Calculating slope of a curve at a single point (where x = 0.5) and testing that slope for significance (slope ~= 0)

Hi there,

I have been fitting a logistic regression model to some data using glmfit:

Data:

dataStims = [(16/36), (16/28), (16/16), (28/16), (36/16)]; % stimulus sizes
dataStims_long = linspace(dataStims(1), dataStims(end), 5000); % make long vector for stimulus sizes - to allow you to make a smoother curve when you fit a curve to your data
dataStims = log(dataStims); % take log to make spacing more even
dataStims_long = log(dataStims_long); % ''

dataResp = [9, 8, 6, 6, 0]; % count of affirmative responses (question: 'stimulus was down?' (options: down/ across)
dataCount = [10, 10, 10, 10, 10]; % count of trials presented

Code:

[~, ~, stats] = glmfit(dataStims', [dataResp' dataCount'], 'binomial', 'logit'); % fit logistic reg

This allows me to get some coefficients for the fit (stats.beta), which are

Output:

stats.beta =

0.4667

-2.1432

Which I understand to be the y-intercept and slope, respectively (someone please correct me if I am wrong). I also believe that the slope value returned is the slope of the line at the y-intercept (y = 0), is this correct too?

I am a bit confused re. the y-intercept, because it looks from the plot like the y-intercept ~= .6 (see image below). That is, the x value when y = 0 is ~= .6. But the y-intercept according to the stats.beta = 0.4667. Could someone let me know if I have misunderstood this?

FYI, I generated the curve using the following

Code:

logitFit_long = glmval(logitCoef, dataStims_long', 'logit');

To my main question: what I would like to do, is get the slope not at the y-intercept, but at the point of the line where x = .5 (marked by a * on the plot below). I want to get the slope at that point, and then test the significance of the slope.

Does anyone know a good way to do this?

I tried the below, which I think does what I intend by getting me the slope at the x = .05 point

Code:

[~, i] = min(abs(logitFit_long-.5)); % find value closest to x = .5

x = dataStims_long'; y = logitFit_long;
index=[i-1 i+1]; % takes the points where you want to calculate your slope (regression)

pse.coeff = polyfit(x(index),y(index),1); % calc coeff
pse.slope_line = polyval(pse.coeff,x); % calc line

Output:

pse.coeff =

-0.5358 0.6167

Which I understand to be the slope and the y-intercept, respectively. Correct?

This looks like what I want from plotting the pse.slope_line – see dashed line below. Looks like it is the slope at x = 0.5, as it intercepts the curve at the * point

What I would then like to do is to test whether this slope is significantly different from 0, i.e., is there a slope on that line or is it flat (slope = 0)?

With the glmfit stats from above, I am able to get the significance of the slope at the y-intercept using stats.t(2) and stats.p(2), for the t and p values of the significance test of the slope (the 2ndbeta value, i.e., the ‘2’).

But I am not sure how I would do it for the slope of the line at this x = .5 point. I tried a few things but they are giving me very odd values and I assume they are incorrect.

Does anyone have any ideas?

Thank you for your help,

Harriet

[Torsten]

Just to start at the beginning:

You have negative values in your original vector dataStims and you take the log of this vector. This will result in complex values for the negative entries. Did you take this into account ?

[HDJ]

Hi Torsten,

My apologies, I had pasted the code incorrectly. The negative values come from logging the data. I am not trying to log negative values, so no complex values.

I have amended my question as above

[HDJ]

Hi Torsten, Did you have a chance to look at my amended comment? Thank you for your help