[Lign251] HW3 Q3

[rlevy at ling.ucsd.edu]

Hi Tristan, Melanie,

Take a look at my response to Kate's query on this one.

Roger

Tristan Davenport wrote:
> Hi,  Melanie and I got the same error as Kathryn.  Here's what we did:
>
> Using the peaks in figure 1 as our values for mu.1 and mu.2, and setting
both sigma variables as square root of the standard deviation of
word.means, we then used the code you gave us in problem 3:
>
>> mu.1 <- 6.35
>> mu.2 <- 6.57
>> sigma.1 <- sqrt(sd(word.means))
>> sigma.2 <- sigma.1
>> c<-.5
>> d2norm <- function(x, mu.1, mu.2, sigma.1, sigma.2, c)
> + { c * dnorm(x,mu.1,sigma.1) + (1-c) * dnorm(x, mu.1, sigma.2)}
>> lines(d2norm)
> Error in as.vector(x, "double") : cannot coerce to vector
>
> If alternatively you do plot(d2norm), you get:
>
>> plot(d2norm)
> Error in x(x) : argument "c" is missing, with no default
>
> ... even though we defined a value for "c" earlier.
>
> In conclusion, this is frustrating.
>
> ~Tristan & Melanie
>
> On 10/21/07, rlevy at ling.ucsd.edu <rlevy at ling.ucsd.edu> wrote:
>> Hi Kate,
>> Could you post a code snippet that gives this error?
>> Thanks!
>> Roger
>> Kathryn Davidson wrote:
>>> In problem 3, I have a perfectly good plot (using plot()) of the
>> function that seems to estimate the histogram. However, I am really
having problems using lines() to overlay it on top of the histogram. It
just keeps saying:
>>> Error in as.vector(x, "double") : cannot coerce to vector
>>> So I guess my function isn't in vector format... but I can't figure
out
>> how to make it one. In Baayen pg. 27 I think he's answering this
question but I can't follow what he's saying.
>>> Any tips?
>>> Kate
>>> On 10/21/07, *rlevy at ling.ucsd.edu <mailto:rlevy at ling.ucsd.edu>*
>> <rlevy at ling.ucsd.edu <mailto:rlevy at ling.ucsd.edu>> wrote:
>>>     Rebecca Colavin wrote:
>>>      > Okay, I give up.
>>>      >
>>>      > The graph in the question generates a distribution where the
>>>     probability
>>>      > mass (=1) is contained under the curve. In short, the y axis is
>>>     (0,9) or
>>>      > so .
>>>      >
>>>      > But when I try to do lines over it with a normal distribution,
>> the y
>>>     values are probabilities. The y axis for this is (0,1).
>>>     Hi Rebecca,
>>>     Because the normal distribution is a continuous probability
density,
>> the
>>>     y values (for the density) are not bounded above by 1.  Remember,
it
>> is
>>>     the *area under the curve* that must total 1.  If the scale of the
x
>>  axis
>>>     is small, p(x) can easily surpass 1.
>>>     As an example, try:
>>>      > x <- seq(-5,5,by=0.01)
>>>      > plot(x,dnorm(x,0,0.1),type="l")
>>>     This normal density peaks around p(x) = 4.
>>>     Using hist() with prob=T also is an estimation of density, which
is
>> why
>>>     the hist() y axis also surpasses 1.
>>>     Roger
>>>     --
>>>     Roger Levy                      Email: rlevy at ucsd.edu
>>>     <mailto:rlevy at ucsd.edu>
>>>     Assistant Professor             Phone: 858-534-7219
>>>     Department of Linguistics       Fax:   858-534-4789
>>>     UC San Diego                    Web:   http://ling.ucsd.edu/~rlevy
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>> --
>> Roger Levy                      Email: rlevy at ucsd.edu
>> Assistant Professor             Phone: 858-534-7219
>> Department of Linguistics       Fax:   858-534-4789
>> UC San Diego                    Web:   http://ling.ucsd.edu/~rlevy
_______________________________________________
>> Lign251 mailing list
>> Lign251 at ling.ucsd.edu
>> http://pidgin.ucsd.edu/mailman/listinfo/lign251
>
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-- 

Roger Levy                      Email: rlevy at ucsd.edu
Assistant Professor             Phone: 858-534-7219
Department of Linguistics       Fax:   858-534-4789
UC San Diego                    Web:   http://ling.ucsd.edu/~rlevy