Calculating Prediction Interval

I have the following data located here. I am attempting to calculate the 95% confidence interval on the mean purity when the hydrocarbon percentage is 1.0. In R, I enter the following.

> predict(purity.lm, newdata=list(hydro=1.0), interval="confidence", level=.95)
fit      lwr      upr
1 89.66431 87.51017 91.81845

However, how can I derive this result myself? I attempted to use the following equation.

$$s_{new}=\sqrt{s^2\left(1+\frac{1}{N}+\frac{(x_{new}-\bar x)^2}{\sum(x_i-\bar x)^2}\right)}$$

And I enter the following in R.

> SSE_line = sum((purity - (77.863 + 11.801*hydro))^2)
> MSE = SSE_line/18
> t.quantiles <- qt(c(.025, .975), 18)
> prediction = B0 + B1*1
> SE_predict = sqrt(MSE)*sqrt(1+1/20+(mean(hydro)-1)^2/sum((hydro - mean(hydro))^2))
> prediction + SE_predict*t.quantiles
 81.80716 97.52146

My results are different from R's predict function. What am I misunderstanding about prediction intervals?

• How are you calculating the MSE in your code? – user25658 Sep 4 '13 at 5:25
• I added the calculation to the post. – idealistikz Sep 4 '13 at 5:29
• as MMJ suggested you should try predict(purity.lm, newdata=list(hydro=1.0), interval="prediction", level=.95) – vinux Sep 4 '13 at 9:30

Confidence interval $$s_{new}=\sqrt{s^2\left(\frac{1}{N}+\frac{(x_{new}-\bar x)^2}{\sum(x_i-\bar x)^2}\right)}$$
Prediction interval $$s_{new}=\sqrt{s^2\left(1+\frac{1}{N}+\frac{(x_{new}-\bar x)^2}{\sum(x_i-\bar x)^2}\right)}$$