Most powerful test involving Poisson

In summary, the problem involves testing the hypothesis that a used car salesman has a probability of 0.10 or 0.25 of making at least one sale per day, based on a random sample of n days. The most powerful test is found by applying the single likelihood ratio test, which rejects Ho if the sum of the sample values is less than or equal to a certain threshold, determined by the significance level alpha. However, this method may not be accurate since the hypotheses involve the probability p and not the parameter lambda. To solve this problem, one can examine the relationship between p and lambda, consider the likelihood of each hypothesis, and derive the power of the test from the likelihood ratio test.
  • #1
safina
28
0

Homework Statement


The number of sales made by a used car salesman, per day, is a Poisson random variable with parameter [tex]\lambda[/tex]. Given a random sample of the number of sales he made on n days, what is the most powerful test of the hypothesis Ho: p = 0.10 versus Ha: p = 0.25, where p is the probability he makes at least one sale (per day)?


Homework Equations


[tex]f\left(x;\lambda\right) = \frac{e^{-\lambda}\lambda^{x}}{x!}[/tex]


The Attempt at a Solution


I applied the single likelihood ratio test which Rejects Ho if [tex]\lambda[/tex] [tex]\leq[/tex] k which I found equivalent in saying to reject Ho if [tex]\sum Xi[/tex] [tex]\leq[/tex] k' where k' is given by [tex]P\left[\sum Xi \leq k'\right] = \alpha[/tex]
But it seems not correct since the hypotheses involve p and not the parameter [tex]\lambda[/tex]. Please help me solve this problem.
 
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  • #2
what are k & k'?

just a few ideas to get you started:
- first I'd look at how p is related to lambda
- then i would look at the likelihood of each hypothesis
- consider how to derive the power of the test
 

Related to Most powerful test involving Poisson

1. What is the most powerful test involving Poisson?

The most powerful test involving Poisson is the likelihood ratio test. This test is used to compare two or more Poisson distributions and determine if they are significantly different from each other.

2. How does the likelihood ratio test work?

The likelihood ratio test works by comparing the likelihood of the observed data under the null hypothesis (no difference between the Poisson distributions) to the likelihood under the alternative hypothesis (there is a difference between the Poisson distributions). The test statistic is then calculated and compared to a critical value to determine if the null hypothesis should be rejected.

3. When should the likelihood ratio test be used?

The likelihood ratio test should be used when comparing two or more Poisson distributions that are not normally distributed and have limited sample size. It is a more powerful alternative to the chi-square test in these situations.

4. What are the assumptions of the likelihood ratio test?

The assumptions of the likelihood ratio test include: 1) the data is independent, 2) the data is from a Poisson distribution, 3) the rates of the Poisson distributions being compared are constant, and 4) the sample size is not too small.

5. What are the advantages of using the likelihood ratio test?

The advantages of using the likelihood ratio test include: 1) it is more powerful than other tests for comparing Poisson distributions, 2) it can be used with small sample sizes, 3) it does not require the data to be normally distributed, and 4) it allows for multiple comparisons between Poisson distributions.

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