On Sat, 10 Nov 2001 11:34:15 -0500 (EST)
Donald Burrill <[EMAIL PROTECTED]> wrote:
> You persist in repeating your original request in your original phrasing,
> with no elaboration(s) that might resolve the ambiguities therein.
Thanks very much for a completely unhelpful reply!
> > Thanks for your reply. Ummm, unfortunately I don't understand this :o)
>
> Not surprising.
No it's not, when you consider that I have only been studying statistics for 5 weeks.
Why do you feel the need to be so rude?
>
> > I am by no means a mathematician. I am studying psychology and 1/4 of
> > my course is statistics *for psychology*, ie it's pretty basic without
> > any of the advanced stuff (I hope!).
>
> A pity, if true. Adequate practice of psychology requires considerably
> more than a minimum knowledge -- and understanding! -- of statistics.
So you're saying that in order to carry out statistical analyses of psychological data
I need to be able to understand "t = 1/(1+pX) where p=.33267 then P(x) = 1-Z(x)*[a_1*t
+ a_2*t^2 + a_3*t^3] where a_1 = 0.4361836 a_2 = -0.1201676 a_3 = 0.9372980"? I hardly
think so.
> > All I want to know, for interest's sake, is how one calculates the
> > mean to z,
>
> Yes, you said that before. In the same words. For the sake of
> (possibly) furthering the conversation, I will assume that what you
> meant was something like "Given a value x of a variable X, which has a
> known mean, how does one convert x to z?"
No, that's not what I said. Re-read my original post and try again. Or don't.
> (Your language admits of
> several other possible meanings, but I'll leave it to you to clarify what
> you intended, if it wasn't what I've conjectured (and if you can).)
I think what I said was clear enough. Other people managed to understand it.
>
> The formula you request, for this purpose, converts x to z:
>
> z = (x - mean)/sd
>
> where "sd" is the known standard deviation of the variable X.
> Now, I'm sure your statistics instruction includes this equation; it
> follows that the question you really want to ask is (probably) something
> else. In which case we all await with interest your clarification.
>
> > larger proportion and smaller proportion of a standardised score,
> > without having to read through a long list of numbers.
>
> Hmm. Numbers scare you, do they?
No, I would just like to be able to work it out for myself without having to carry
round reams of paper listing every possible z score. If I know the method of doing it
myself, I can work it out on my Palm Pilot with a Basic program. That's why I want the
equation in the first place.
>
> There are essentially three ways
> of going about this part:
>
> 1. Look the proportions up in a table of the standard normal
> distribution, which by your account you are apparently too lazy to do.
No I'm not, I'm just trying to have less paper to look through.
> Sounds as though you're being inefficient, by the way: there's no need
> to "read through a long list of numbers", only to look up a single
> number in the table (the other proportion you can get by subtracting
> from 1.)
For christ sake, you know what I mean. Do you really think I meant that I would have
to read every single number until I got to the one I wanted? I don't think so. You're
just being pedantic.
>
> 2. Use convenient statistical software (MINITAB, SAS, SPSS, a TI-83
> calculator, etc.) to calculate the proportions by numerical
> approximation. This of course does not satisfy your request for
> "the formulae".
This does not exist for the Palm Pilot which is why I want the formula so I can write
the programme myself.
>
> 3. Start with the mathematical expression for the density function of a
> standard normal distribution, and integrate it from minus infinity to z.
> Which is what Rich was referring to when he asked if you wanted the
> calculus. Again, by your account you haven't the mathematics for this;
> especially as the integral in question does not exist in closed form.
> (Which, of course, is precisely the reason why tables were constructed
> in the first place, to avoid a _very_ tedious computational chore every
> time one had a value of z for which proportions, or probabilities, were
> needed.)
Fair enough, but of course the task would not be tedious if carried out by a computer.
>
> > Forgive me if that was covered in your FAQ, but I couldn't see
> > it! Perhaps you could point me in the direction of the formulae?
>
> Forgive me if my candour is uncomfortable, but this sounds to me very
> like asking a sorcerer for the spell(s) you think he uses. Do you want a
> magic wand also, and perhaps a cloak of invisibility?
You are being ridiculous. All I want to know is if there exists an equation to work
out the things I have said. Clearly there is, otherwise we would not have the tables
of z-scores.
>
> I am reminded of the time, years ago, .......
This is irrelevant.
You are obviously well educated in the field of statistics, so if you have a helpful
reply I would like to hear it. Otherwise please don't reply.
Mark
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