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I have a dataset that describe the number of passengers that fly with an airline per day. The airline guesses that on average 1300 passengers fly per day and I want to test this hypothesis using a likelihood ratio test.

My first thought was to set the hypotheses: H0: λ = 1300 and H1: λ <>1300. According to wikipedia the likelihood ratio test is: -2ln(likelihood for null model) + 2ln(likelihood for alternative model).

Using R I can find the likelihood of the null hypothesis but how can I calculate the likelihood for the alternative hypothesis in a Poisson distribution?

Thanks

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  • $\begingroup$ Poisson assumes the number of passengers is the aggregation of 1300 independent but identical events. Do you think it's possible that trips are not iid... such as the first day of spring break? $\endgroup$
    – RegressForward
    Commented Jun 10, 2015 at 16:37
  • $\begingroup$ Is this for a class? $\endgroup$
    – Glen_b
    Commented Jun 11, 2015 at 11:27

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how can I calculate the likelihood for the alternative hypothesis in a Poisson distribution?

You find the MLE of $\lambda$ and then compute the log-likelihood at that value of $\lambda$ (and multiply by -2)

[However, some simplifications may be made.]

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  • $\begingroup$ Thanks for the help! What I did was to take λ = 1300 as the likelihood of the null hypothesis and then I found the likelihood of the alternative by summing all the values in my dataset and dividing by the size of it. $\endgroup$
    – Dr.Fykos
    Commented Jun 11, 2015 at 12:14
  • $\begingroup$ No, the likelihood is not the parameter estimate. You evaluate the likelihood function at the two parameter estimates. $\endgroup$
    – Glen_b
    Commented Jun 11, 2015 at 12:25
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Your likelihoods should be maximized.

Assuming you have a list $\left\{ n_i \right\}$ of passengers each day, for the null hypothesis, taking $\lambda_0=1300$, you calculate $\hat{\cal L_0} = \prod \frac{e^{-{\lambda_0}} \lambda_O^{n_i}}{n_i!}$.

For the alternate hypothesis, ${\cal L_1} = \prod \frac{e^{-{\lambda}} \lambda^{n_i}}{n_i!}$. Maximizing this leads to the estimator $\hat{\mu}$ equal to the average of the $n_i$; you plug the estimator back into the likelihood to obtain $\hat{\cal L_1}$. Using the logarithm makes things easier.

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