1}: \theta \neq \theta^{\prime}\) leads to the inequality \(\left|\bar{x}-\theta^{\prime}\right| \geq c\). (a) Is this a uniformly most powerful test of \(H_{0}\) against \(H_{1} ?\) (b) Is this a uniformly most powerful unbiased test of \(H_{0}\) against \(H_{1}\) ?

Let \(X\) have the pmf \(f(x ; \theta)=\theta^{x}(1-\theta)^{1-x}, x=0,1\), zero elsewhere. We test the simple hypothesis \(H_{0}: \theta=\frac{1}{4}\) against the alternative composite hypothesis \(H_{1}: \theta<\frac{1}{4}\) by taking a random sample of size 10 and rejecting \(H_{0}: \theta=\frac{1}{4}\) if and only if the observed values \(x_{1}, x_{2}, \ldots, x_{10}\) of the sample observations are such that \(\sum_{1}^{10} x_{i} \leq 1\). Find the power function \(\gamma(\theta), 0<\theta \leq \frac{1}{4}\), of this test.

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