Need help verifying I have the correct polyfit line for my function.

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Question: Ps7.mat contain scores for 100 subjects on two different tests. Use the linear regression function you wrote in class to fit a linear model to these data. Plot the data using a scatter plot and plot the fitted model with a line. Report the estimated parameters a and b in the figure title. Then, use polyfit to find the best-fit line predicting score2 from score1. Your coefficients should be the same as the ones you found using your linear regression function in problem 2. List these coefficients in a comment or show them on the figure.
Below I have included my regression line, my a (named alpha) and b (named beta) parameters (that I hope are correct), and what I plugged in for polyfit. I am a bit concerned because my polyfit gives me 0 -0.1606 , and my b is also equal to -0.1606 so I want to make sure that is correct, and if not where I'm going wrong.
load('ps7.mat')
function [b,a,rsq] = regressline(x,y)
n=length(x);
xbar=mean(x);
ybar=mean(y);
sigxsq= sum(x.^2) - (sum(x)^2)/n;
sigxy=sum(x.*y)-(sum(x)*sum(y))/n;
beta=sigxy./sigxsq; % = -0.1606
alpha=ybar-beta*xbar;% = -0.5841
plot(x,y,'o')
hold on
title('alpha=0.5841, beta=-0.1606')
plot(x,(beta*x) + alpha, 'r')
end
p=polyfit(alpha,beta,1)
plot(p)
%p=0 -0.1606
Thanks!

Respuesta aceptada

Walter Roberson
Walter Roberson el 18 de Nov. de 2021
No you should polyfit x and y.
  2 comentarios
Kristin Aldridge
Kristin Aldridge el 18 de Nov. de 2021
So it would be
p=polyfit(x,y,1)
plot(p)
%p=[-0.1606,0.5841]
Walter Roberson
Walter Roberson el 18 de Nov. de 2021
Do not plot() the output of polyfit. If you want to plot the prediction line then use polyval to create projected locations and plot() those

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