Regression and confidence intervals

[SirEllwood]

In a study of
the effect of temperature (x) on yield (y) of a chemical process, the following data were obtained:

x 25 26 27 28 29 30 31 32 33 34 35
y 11 15 14 17 20 18 19 23 24 23 28

(a) Plot the data and superimpose the linear regression of y on x fitted by least squares.
(b) Find the 95% confidence intervals for beta0 and beta1.
(c) Predict the mean value of y at x = 30.5 and give an associated 95% confidence interval.
(d) Calculate a tolerance interval for y at x = 30.5 and give an associated 95% confidence interval.
(e) Why is the prediction interval smaller than the tolerance interval?

The problem im having with the above question is that iv never had to work out these things by hand before and show my working.... iv always been able to use R..... but now i want to understand the manual method

 

[Deleted member 4993]

In a study of
the effect of temperature (x) on yield (y) of a chemical process, the following data were obtained:

x 25 26 27 28 29 30 31 32 33 34 35
y 11 15 14 17 20 18 19 23 24 23 28

(a) Plot the data and superimpose the linear regression of y on x fitted by least squares.
(b) Find the 95% confidence intervals for beta0 and beta1.
(c) Predict the mean value of y at x = 30.5 and give an associated 95% confidence interval.
(d) Calculate a tolerance interval for y at x = 30.5 and give an associated 95% confidence interval.
(e) Why is the prediction interval smaller than the tolerance interval?

The problem im having with the above question is that iv never had to work out these things by hand before and show my working.... iv always been able to use R..... but now i want to understand the manual method

Do you know the definitions of :

1) beta0 and beta1

2) tolerance interval

3) confidence interval

Please share your work with us, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[SirEllwood]

Ok i have managed tow ork through the first part to get Sxx = 110 and Sxy = 158, so hence Beta = 1.436 and alpha = -23.81

For the second part, i udnerstand that a confidence interval is "beta +/- t0.975,n-2 * sqrt(var(beta))" but i have been thrown by having to do two seperate intervals for beta0 and beta1.... im am used to just one for beta(hat) so im confused?

for the third part im assuming i just need to substitue x=30.5 into the linear regression equation so the mean value of y would be 19.9799?.... but again im stuck with the confidence interval?

parts 4 and 5.... i have no idea where to start!

Am i going ok so far?

Thanks