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| AverageJoe... |
 Posted: Fri Aug 21, 2009 4:47 am |
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Guest
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Hi,
suppose the probability for a normal distributed event is 25%.
Someone achieves a hit rate of 34.46% (122 hits out of 354).
How significant is this result, ie. what is the z score for this result?
Is it z = 3.89 ? |
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| Ray Koopman... |
| Posted: Fri Aug 21, 2009 12:23 pm |
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Guest
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On Aug 21, 3:47 am, AverageJoe <no... at (no spam) amitrader.com> wrote:
[quote:7a4e216b7b]Hi,
suppose the probability for a normal distributed event is 25%.
Someone achieves a hit rate of 34.46% (122 hits out of 354).
How significant is this result, ie. what is the z score for this result?
Is it z = 3.89 ?
[/quote:7a4e216b7b]
If a variable is normally distributed then the probability of any
particular value is zero. Only intervals have a nonzero probability.
When do you mean by an event? |
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| Stan Brown... |
| Posted: Fri Aug 21, 2009 5:37 pm |
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Fri, 21 Aug 2009 12:47:00 +0200 from AverageJoe
<nospa at (no spam) amitrader.com>:
[quote:4267ff1b2e]
Hi,
suppose the probability for a normal distributed event is 25%.
Someone achieves a hit rate of 34.46% (122 hits out of 354).
How significant is this result, ie. what is the z score for this result?
Is it z = 3.89 ?
[/quote:4267ff1b2e]
Standard error = sqrt( p * (1-p) / n)
= sqrt( .25 * .75 / 354)
= about .023
z = (phat - p) / standard error
= (.3446-.25).023 = 4.54
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
Shikata ga nai... |
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| AverageJoe... |
| Posted: Sat Aug 22, 2009 3:36 am |
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Guest
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Ray Koopman wrote:
[quote:050e299d07]On Aug 21, 3:47 am, AverageJoe <no... at (no spam) amitrader.com> wrote:
Hi,
suppose the probability for a normal distributed event is 25%.
Someone achieves a hit rate of 34.46% (122 hits out of 354).
How significant is this result, ie. what is the z score for this result?
Is it z = 3.89 ?
If a variable is normally distributed then the probability of any
particular value is zero. Only intervals have a nonzero probability.
When do you mean by an event?
[/quote:050e299d07]
Hi, it has been solved. See the other reply here,
and here for more details: http://groups.google.com/group/sci.math/browse_thread/thread/8d6547beb0443a8e |
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| AverageJoe... |
| Posted: Sat Aug 22, 2009 3:44 am |
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Guest
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Stan Brown wrote:
[quote:7dbc9b4ae2]Fri, 21 Aug 2009 12:47:00 +0200 from AverageJoe
nospa at (no spam) amitrader.com>:
Hi,
suppose the probability for a normal distributed event is 25%.
Someone achieves a hit rate of 34.46% (122 hits out of 354).
How significant is this result, ie. what is the z score for this result?
Is it z = 3.89 ?
Standard error = sqrt( p * (1-p) / n)
= sqrt( .25 * .75 / 354)
= about .023
z = (phat - p) / standard error
= (.3446-.25).023 = 4.54
[/quote:7dbc9b4ae2]
Thx, the result of the above calc should be about 4.11 instead of 4.54 IMHO.
And, why did you call it "Standard error"? Shouldn't that be rather the "Standard deviation"?
The eq for the std error is StdErr = StdDev / sqrt(n)
according to http://en.wikipedia.org/wiki/Standard_error_(statistics)
Nevertheless it helped much, thx. |
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| Bruce Weaver... |
| Posted: Sat Aug 22, 2009 7:00 am |
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AverageJoe wrote:
[quote:7e2f9b99f8]Stan Brown wrote:
Fri, 21 Aug 2009 12:47:00 +0200 from AverageJoe <nospa at (no spam) amitrader.com>:
Hi,
suppose the probability for a normal distributed event is 25%.
Someone achieves a hit rate of 34.46% (122 hits out of 354).
How significant is this result, ie. what is the z score for this result?
Is it z = 3.89 ?
Standard error = sqrt( p * (1-p) / n)
= sqrt( .25 * .75 / 354)
= about .023
z = (phat - p) / standard error
= (.3446-.25).023 = 4.54
Thx, the result of the above calc should be about 4.11 instead of 4.54
IMHO.
And, why did you call it "Standard error"? Shouldn't that be rather the
"Standard deviation"?
The eq for the std error is StdErr = StdDev / sqrt(n)
according to http://en.wikipedia.org/wiki/Standard_error_(statistics)
Nevertheless it helped much, thx.
[/quote:7e2f9b99f8]
The standard error of a statistic *is* a standard deviation--it is
the standard deviation of the sampling distribution of that
statistic.
How you compute the SE depends on what the statistic is. For
example, if the statistic is the sample mean, then SE = SD /
SQRT(n). But if the statistic is the log of the odds ratio from a
2x2 table, SE = SQRT(1/a + 1/b + 1/c + 1/d). And if the statistic
is the sample proportion (p), then the (large-sample) SE is given
by the formula Stan shows above.
--
Bruce Weaver
bweaver at (no spam) lakeheadu.ca
http://sites.google.com/a/lakeheadu.ca/bweaver/
"When all else fails, RTFM." |
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| Stan Brown... |
| Posted: Sat Aug 22, 2009 8:09 pm |
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Guest
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Sat, 22 Aug 2009 11:44:38 +0200 from AverageJoe
<nospa at (no spam) amitrader.com>:
[quote:d84356ff83]
Stan Brown wrote:
Fri, 21 Aug 2009 12:47:00 +0200 from AverageJoe
nospa at (no spam) amitrader.com>:
Hi,
suppose the probability for a normal distributed event is 25%.
Someone achieves a hit rate of 34.46% (122 hits out of 354).
How significant is this result, ie. what is the z score for this result?
Is it z = 3.89 ?
Standard error = sqrt( p * (1-p) / n)
= sqrt( .25 * .75 / 354)
= about .023
z = (phat - p) / standard error
= (.3446-.25)/.023 = 4.54
Thx, the result of the above calc should be about 4.11 instead of 4.54 IMHO.
[/quote:d84356ff83]
Yes, you're right. I'm not sure how I got it wrong before.
[quote:d84356ff83]And, why did you call it "Standard error"? Shouldn't that be rather
the "Standard deviation"?
The eq for the std error is StdErr = StdDev / sqrt(n)
[/quote:d84356ff83]
Indeed it is, and that's the equation I used. The standard deviation
of a binomial population is sqrt(p*(1-p)), and dividing by sqrt(n)
gives the computation I used.
But a standard error *is* a standard deviation: it's the standard
deviation of the distribution of sample proportions. In chapter 1 of
your textbook you should have learned about sampling error, the
phenomenon that different samples typically do not give the same
statistic. (Of course sampling error is not an error or mistake,
just the name from the natural variability from one random sample to
the next. Sorry, I didn't create the terminology.)
The standard deviation of the sampling distribution of sample
proportions, or standard error of the proportion, is a yardstick for
the amount of variability to be expected in the sample proportion
phat of different samples.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
Shikata ga nai... |
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| Stan Brown... |
| Posted: Sun Aug 23, 2009 10:09 am |
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Guest
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Sat, 22 Aug 2009 22:09:01 -0400 from Stan Brown
<the_stan_brown at (no spam) fastmail.fm>:
[quote:2c1c98dc00]
Sat, 22 Aug 2009 11:44:38 +0200 from AverageJoe
nospa at (no spam) amitrader.com>:
Stan Brown wrote:
z = (phat - p) / standard error
= (.3446-.25)/.023 = 4.54
Thx, the result of the above calc should be about 4.11 instead of 4.54 IMHO.
Yes, you're right. I'm not sure how I got it wrong before.
[/quote:2c1c98dc00]
But at least I showed my work, so that you could see the method I
used and determine that I had made an arithmetic mistake. :-)
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
Shikata ga nai... |
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