14 Feb
2009
14 Feb
'09
12:17 p.m.
Hi all,
The wording on questions like this always confuses me. We are asked, "If a
researcher interviews 5 separate individuals [out of the 25 in the village]
randomly selected with equal probability, what is the probability that she
will not talk to a single supporter of the government?"
I seems like you could treat this two ways:
1. What is the probability of getting 0 successes in 5 draws from a
population of arbitrary size but fixed probability of success? If you did it
this way, you'd use the binomial distribution with n = 5, k = 0 and pi =
6/25.
2. What is the probability of getting 5 sequential failures in 5 draws from
a population of 25 where there are 6 potential "successes"? If you did it
this way, you'd use the hypergeometric distribution with k = 0, N = 25, m =
6 and n = 5.
The second approach feels more natural, but is it?
John-Paul
14 Feb
14 Feb
12:23 p.m.
I would think the 2nd approach, too. Each draw alters the prob of success
for the next draw a significant enough amount.
Btw-in general would this be a violation of independence?
_____
From: gov2001-l-bounces at lists.fas.harvard.edu
[mailto:gov2001-l-bounces at lists.fas.harvard.edu] On Behalf Of John-Paul
Ferguson
Sent: Saturday, February 14, 2009 12:17 PM
To: gov2001-l at lists.fas.harvard.edu
Subject: [gov2001-l] Clarification on pset 2, question 2
Hi all,
The wording on questions like this always confuses me. We are asked, "If a
researcher interviews 5 separate individuals [out of the 25 in the village]
randomly selected with equal probability, what is the probability that she
will not talk to a single supporter of the government?"
I seems like you could treat this two ways:
1. What is the probability of getting 0 successes in 5 draws from a
population of arbitrary size but fixed probability of success? If you did it
this way, you'd use the binomial distribution with n = 5, k = 0 and pi =
6/25.
2. What is the probability of getting 5 sequential failures in 5 draws from
a population of 25 where there are 6 potential "successes"? If you did it
this way, you'd use the hypergeometric distribution with k = 0, N = 25, m =
6 and n = 5.
The second approach feels more natural, but is it?
John-Paul
1:05 p.m.
It seems like it has to be a violation of independence, because the order
matters here; the probability of a second success is conditional on the
probability of the first success.
But that assumes that we interview one person, then pick another person and
interview them and so on. If we select five people *at the start*, the
probability that all five are not government supporters would seem to follow
a binomial distribution.
John-Paul
On Sat, Feb 14, 2009 at 12:23 PM, Natasha Khouri
<natasha.khouri at gmail.com>wrote:
I would think the 2nd approach, too. Each draw
alters the prob of
success for the next draw a significant enough amount.
Btw?in general would this be a violation of independence?
------------------------------
*From:* gov2001-l-bounces at lists.fas.harvard.edu [mailto:
gov2001-l-bounces at lists.fas.harvard.edu] *On Behalf Of *John-Paul Ferguson
*Sent:* Saturday, February 14, 2009 12:17 PM
*To:* gov2001-l at lists.fas.harvard.edu
*Subject:* [gov2001-l] Clarification on pset 2, question 2
Hi all,
The wording on questions like this always confuses me. We are asked, "If a
researcher interviews 5 separate individuals [out of the 25 in the village]
randomly selected with equal probability, what is the probability that she
will not talk to a single supporter of the government?"
I seems like you could treat this two ways:
1. What is the probability of getting 0 successes in 5 draws from a
population of arbitrary size but fixed probability of success? If you did it
this way, you'd use the binomial distribution with n = 5, k = 0 and pi =
6/25.
2. What is the probability of getting 5 sequential failures in 5 draws from
a population of 25 where there are 6 potential "successes"? If you did it
this way, you'd use the hypergeometric distribution with k = 0, N = 25, m =
6 and n = 5.
The second approach feels more natural, but is it?
John-Paul
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
1:50 p.m.
Hunh! This is one where writing out the rules for the program makes it clear
which distribution you should use; and the results confirm it.
John-Paul
On Sat, Feb 14, 2009 at 1:05 PM, John-Paul Ferguson <jpferg at mit.edu> wrote:
It seems like it has to be a violation of
independence, because the order
matters here; the probability of a second success is conditional on the
probability of the first success.
But that assumes that we interview one person, then pick another person and
interview them and so on. If we select five people *at the start*, the
probability that all five are not government supporters would seem to follow
a binomial distribution.
John-Paul
On Sat, Feb 14, 2009 at 12:23 PM, Natasha Khouri <natasha.khouri at gmail.com
wrote:
I would think the 2nd approach, too. Each draw
alters the prob of
success for the next draw a significant enough amount.
Btw?in general would this be a violation of independence?
------------------------------
*From:* gov2001-l-bounces at lists.fas.harvard.edu [mailto:
gov2001-l-bounces at lists.fas.harvard.edu] *On Behalf Of *John-Paul
Ferguson
*Sent:* Saturday, February 14, 2009 12:17 PM
*To:* gov2001-l at lists.fas.harvard.edu
*Subject:* [gov2001-l] Clarification on pset 2, question 2
Hi all,
The wording on questions like this always confuses me. We are asked, "If a
researcher interviews 5 separate individuals [out of the 25 in the village]
randomly selected with equal probability, what is the probability that she
will not talk to a single supporter of the government?"
I seems like you could treat this two ways:
1. What is the probability of getting 0 successes in 5 draws from a
population of arbitrary size but fixed probability of success? If you did it
this way, you'd use the binomial distribution with n = 5, k = 0 and pi =
6/25.
2. What is the probability of getting 5 sequential failures in 5 draws
from a population of 25 where there are 6 potential "successes"? If you did
it this way, you'd use the hypergeometric distribution with k = 0, N = 25, m
= 6 and n = 5.
The second approach feels more natural, but is it?
John-Paul
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
1:59 p.m.
Independence is violated regardless of whether we pick people at once or in
order, simply because we cannot pick the same person twice. This is also
related to the "finite sample correction" that Gary talked about in a
previous email. The finite sample correction adjusts for the fact that our
dataset is a random sample from a population except that we do not sample an
observation twice, which is what we really should be doing if we want
unbiased estimates.
On Sat, Feb 14, 2009 at 1:05 PM, John-Paul Ferguson <jpferg at mit.edu> wrote:
It seems like it has to be a violation of
independence, because the order
matters here; the probability of a second success is conditional on the
probability of the first success.
But that assumes that we interview one person, then pick another person and
interview them and so on. If we select five people *at the start*, the
probability that all five are not government supporters would seem to follow
a binomial distribution.
John-Paul
On Sat, Feb 14, 2009 at 12:23 PM, Natasha Khouri <natasha.khouri at gmail.com
--
Patrick Lam
Department of Government and Institute for Quantitative Social Science,
Harvard University
http://www.people.fas.harvard.edu/~plam
wrote:
I would think the 2nd approach, too. Each draw
alters the prob of
success for the next draw a significant enough amount.
Btw?in general would this be a violation of independence?
------------------------------
*From:* gov2001-l-bounces at lists.fas.harvard.edu [mailto:
gov2001-l-bounces at lists.fas.harvard.edu] *On Behalf Of *John-Paul
Ferguson
*Sent:* Saturday, February 14, 2009 12:17 PM
*To:* gov2001-l at lists.fas.harvard.edu
*Subject:* [gov2001-l] Clarification on pset 2, question 2
Hi all,
The wording on questions like this always confuses me. We are asked, "If a
researcher interviews 5 separate individuals [out of the 25 in the village]
randomly selected with equal probability, what is the probability that she
will not talk to a single supporter of the government?"
I seems like you could treat this two ways:
1. What is the probability of getting 0 successes in 5 draws from a
population of arbitrary size but fixed probability of success? If you did it
this way, you'd use the binomial distribution with n = 5, k = 0 and pi =
6/25.
2. What is the probability of getting 5 sequential failures in 5 draws
from a population of 25 where there are 6 potential "successes"? If you did
it this way, you'd use the hypergeometric distribution with k = 0, N = 25, m
= 6 and n = 5.
The second approach feels more natural, but is it?
John-Paul
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
1:15 p.m.
I would think about it the second way, since we do not interview the same
person twice.
That would violate the principle of independence.
On Sat, Feb 14, 2009 at 12:23 PM, Natasha Khouri
<natasha.khouri at gmail.com>wrote:
I would think the 2nd approach, too. Each draw
alters the prob of
success for the next draw a significant enough amount.
Btw?in general would this be a violation of independence?
------------------------------
*From:* gov2001-l-bounces at lists.fas.harvard.edu [mailto:
gov2001-l-bounces at lists.fas.harvard.edu] *On Behalf Of *John-Paul Ferguson
*Sent:* Saturday, February 14, 2009 12:17 PM
*To:* gov2001-l at lists.fas.harvard.edu
*Subject:* [gov2001-l] Clarification on pset 2, question 2
Hi all,
The wording on questions like this always confuses me. We are asked, "If a
researcher interviews 5 separate individuals [out of the 25 in the village]
randomly selected with equal probability, what is the probability that she
will not talk to a single supporter of the government?"
I seems like you could treat this two ways:
1. What is the probability of getting 0 successes in 5 draws from a
population of arbitrary size but fixed probability of success? If you did it
this way, you'd use the binomial distribution with n = 5, k = 0 and pi =
6/25.
2. What is the probability of getting 5 sequential failures in 5 draws from
a population of 25 where there are 6 potential "successes"? If you did it
this way, you'd use the hypergeometric distribution with k = 0, N = 25, m =
6 and n = 5.
The second approach feels more natural, but is it?
John-Paul
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
--
Patrick Lam
Department of Government and Institute for Quantitative Social Science,
Harvard University
http://www.people.fas.harvard.edu/~plam
5571
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gov2001@lists.gking.harvard.edu
5 comments
3 participants
participants (3)
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jpferg＠mit.edu -
natasha.khouri＠gmail.com -
plam＠fas.harvard.edu