[Lign251] HW3 Q3

[Tristan Davenport]

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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> >
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> >
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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
>
>
>
>
>
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