First , sorry for my last post there was a picture attachment that I thought
would paste as text.

My problem is how do I get the answer I need it has puzzled me for days,the
following formula gives me 15.9% which is correct
=1-NORMDIST(1,0,1,TRUE)

the number one is derived from the z test which I can not get to work and
any help would be great,
in the table below

WINS BETS
55 100 = Z TEST = 1 = << =1-NORMDIST(1,0,1,TRUE) WHICH GIVES THE
15.9%

IF THERE WERE
WINS BETS
275 500 = Z TEST = .854 ( THIS IS THE NUMBER I CANT GET ) = 1.3%
550 1000= Z TEST = .842 ( ""
"") = .1%

THE PARAMETERS FOR THE TEST ARE

A = NUMBER OF WINNING SELECTIONS
B = NUMBER OF BETS MADE
C = A 5% EDGE ON A 50% LINE

SEE ARTICLE BELOW
---------------------------------------------------------------------------------------------
What we do when testing such information is assume that each bet that he
makes has a 50% chance of winning.
Because he is betting on the line for each bet, this is of course a good
assumption. Naturally he would want this
to be greater than 50% so that he can make a profit, but for the time being
we will assume that it is 50%.

Now the punter has recorded a success rate of 55% of his bets, thus meaning
he is 5% above the average if he
was just blindly betting on any bet. To test whether this is due to luck or
wise betting we have to use what is
called the "binomial test" or the "one sample z-test for proportions".
------------------------------------------------------------------------------------------------------------------------------------------------

If you can help with the excel solution the above it would be great, I have
tried the z test function and can get the 1
but when I apply the other examples it does not give me the answer I need
and am doing something wrong.

regards
Pete

--
(][ This Email has been scanned by Norton AntiVirus. ][)

Most participants will not not do your homework for you, but will help
you figure out how to do it yourself.

Look back at your lecture notes/text. What is the mean and variance of
a binomial variable? What does that reduce to under the assumption that
p=0.5? Now, perform a normal theory test on the observed number of wins
under the assumption that the true mean and variance are those values
from the binomial distribution.

Jerry

OZDOC1050 wrote:

First , sorry for my last post there was a picture attachment that I thought
would paste as text.

My problem is how do I get the answer I need it has puzzled me for days,the
following formula gives me 15.9% which is correct
=1-NORMDIST(1,0,1,TRUE)

the number one is derived from the z test which I can not get to work and
any help would be great,
in the table below

WINS BETS
55 100 = Z TEST = 1 = << =1-NORMDIST(1,0,1,TRUE) WHICH GIVES THE
15.9%

IF THERE WERE
WINS BETS
275 500 = Z TEST = .854 ( THIS IS THE NUMBER I CANT GET ) = 1.3%
550 1000= Z TEST = .842 ( ""
"") = .1%

THE PARAMETERS FOR THE TEST ARE

A = NUMBER OF WINNING SELECTIONS
B = NUMBER OF BETS MADE
C = A 5% EDGE ON A 50% LINE

SEE ARTICLE BELOW
---------------------------------------------------------------------------------------------
What we do when testing such information is assume that each bet that he
makes has a 50% chance of winning.
Because he is betting on the line for each bet, this is of course a good
assumption. Naturally he would want this
to be greater than 50% so that he can make a profit, but for the time being
we will assume that it is 50%.

Now the punter has recorded a success rate of 55% of his bets, thus meaning
he is 5% above the average if he
was just blindly betting on any bet. To test whether this is due to luck or
wise betting we have to use what is
called the "binomial test" or the "one sample z-test for proportions".
------------------------------------------------------------------------------------------------------------------------------------------------

If you can help with the excel solution the above it would be great, I have
tried the z test function and can get the 1
but when I apply the other examples it does not give me the answer I need
and am doing something wrong.

regards
Pete

Thanks For your help Jerry, I'm way past home work and have been helping out
in the
forums in some way for years but this one has me stuck
perhaps im to old, any way I see by you clue it has something to do with the
..05 p
but really all this stuff is Dutch to me, so if you could help out further
it would be great
regards
Pete

--
(][ This Email has been scanned by Norton AntiVirus. ][)
"Jerry W. Lewis" wrote in message
...
Most participants will not not do your homework for you, but will help you
figure out how to do it yourself.

Look back at your lecture notes/text. What is the mean and variance of a
binomial variable? What does that reduce to under the assumption that
p=0.5? Now, perform a normal theory test on the observed number of wins
under the assumption that the true mean and variance are those values from
the binomial distribution.

Jerry

OZDOC1050 wrote:

First , sorry for my last post there was a picture attachment that I
thought would paste as text.

My problem is how do I get the answer I need it has puzzled me for
days,the following formula gives me 15.9% which is correct
=1-NORMDIST(1,0,1,TRUE)

the number one is derived from the z test which I can not get to work and
any help would be great,
in the table below

WINS BETS
55 100 = Z TEST = 1 = << =1-NORMDIST(1,0,1,TRUE) WHICH GIVES THE
15.9%

IF THERE WERE
WINS BETS
275 500 = Z TEST = .854 ( THIS IS THE NUMBER I CANT GET ) = 1.3%
550 1000= Z TEST = .842 ( "" "") = .1%

THE PARAMETERS FOR THE TEST ARE

A = NUMBER OF WINNING SELECTIONS
B = NUMBER OF BETS MADE
C = A 5% EDGE ON A 50% LINE

SEE ARTICLE BELOW
---------------------------------------------------------------------------------------------
What we do when testing such information is assume that each bet that he
makes has a 50% chance of winning.
Because he is betting on the line for each bet, this is of course a good
assumption. Naturally he would want this
to be greater than 50% so that he can make a profit, but for the time
being we will assume that it is 50%.

Now the punter has recorded a success rate of 55% of his bets, thus
meaning he is 5% above the average if he
was just blindly betting on any bet. To test whether this is due to luck
or wise betting we have to use what is
called the "binomial test" or the "one sample z-test for proportions".
------------------------------------------------------------------------------------------------------------------------------------------------

If you can help with the excel solution the above it would be great, I
have tried the z test function and can get the 1
but when I apply the other examples it does not give me the answer I need
and am doing something wrong.

regards
Pete

OZDOC1050 wrote:

Thanks For your help Jerry, I'm way past home work and have been helping out
in the
forums in some way for years but this one has me stuck
perhaps im to old, any way I see by you clue it has something to do with the
.05 p
but really all this stuff is Dutch to me, so if you could help out further
it would be great
regards
Pete


For a binomial random variable, mean = n*p; variance = n*p*(1-p). The
general form of a z-test is
(x - mu(x)) / sigma(x)
where the standard deviation, sigma, is the square root of the variance.

The Excel worksheet function ZTEST looks different because it takes x to
be a sample average, but its sigma argument is the standard deviation of
a single observation, not the standard deviation of of the average,
hence the extra sqrt(n) factor.

You can save typing by using the fact that
NORMSDIST(x) = NORMDIST(x,1,0,TRUE)

Jerry