You and a friend will play a game: He pays you 5 dollars if the outcome of a dic
ID: 1841799 • Letter: Y
Question
You and a friend will play a game: He pays you 5 dollars if the outcome of a dice roll is 6. If the outcome of the dice roll is anything else, you pay him 1 dollar. You did the math and this game is fair only if Pr[X = 6] = 1/6 (where X is the outcome of the six-faced dice roll). So, before you agreed on playing the game, you decided to estimate this probability. Let X = {X_1, X_2, ..., X_n} be a sample of N independent draws from the distribution of X (independent repetitions of your experiment of rolling the six-faced dice). Use the analogy principle to propose an estimator for Pr[X = 6] from a sample of size N. Compute your estimate of Pr[X = 6] for this dataset: X = {1, 3, 6, 1, 2, 5, 3, 6, 2, 3} Use the Central Limit Theorem to provide a confidence interval for the Pr[X = 6] that contains the true probability 95% of the times in large samples. Test the null hypotesis H_0 that Pr[X = 6] 1/6 against the alternative that Pr[X = 6] notequalto 1/6 (you can use the Central Limit Theorem to approximate the distribution of your test statistic). Do you acept or reject the null hypothesis using this dataset? Write your expected profit from playing this game as a function of Pr[X = 6]. Show that your expected profit playing this game is equal to zero if the null hypothesis is true.Explanation / Answer
1) Analogy principle : use a function instead of say random generators to estimate the throw of a dice. This is central to probability theory
We need to find a paramater b st T(P,b) =0, P is a smaller population from the larger population of all possible throws, b a parameter which satisies the probaility of getting say a 6 . This reduces to a small sample from a larger distrbution, and if the die is unbiased, the parameter b remains at 1/6. So take b =1/6 which should remain the probability of getting any particlar face on a throw.
b)given the smaller data set : X( 1,3,6,1,2,5,3,6,2,3) out of 10 events, 6 occurs 2 times so the parameter here is 2/10 = 1/5
c) the true probability 1/6 = .1667, the observed is .2, sd 1.778, mean 3.2, true mean for random (1+2+3+4+5+6)/6= 21/6 =3.5
probability difference |.1667-.2|
A)Diff/(sd/n)0.5 = .0333/.421 = .079 compare with t value for n=9
B) compare diff in means (3.5-3.2)/ sqrt(npq) = .3/1.179 =.025
whichever way it is seen, the deviation wrt sd is small, and not significant to 95% ( 1 sd ).
C) expected profit :1/6 of the time
for small sample get 1/X : profit or loss varies as !/X-1/6
as X tends to large number 1/X-->1/6 so the profit or loss tends to zero
net gain 0
so within
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.