MLE derivation was fine, lambda-hat equals x-bar, straightforward log-likelihood stuff.
Start by stating the Poisson MLE for λ is the sample mean, then derive its asymptotic variance using the Fisher information. Use the plug-in estimator for variance and construct the Wald interval with the normal approximation, ensuring to mention the delta method if needed for transformations.
Pro tip: Emphasize that the Poisson model assumes mean equals variance; here the sample variance (16.2) is close to the mean (14.5), so the model is reasonable. Also, note that the Wald interval can be improved with transformations like log or square root for better coverage, but stick to the question's request for Wald.
For independent Poisson counts, the MLE for λ is the sample mean, so λ_hat = 14.5.
The Fisher information for a single observation is 1/λ, so for n=200, the asymptotic variance of λ_hat is λ/n. Plug in λ_hat to get 14.5/200 = 0.0725.
The standard error is the square root of the variance: sqrt(0.0725) ≈ 0.2693.
Use the formula λ_hat ± 1.96 * SE. This gives 14.5 ± 1.96*0.2693, resulting in approximately (13.972, 15.028).
Interpret the interval and note that the Poisson assumption seems plausible since sample variance (16.2) is close to mean (14.5).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This is where I actually blanked for a second.
First, clarify that the test for equidispersion versus overdispersion in a Poisson context is based on the dispersion index (variance-to-mean ratio). Then, compute the test statistic as (n-1)*sample_variance/sample_mean, which follows a chi-square distribution with n-1 degrees of freedom under the null hypothesis of equidispersion. Finally, compare the test statistic to the critical value from the chi-square distribution at alpha=0.05 to make a decision.
Pro tip: When presenting the decision, explicitly state the null and alternative hypotheses: H0: variance = mean (equidispersion) vs. H1: variance > mean (overdispersion). Also, mention that this test assumes the sample size is large enough for the chi-square approximation to hold.
Define the null hypothesis (equidispersion: variance equals mean) and the alternative hypothesis (overdispersion: variance greater than mean).
Calculate the dispersion test statistic as (n-1)*sample_variance/sample_mean, where n is the sample size.
Under the null hypothesis, the test statistic follows a chi-square distribution with n-1 degrees of freedom.
At alpha=0.05, find the critical value from the chi-square distribution (or compute the p-value) and compare it to the test statistic. Reject H0 if the test statistic exceeds the critical value (or if p-value < 0.05), indicating overdispersion.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, recall that for a Poisson(λ=120), the mean and variance are both 120, so the standard deviation is √120 ≈ 10.954. Apply the continuity correction by converting the discrete bounds to continuous intervals: P(100 ≤ X ≤ 140) ≈ P(99.5 < Y < 140.5) where Y ~ Normal(120, 120). Then standardize the bounds and express the probability as a difference of standard normal CDFs, explaining why the correction improves accuracy.
Pro tip: Mention that without continuity correction, the approximation would use P(100 < Y < 140), which shifts both bounds by 0.5 and can lead to noticeable error, especially near the tails. Also, note that the correction is most impactful when the interval is narrow or when λ is small, but even for λ=120 it improves the approximation by accounting for the discrete nature of Poisson.
State that X ~ Poisson(λ=120), so mean μ = λ = 120 and variance σ² = λ = 120, hence σ = √120 ≈ 10.954. Confirm that λ is large enough for a good normal approximation.
Convert the discrete bounds to continuous: P(100 ≤ X ≤ 140) ≈ P(99.5 < Y < 140.5) where Y ~ N(120, 120). Explain that this adjustment accounts for the fact that a discrete value k corresponds to the interval [k-0.5, k+0.5] on the continuous scale.
Compute z-scores: z1 = (99.5 - 120)/√120 ≈ -1.872, z2 = (140.5 - 120)/√120 ≈ 1.872. Write the probability as Φ(1.872) - Φ(-1.872) = 2Φ(1.872) - 1.
Write the probability as ∫_{99.5}^{140.5} (1/√(2π·120)) exp(-(x-120)²/(2·120)) dx. This is the exact normal integral with continuity correction.
Discuss that without it, the approximation would use P(100 < Y < 140), which ignores the 0.5 shift and can underestimate or overestimate the probability. The correction improves accuracy by aligning the discrete distribution with the continuous one, especially for small intervals or when λ is not very large.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Rushed this one because we were running low on time.
Start by stating the conditions for a valid Normal approximation to the Poisson distribution, emphasizing that both the mean and variance must be equal and sufficiently large. Then explain when the approximation breaks down, particularly when the variance exceeds the mean (overdispersion). Finally, introduce the Negative Binomial as an alternative model and mention a diagnostic like the dispersion test or a plot of residuals.
Pro tip: Demonstrate practical experience by noting that in real-world data, overdispersion is common and can lead to underestimated standard errors if ignored. Mention that Apple often deals with large-scale user data where such nuances matter for reliable inference.
Explain that the Poisson distribution can be approximated by a Normal distribution when the mean (λ) is large (typically λ > 20) and events are independent. The Normal approximation uses mean = λ and variance = λ.
Discuss that the approximation fails when λ is small (e.g., λ < 5) due to skewness and discreteness, and when the variance is much larger than the mean (overdispersion), indicating the Poisson assumption is violated.
Introduce the Negative Binomial distribution as a flexible alternative that accounts for overdispersion by allowing variance to exceed the mean. Mention that it reduces to Poisson when the dispersion parameter is zero.
Suggest a diagnostic such as the dispersion test (e.g., Cameron & Trivedi test) or examining the ratio of residual deviance to degrees of freedom. A ratio significantly greater than 1 indicates overdispersion.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.