Total responses: 103
A random number
===============
Stem plot (step 1):
0 | 3 2 8 9 7 9 7 π 7 8 8 8 9 5 1 0 7
1 | 9 1 5 1 6 4 1 0 3 1 6 1 4 1 1 8 6 2
2 | 2 3 5 3 2 1
3 | 6 4 4 6 3 4 6 3 6
4 | 7 1 4 2 8
5 | 5 7 0 6 2 ɑ 0 6
6 | 9 0 3 9 9 3 6 7 9 9 β 3 6 1 6 7 7
7 | 4 6 7 4 7 7 2 3 6 2
8 | 9 7 8 5 8 1 8
9 | 7 6 9 0 5
10 | 0
(Note: π = 3.141592..., ɑ = 6.789, β = 9.7)
Step plot, final (after sorting):
0 | 0 1 2 3 π 5 7 7 7 7 8 8 8 8 9 9 9
1 | 0 1 1 1 1 1 1 1 2 3 4 4 5 6 6 6 8 9
2 | 1 2 2 3 3 5
3 | 3 3 4 4 4 6 6 6 6
4 | 1 2 4 7 8
5 | 0 0 2 5 6 6 ɑ 7
6 | 0 1 3 3 3 6 6 6 7 7 7 9 9 9 9 9 β
7 | 2 2 3 4 4 6 6 7 7 7
8 | 1 5 7 8 8 8 9
9 | 0 5 6 7 9
10| 0
Just looking at the data, it doesn't look all that different from what we'd get
if we used a computer to pick random integers from 0-100. Here's an example of
doing that (obviously every run will look different):
0 | 2 2 3 5 6 6 7 7 7 8 9
1 | 0 0 0 1 2 8
2 | 0 0 1 4 5 5 6
3 | 0 1 1 2 2 2 2 3 3 3 4 4 5 6 7 8 8
4 | 2 3 3 5 5 5 7 8 9
5 | 0 0 1 2 4 4 4 4 5 5 6 8 8 9 9 9 9
6 | 0 1 2 3 3 4 5 6 6 6 8
7 | 0 1 3 6
8 | 1 2 4 4 4 4 6 7 8 8 9
9 | 2 2 3 4 4 7 7 8 9
10| 0
But note that there are some considerably overrepresented numbers in the list:
most notably 11 (7 times). The highest repeated count in the computer sample is
4.
However if we compute the mean (we'll learn how to do that soon), we get 43.68,
but a computer drawing the random numbers would only give us a mean at least
that low about 1.4% of the time.
So we are, as a group, a little biased towards choosing numbers that slightly
favour lower numbers to large ones.
Imaginary coin tossing
======================
1. Responses base on H/T counts:
Responses with 0H/5T: 1 = 0.97%
Responses with 1H/4T: 5 = 4.85%
Responses with 2H/3T: 36 = 34.95%
Responses with 3H/2T: 50 = 48.54%
Responses with 4H/1T: 10 = 9.71%
Responses with 5H/0T: 1 = 0.97%
2. Number of heads observed in each position.
Total heads: 272 of 515 flips = 52.8%
Responses with H first: 80 = 77.7%
Responses with H second: 54 = 52.4%
Responses with H third: 34 = 33.0%
Responses with H fourth: 53 = 51.5%
Responses with H fifth: 51 = 49.5%
3. The number of occurrences of all the specific sequences of flips are:
Seq. # Seq. # Seq. # Seq. #
======= ======== ======= =======
TTTTT 1 TTTHT 0 TTTTH 1 TTTHH 0
HTTTT 1 HTTHT 12 HTTTH 5 HTTHH 7
THTTT 1 THTHT 0 THTTH 4 THTHH 0
HHTTT 4 HHTHT 20 HHTTH 8 HHTHH 5
TTHTT 2 TTHHT 0 TTHTH 7 TTHHH 1
HTHTT 2 HTHHT 4 HTHTH 5 HTHHH 1
THHTT 2 THHHT 1 THHTH 3 THHHH 0
HHHTT 1 HHHHT 1 HHHTH 3 HHHHH 1
How good are ECON 250 students at flipping imaginary fair coins?
(Don't worry for now about where all of the below probabilities come from; we'll
learn sometime between the first and second midterms how to calculate these
probabilities).
1. For comparison, the probabilities of these outcomes for a fair coin toss are:
0H/5T: 0.03125
1H/4T: 0.15625
2H/3T: 0.3125
3H/2T: 0.3125
4H/1T: 0.15625
5H/0T: 0.03125
Our imagined distribution looks nothing like this: we seems clearly biased
towards heads, and we're too heavily biased towards roughly even (3/2 and 2/3)
sequences rather than more unequal sequences (4/1, 1/4, 5/0, 0/5)
2. Number of heads in any position (and overall) should be close to 0.5 (and
will generally get closer and closer as we do more flips).
Overall, we do pretty well here (0.53 is pretty close to 0.5), and similarly the
2nd, 4th, and 5th flips turn up heads the right number of times. The first
flip, however, is heads more than 3/4 of the time, while the third flip is heads
only 1/3 of the time.
3. Specific sequences. For a fair coin flipped 5 times, *any* specific sequence
of flips will have the same probability of 1/32 = 0.03125. In a fixed number of
samples, like this, you would You see this some variation of course, not not to
the level we see here (this is most noticeable in the high responses for the
specific HHTHT and HTTHT sequences, but also noticeably large in some of the
others.)