ANSWERS TO ACTIVITIES
TOPIC 16 - Learning About a Proportion
Activity 16-1
(a) .75
(b) .5
(c)� Table after observing 1st data result = "acceptable" should look like
|
MODEL |
PRIOR |
LIKE |
PRODUCT |
POST |
|
p=.75 |
.9 |
.75 |
.675 |
.931 |
|
p=.5 |
.1 |
.5 |
.050 |
.069 |
|
SUM |
� |
|
.725 |
1.000 |
(g) Completed table after
observing 2nd data result = "defective":
|
MODEL |
PRIOR |
LIKE |
PRODUCT |
POST |
|
p=.75 |
.931 |
.25 |
.233 |
.869 |
|
p=.5 |
.069 |
.5 |
.035 |
.131 |
|
SUM |
� |
|
.268 |
1.000 |
(h) Completed table after
observing 3rd data result = "defective":
|
MODEL |
PRIOR |
LIKE |
PRODUCT |
POST |
|
p=.75 |
.869 |
.25 |
.217 |
.770 |
|
p=.5 |
.131 |
.5 |
.065 |
.230 |
|
SUM |
� |
|
.282 |
1.000 |
(g) Completed table after
observing 4th data result = "defective":
|
MODEL |
PRIOR |
LIKE |
PRODUCT |
POST |
|
p=.75 |
.770 |
.25 |
.192 |
.627 |
|
p=.5 |
.230 |
.5 |
.115 |
.373 |
|
SUM |
� |
|
.307 |
1.000 |
(j) Likelihood for p=.75: LIKE =
.75 x .25 x .25 x .25 = .0117
���� Likelihood for p = .25: LIKE = .5 x .5 x .5 x .5 = .0625
|
MODEL |
PRIOR |
LIKE |
PRODUCT |
POST |
|
p=.75 |
.9 |
.0117 |
.105 |
.629 |
|
p=.5 |
.1 |
.0625 |
.062 |
.371 |
|
SUM |
� |
|
.167 |
1.000 |
Activity 16-2
(a) Likelihood of (H,H,O,O) for p=.3 is LIKE = .3 x .3 x .7 x .7 = .0441
������� Likelihood of (H,H,O,O) for p = .4 is LIKE = .4 x .4 x .6 x .6 = .0576
(b)����������
|
�MODEL |
PRIOR |
LIKE |
PRODUCT |
POST |
|
p = .2 |
.4 |
.0256 |
.0102 |
.278 |
|
p = .3 |
.59 |
.0441 |
.0260 |
.706 |
|
p = .4 |
.01 |
.0576 |
.0006 |
.016 |
|
SUM |
|
|
.0368 |
1.000 |
Activity 16-3
Let's suppose I obtained 2 heads in 5 spins -- I have the following posterior probabilities:
|
����� p������� |
post |
|
������� 0 ���� |
0 |
|
��� 0.1000 |
0.0440 |
|
��� 0.2000 |
0.1230 |
|
��� 0.3000 |
0.1850 |
|
��� 0.4000 |
0.2070 |
|
��� 0.5000 |
0.1880 |
|
��� 0.6000 |
0.1380 |
|
��� 0.7000 |
0.0790 |
|
��� 0.8000 |
0.0310 |
|
��� 0.9000 |
0.0050 |
|
0 |
post |
(b) The most likely value of p is .4 -- this value has a probability of .207.
(c) �The probability that p is equal to .7 is .079
(d) The probability that p is greater than 5
= prob (p = .6, .7, .8, .9, 1)
= .138 + .079 + .031 + .005 + 0 =
.253
(e) ordering values of p by most likely to least likely
|
����� p������� |
post |
cum prob |
|
.4 |
0.2070 |
0.2070 |
|
.5 |
0.1880 |
0.3950 |
|
.3 |
0.1850 |
0.5800 |
|
.6 |
0.1380 |
0.7180 |
|
.2 |
0.1230 |
0.8410 |
|
.7 |
0.0790 |
0.9200 |
|
.1 |
0.0440 |
0.9640 |
|
.8 |
0.0310 |
0.9950 |
|
.9 |
0.0050 |
1.0000 |
|
1 |
0 |
1.0000 |
|
0 |
0 |
1.0000 |
(f) 50% probability set is {.3, .4, .5}
90% probability set is {.2, .3, .4, .5, .6, .7}
Activity 16-4
(c) 16/22 = .727
(d)
No, since you are only considering couples with different
ages.
(e)
The most likely values are p = .71, .72, .73, .74
(f) Prob(p is larger than .5) = 1 - Prob(p is .5 or smaller)
= 1 - Prob(p is 0, .01, .02, �, .5)
= 1 - [.001 + .001 + .001 + .001 + .002 + .002 + .003 + .003 + .004]
= 1 - .018 = .982 .
(g) I put in the 15 most likely values of p and the associated probabilities in the table below:
|
PROPORTION p |
PROBABILITY |
|
.71, .72, .73, .74 |
.043, .043, .043, .043 |
|
.70, .75� |
.042, .042 |
|
.76, .69���� |
.041, .040 |
|
.68, .77, .67, .78 |
.039, .039, .037, .037 |
|
.66, .79, .65�� |
.034, .034, .032 |
������������ The
total probability on the right is .043 + .043 + �+ .032 = .589
����������� So the
set of values {.71, .72, �, .65} has total probability .589
(h) The proportion p lies between .65 and .79 with probability .589 - that is, [.65, .79] is a 58.9% probability interval.
���� Activity 16-5
(a) If� I am very skeptical that Frank has ESP, I would assign the model p = .5 a small probability.
My prior might be
|
MODEL |
PRIOR |
|
p = .25�������� |
.9�������� |
|
p = .5���������� |
.1�������� |
Your prior could be different, but the model p = .5 should be given a small probability.
(c) data is C = Frank got card correct
|
MODEL |
PRIOR |
��� LIKE��� |
PRODUCT |
POSTERIOR |
|
p = .25�������� |
.9�������� |
.25��������� |
.225 |
.818 |
|
p = .5���������� |
.1�������� |
.5����������� |
.050������������� |
.182 |
|
SUM |
|
|
.275 |
1.000 |
(d) Frank got second card wrong (W)
|
MODEL |
PRIOR |
��� LIKE��� |
PRODUCT |
POSTERIOR |
|
p = .25�������� |
.818�������� |
.75��������� |
.614 |
.871 |
|
p = .5���������� |
.182�������� |
.5����������� |
.091 |
.129 |
|
SUM |
|
|
.705 |
1.000 |
��� Activity 16-6
(a)
|
MODEL |
PRIOR |
|
p=.5 |
1/3 = .33 |
|
p=.67 |
1/3 = .33 |
|
p=.83 |
1/3 = .33 |
(b) Data is G, G, G, M
�� LIKELIHOODS:
�������� if p = .5, prob(G, G, G, M) = .5 x .5 x .5 x .5 = .0625
�������� if p = .67, prob(G, G, G, M) = .67 x .67 x .67 x .33 = .0993
�������� if p = .83, prob(G, G, G, M) = .83 x .83 x .83 x .17 =� .0972
(d)
|
MODEL |
PRIOR |
LIKE |
PRODUCT |
POSTERIOR |
|
p=.5 |
1/3 = .33 |
.0625������� |
.0206���������� |
.241 |
|
p=.67 |
1/3 = .33 |
.0993������� |
.0328���������� |
.383 |
|
p=.83 |
1/3 = .33 |
.0972������� |
.0321���������� |
.375 |
|
SUM |
|
|
.0855 |
|
� Activity 16-8
(d) Of the 20 pages, 6 contain ads, so I guess p is in the neighborhood of 6/20 = .3
(e)�� The 3 most likely values of p are .2, .3, .4
(f)� The probability that p is either
.2, .3, .4� = .229 + .403 + .261 = .893
(g)� Prob( p is at least .5) = Prob( p
is .5, .6, �, 1) = .078 + .010 + 0 + �+ 0 = .088.