Been bugging me for a while. Havent found much help online
I have just 2 values
X- 20.53
Y- 3.23
No more data. 
I need to write a valid sentence to somehow state that Y is
STATISTICALLY lower than X. How would I go about proving and/or
stating it?
Thanks.

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Clarification of Question by mathstats-ga on 07 Mar 2005 11:37 PST

I didnt mean to state that I just need ONE valid sentence to state the
same. If we need to use 5 or 10 sentences to state it, it wouldnt
matter.. What I meant to ask was if there was a valid way to state
and/or prove that Y is STATISTICALLY lower than X.

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Request for Question Clarification by pafalafa-ga on 07 Mar 2005 12:51 PST

Why not simply say:  Y is lower than X...?

"Statistics" generally refers to properties of large collections of numbers.  

The average height of a room full of 50 people is a statistic.  

But if you're just comparing the heights of two people, it's perfectly
usual to say Y is taller than X, without having to invoke any
statistics.

Is this helpful at all?

pafalafa-ga

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Clarification of Question by mathstats-ga on 07 Mar 2005 13:40 PST

Saying lower makes my statement less validatory. I have no option but
to make a reasonable argument as to Y was decreased when compared to
X.

If that is absolutely not possible with one value each for X and Y,
what if I had 2 values for X, Y and Z and the levels were
X= 20.53 and 21.64
Y= 3.23 and 2.61
Z= 0.11 and 0.08
The problem is I dont know if the second set of values are entirely
accurate, which is why I was hoping to get it with the 1st set.. But
lets assume that I did have this second set of values. Now how do I
validatorily prove and/or state that Z<Y<X ?

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Clarification of Question by mathstats-ga on 07 Mar 2005 13:49 PST

For only one observation, Volterwd-ga suggested a non parametric inference.  

If this holds true and can be explained to me in dumb english, perhaps
using the data I gave, I think it would answer my question, would it
not? (if X= 20.53, Y= 3.23, Z= 0.11, how to show that X>Y>Z)

"Our null hypothesis is H0: median of X distribution = median of Y
distribution.  Our alternative hypothesis is H1: median of X
distribution > median of Y distribution.

Now under the null hypothesis we have P(X>Y)=a.  We would reject H0 of
course if X>Y and accept H0 if X<=Y.

The two values of interest would be the power and level of significance.

alpha = a
beta > a

These numbers would require information on the actual distributions
but we know that 3/4>a>0"

Unless you know the formal error or uncertainty associated with each
of your two values, you cannot say whether Y is significantly (in the
sense that word is used in statistics) smaller than X.

What if I had 2 values for X, Y and Z
X= 20.53 and 21.64
Y= 3.23 and 2.61
Z= 0.11 and 0.08
The problem is I dont know if the second set of values are entirely
accurate, which is why I was hoping to get it with the 1st set.. But
lets assume that I did have this second set of values. Now how do I
statistically prove and/or state that Z<Y<X ?

20.53 what.
% chance of succes?
Standard deviations?
Cost of 10 hamburgers and a soda?
Etheopians it takes to equal Rose Anne's weight?

How big is the sample size?
Any z-index?

"statistically smaller"? What exactly does that mean? "Statistically
smaller" as a phrase only has relevance if there's a z-index,
otherwise calling it smaller will suffice, unless you mean probability
of it's occurance.


If it's graphical rather than variable inputs (x and y), then we have
to ask what does the y axis stand for and what does the x axis stand
for to understand the point on the graph. It wouldn't make sense as a
graph though because then one could never be bigger than another.


x and y are statistically equal, given a sample size of two, because
they both have the same chance of occuring.

Ok... i would like to point out 1 problem.  Your question is not well
defined.  For example a well defined question would be... "Is there
evidence that the mean of X is lower than the mean of Y?" or "Is the
mean of X statistically lower than the mean of Y?"

And i should point you cant 'prove' anything with statistics... avoid that word.

So first off... lower case letters tend to represent OUTCOMES so
we have 
x- 20.53
y- 3.23
Now its quite obvious x<y... no statistical methods are required for
that... but perhaps you want to discuss whether the mean of the random
variable X is greater than the mean of the random variable Y.  Asking
if X>Y (the randomvariables) makes no sense.

Now since you only have one observation of each variable you will not
be able to make any parametric inference.

So lets make some non parametric inference.  

Our null hypothesis is H0: median of X distribution = median of Y
distribution.  Our alternative hypothesis is H1: median of X
distribution > median of Y distribution.

Now under the null hypothesis we have P(X>Y)=a.  We would reject H0 of
course if X>Y and accept H0 if X<=Y.

The two values of interest would be the power and level of significance.

alpha = a
beta > a

These numbers would require information on the actual distributions
but we know that 3/4>a>0

That is the best you can do.

mathstats

in response to your question for clarification

Since you have your data you know that x>y>z, thats not supposed to be
the question.  The question has to be phrases in terms of the random
variable... not the output.  You actually have the numbers so if you
simply want to say the output x is greater than the output y, you can.

So assume you want to talk about the median of the distribution i will
try and rephrase it.

Assume the medain of the distribution of X and the distribution of Y are the same.

Our criteria for deciding will be if the outcomes x and y follow x > y... 

Then we will have P(x>y) = a , where 0<a<3/4.  So if we observed x>y
(which we did) we reject the hypothesis that the medians are the same
and conclude that the median of X is greater than the median of Y.

Now the next part is of interest to statisticians.  

alpha (a) represents how often you will be wrong when null hypothesis
(the medians are the same) is true.  So if a is 0.5 you will wrongly
assume that the median of X is greater than the median of Y 50% of the
time.

Beta (B) represents how often you will be right when the alternative
hypothesis is true (median of X is greater than median of Y)

Do you actually want me to write a sentence?

How about this... 

Based on the data we have and using a non parametric procedure using
the medians.  We can be at least 25% confident that the mean of X is
greater than the mean of Y.

(25% is low... standard lowest is 90-95%, but since u have one
number... thats all you can do)

Sorry that should be MEDIAN

Based on the data we have and using a non parametric procedure using
the medians.  We can be at least 25% confident that the MEDIAN of X is
greater than the MEDIAN of Y.

I stand corrected... you can not make a confidence statement... i
incorrectly thought that 0<a<3/4... but constructed a situation where
3/4<a<1...

So you can say the same thing but make no confidence statement...
which of course renders the whole thing pointless... but well... it is
only 1 data point.

So rethinking the whole problem... i must say i believe you CAN NOT
make a valid statement simply with 1 data point without some prior
knowledge of the distributions.

You can say that x > y gives some evidence to say the MEAN or MEDIAN
of X > Y... but you can make no confidence statement... thus makes the
remarks pointless.

However if you had to make a statement, simply say something like...

and this will be my final answer

GIVEN THE DATA AT HAND, IT IS ONLY POSSIBLE TO SAY THAT THERE IS SOME
EVIDENCE THAT THERE IS A POSITIVE SHIFT IN SOME SENSE, IN THE
DISTRIBUTION OF X RELATIVE TO THE DISTRIBUTION OF Y.

Ok... after not being able to sleep and trying to remember my non
parametrics course i took 4 years ago.i will correct some
statements... btw users should be able to delete comments... how lame

You CAN make inference about the MEDIAN if you make this assumption. 
That the distribution of X and Y are similar up to a horizontal
shift...

If you assume this then P(x<y)=1/2 under the null hypothesis.

So your sentence could be...

Under the assumption that the distribution of X and Y are identical up
to a shift of delta (x-y) then we can test the hypothesis H0: delta =
0 vs H1: delta > 0 by accepting H0 if x<=y and rejecting H1 if x>y. 
We have that the probability of type 1 error (wrongly rejecting H0
when its true) is 1/2 and the power of the test is greater than 1/2
but depends on delta and the distributions.  So in our case we can say
that the median of X is greater than the median of Y with confidence
50% under the previously mentioned assumption.

1 more comment... you need to assume continuous distributions...
otherwise your back to square one... so you are assuming continuous
identical distributions up to median shift

volterwd-ga:
This is GREAT!! Thanks so much... Only I'm really not that great in
statistics.. I know math to some extent but am challenged much when it
comes to stats..
So if I have A=20, B=10, C=7, D=1, with your non-parametric
statements, is it right for me to safely conclude that we can be at
least 50% confident that the median of A is greater than the mean of B
and the median of B is greater than the median of C and the median of
C is greater than the median of D and the median of A is greater than
the median of D.

To state a valid statement, I would therefore have to say that

"Under the assumption that the continous distribution of a,b,c and d
are identical up to a shift of delta, we can test the hypothesis H0:
delta =0 vs H1: delta > 0 by accepting H0 if a<=b and rejecting H1 if
a>b. Therefore the probability of type 1 error is 1/2 and the power of
the test is greater than 1/2. The median of A is therefore greater
than the median of B (or C, or D) with confidence of 50%."

volterwd-ga
one more question.

say if i was measuring weights of 3 students in a class, student a,
student b and student c over a period of say 1 semester
Student A goes from 150 pounds to 130 pounds to 120 pounds to 100
pounds to 98 pounds
Student B goes from 147 pounds to 122 pounds to 120 pounds to 118
pounds to 116 pounds
Student C goes from 142 pounds to 157 pounds to 169 pounds to 172
pounds to 190 pounds

How do I validly (statistically) state that
a) Student A statistically decreased from 150 --> 130--> 120--> 100--> 98
b) Decrease in Student A was more significant that Student B
c) Student C statistically increased from 142 to 190 pounds

Finally how do I thank you for your GREAT help? I wish you could at
least accept this money posted here.

with regards to the ABCD... yeah its 50%... (which is low btw) thats
only for PAIRWISE comparisons...

with regards to the weights... 

These really dont seem like statistical problems... its quite obvious
they gained a large portion of weight... but lets do this a
statistical way

for A (and C)... if the person wasnt losing weight... ie... their
weight was just fluctuating.. you would expect their weight to go up
just as much as down correct?

so lets assume that the weight loss/gain is purely random... then what
is the chance of losing weight 4 times in a row?  well its about
1/32... (it would probably be smaller since when your losing weight it
should be harder to keep losing)

with regards to B... you need to define losing weight better... as a
fraction or by total lbs?

So we have a test that the person is losing weight... null hypothesis
is random fluctuation in weight... alternative is losing

Basically we are using a non-parametric test called the sign test...
(you could use a signed rank test but since they are all going down
they will give the same result)

So the probability of down down down down when fluctuating should be
about 1/32... (most likely less) and this is the a value i was talking
about before...

so this is 0.03125... which at a standard level of 5% says that there
is significant evidence that they are losing weight...

the same goes for D just change down and losing to up and gaining...

bad math make 1/2^4 = 1/16 = 0.0625

still significant at 10%... should be good enough especially
considering like i said it should be smaller then that number