Question

1. The following is data obtained from an experiment, Do this by hand and using MATLAE 2.5 a. Determine the equation ofthe least squares line of the form y-As + Ax b. Determine the significance of the fit What is the R2? c. Determine the confidence intervals for the βο and A. Also use bootstrap analysis to determine the confidence intervals of the regression coefficient. d. Test the hypothesis that B 5 e. What are the mean response corresponding tox 3.75? What is the confidence interval associated with the mean response? f. What is the response corresponding to x = 3.75? What is the confidence interval associated with this response? What is the correlation coefficient? g. h. Analyze the residuals. Determine the correlation coefficient and its 95% confidence interval. Also conduct a bootstrap analysis.

Answer 1

a) Using the following MATLAB command we can get information about X and Y x2.0:0.5:8.0 y-15 81 18 16 18 18 25 21 31 37 31 39]; C -cov (x ,y); vx var (x); mx - mean (x); my mean (y); C, vx ,mx, my Producng the following output: С 17.458 3.7917 my-22.154 From this we can say that the least square estimates of Po, Bi are respectively A-C/vx-4.6043 and β(,-my-βι ma--0.867. Hence the least square equation would be y0.867 4.6043 x r b) We will use sum of square errors to measure the goodness of fit: e (y-yy ) . ^ 2 ; sse- sum(e) Giving us the SSE- 171.07, which does not indicate a good fit. Now for R2, we will run the following code: e (y-yy ) . ^ 2 ; sse- sum (e); vy- var (y)*13; r2 -1 - (sse/vy) which gives us R2-0.86096. It implies that 86.096% of the variance in y is explained by the linear equation, which is considered as a good fit

c) In order to find the confidence interval for A) , Ả we need to find their standard error(SE) first. The standard error for Bo, B can be obtained by doing following sse For our sample data, SE(A)-2.0252 SE(%) = 10.185 Hence a 5% confidence interval for A) , A would be 4.6043-2.201 × 2.0252 <A < 4.6043 + 2.201 × 2.0252 -0.867 2.201 × 10.185 <A) <-0.867 +2.201 × 10.185 → 0.1468348 <A < 9.0617652 23.284 185 </30 21.550185 Also we can do a bootstrap analysis to obtain the standard error by the following code: resid (y-yy): se std (bootstrp (1000, (bootr) regress (yy+bootr ,x), resid)); d) Observe that A-5 falls in the 5% confidence interval we have calculated above. Hence we can accept the null hypothesis and say that ^i-5