Lady Guessing Coin Flips. There is a famous story involving statistician (Sir Ronald Fisher) and a lady (Muriel Bristol). The lady claimed that, when tasting tea with milk, she can positively determine whether tea or milk had been poured in a cup first. Fisher designs an experiment with 8 cups, 4 in which milk was poured first and 4 in which tea was poured first. Muriel correctly guesses in all 8 cases. This has come to be known as the Lady Tasting Tea Experiment and the example is often used to illustrate Fisher’s exact test.

Now, imagine another lady who claims ESP and the ability to predict results in subsequent flips of a fair coin. A fair coin is flipped 16 times and the lady correctly predicts the outcome 15 times. Let p be the probability of the lady guessing correctly.

Test a precise hypothesis H0 : p = 0.5 (guessing) versus the alternative H1 : p > 0.5 (presence of ESP), in Bayesian fashion.

To test a precise hypothesis in Bayesian fashion, a prior with a point-mass is needed. Assume that the prior on p is:

π(p) = π0 δ0.5 + π1 U(0.5, 1) = 0.95 δ0.5 + 0.05 · 2 · 1(0.5 < p < 1),

where δ0.5 is point-mass at 0.5, and U(a, b) is uniform distribution on (a, b). From the form of the prior we see that prior probabilities of the null and alternative hypotheses are P(H0) = π0 = 0.95 and P(H1) = π1 = 1 − π0 = 0.05. Thus, apriori we are sceptic of lady’s ESP, and favor H0.

(a) Find the posterior probabilities of hypotheses, p0 and p1, and Bayes Factor.

(b) What is the assessment of H1 according to Jeffreys scale, i. e., is the experiment convincing that the lady possesses ESP?

i can answer this question sweetie!

step-by-step explanation:

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497

answer: the resulting cross section is a rectangle. solution: a plane parallel to the base of a triangular prism will intersect a cross section that is the same shape as its bases. so the cross section is a triangle.

step-by-step explanation:

the answer is 21%.