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Applied Statistics
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Entrance to a country can be denied for a number of reasons.

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Entrance to a country can be denied for a number of reasons. When someone arrives by air, and their entry is denied, they get put into a detention hotel before deportation. While in the hotel they can appeal against the deportation decision. If the appeal is accepted they will be allowed to enter the country,

if it’s rejected, they have one more chance for appeal before getting deported. If their second appeal is successful, they can leave the hotel and enter the country, but if this second appeal is unsuccessful again, they will be deported. The possible scenarios can be found in the table below (we don’t consider any other scenarios, i.e. we assume that everyone chooses to appeal): First appeal Second appeal Outcome Successful No second appeal Entrance allowed Rejected Successful Entrance allowed Rejected Rejected Deportation Historic data shows that 70% of people who end up in the detention hotel, are there because they made a minor mistake in the visa application, while 10% get put into the detention hotel because they lied in the visa application (and it’s possible to prove it), while 20% have missing documents. If a minor mistake was the reason behind the initial rejection, then each appeal has a 95% chance of being successful.

However, if the reason was a provable lie or missing documents, then an appeal gets rejected with probability 0.93.

(a) [7 marks] Considering only the first round of appeals, find the probability that a randomly chosen person in the detention hotel made a minor mistake in their application given that their first appeal is approved.

(b) [6 marks] By conditioning on the events {Minor mistake AND First appeal was successful; Minor mistake AND First appeal was rejected; Lie/Missing document AND First appeal was successful; Lie/Missing document AND First appeal was rejected} find the probability that a randomly chosen person from the detention hotel is eventually allowed to enter the country.

(c) [11 marks] R: Simulate the relevant experiments using R to find the required probabilities in part (a) and (b). Hint: After taking a large enough population, where each person is in the detention hotel because of a ‘minor mistake’ with probability 0.7 and they are in the hotel because of a ‘lie/missing document’ with probability 0.3, the outcome of the appeal process can be simulated by using the sample function. Observations Y1, Y2, . . . , Yn are assumed to be independent and identically distributed samples from a data model following a half normal distribution, with probability density function: f(y; θ) = r 2 θπ exp  − y 2 2θ  for θ > 0 and 0 < y < ∞. The mean of this distribution is µ = r 2θ π , and the variance is σ 2 = θ  1 − 2 π  . (Note that here π is not a parameter, it is the usual mathematical constant i.e. 3.14...)

(a) [2 marks] Find the method of moments estimator ˜θ of θ. (b) [5 marks] Is your estimator ˜θ unbiased? If not, then suggest an adjustment to this estimator that would make it unbiased and report your final unbiased estimator. Justify each step of your reasoning. (c) [4 marks] An alternative estimator is ˆθ = 1 n Pn i=1 Y 2 i . Is this estimator unbiased? If not suggest an adjustment that makes it unbiased. Show your working. (d) [5 marks] Using the fact that the random variable X = Y 2/θ has a χ 2 1 distribution, assess whether the estimator ˆθ from part 2c is consistent.

Note that if X ∼ χ 2 k , then E(X) = k and V ar(X) = 2k. (e) [5 marks] We have 120 samples from a half normal distribution with sample mean 3.4. Suggest a suitable estimate of the variance, and construct an approximate 95% confidence interval for the mean of the distribution. (You can use R to find the relevant quantities).

3. Observations Y1, Y2, . . . , Yn are assumed to be independent and identically distributed samples from a data model with probability density function: f(y; θ) = θe y+θ−θe y for y > 0 and θ > 0. (a) [7 marks] Derive the maximum likelihood estimator, ˆθ of θ. Let y1, . . . y8 below correspond to 8 samples from this distribution. 0.570 0.773 0.655 0.824 0.827 0.581 0.799 0.512. (b) [2 marks] Calculate the maximum likelihood estimate for θ based on these data. (c) [4 marks] R: Use R to plot the relevant log-likelihood function for the above data. Is the maximum where you expect it to be?

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