[R] confidence interval in "predict.lm"

[Martin Maechler]

You are looking for some (most?) statisticians call
 ``prediction interval''

==> just give "prediction" instead of "confidence" :

> predict(mod,data.frame(temp = 45),level = .95,
 +         interval = "prediction", se.fit = TRUE)

$fit
          fit      lwr     upr
[1,] 32.96429 26.48597 39.4426

$se.fit
[1] 0.9148715

$df
[1] 5

$residual.scale
[1] 2.348252



>>>>> "Fred" == Fred Mellender <fredm at frontiernet.net>
>>>>>     on Fri, 15 Nov 2002 11:43:28 -0500 writes:

    Fred> I am studying statistics using R and a book
    Fred> "Understandable Statistics", by Brase and Brase.  The
    Fred> book has two worked examples for calculating a
    Fred> confidence interval around a predicted value from a
    Fred> linear model.  The answers to the two examples in the
    Fred> book differ from those I get from R.  The regression
    Fred> line, the standard error, and the predicted value in R
    Fred> and the book all agree for the examples.  Hence I
    Fred> gather that R and the book use different formula to
    Fred> calculate the confidence interval.  Could someone
    Fred> explain why the difference exists, and which
    Fred> function(s) in R I might use to get the answers in the
    Fred> book, and (perhaps) an explanation as to which method
    Fred> to use in various situations).

    Fred> The example:

    >> x<-c(10,20,30,40,50,60,70) y<-c(17,21,25,28,33,40,49) dat
    >> <- data.frame(temp=x,amnt=y)
    Fred>   temp amnt 1 10 17 2 20 21 3 30 25 4 40 28 5 50 33 6
    Fred> 60 40 7 70 49

    Fred> being a table of temperatures (temp) and the
    Fred> corresponding amounts of copper sulfate that disolve
    Fred> in 100g of water at that temperature.

    Fred> The regression line:

    >> mod <- lm(amnt ~ temp,dat) summary(mod)

    Fred> Call: lm(formula = amnt ~ temp, data = dat)

    Fred> Residuals: 1 2 3 4 5 6 7 1.7857 0.7143 -0.3571 -2.4286
    Fred> -2.5000 -0.5714 3.3571

    Fred> Coefficients: Estimate Std. Error t value Pr(>|t|)
    Fred> (Intercept) 10.14286 1.98463 5.111 0.00374 ** temp
    Fred> 0.50714 0.04438 11.428 8.98e-05 *** --- Signif. codes:
    Fred> 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1

    Fred> Residual standard error: 2.348 on 5 degrees of freedom
    Fred> Multiple R-Squared: 0.9631, Adjusted R-squared: 0.9558
    Fred> F-statistic: 130.6 on 1 and 5 DF, p-value: 8.985e-05

    Fred> The .95 confidence interval for a temperature of 45
    Fred> degrees:
    >>
    Fred> foo<-predict(mod,data.frame(temp=45),level=.95,interval="confidence",se.fit=
    Fred> T)
    >> foo
    Fred> $fit fit lwr upr [1,] 32.96429 30.61253 35.31604

    Fred> $se.fit [1] 0.9148715

    Fred> $df [1] 5

    Fred> $residual.scale [1] 2.348252

    Fred> The book gives the confidence interval as 26.5 <= y <=
    Fred> 39.5.  The book defines the confidence interval
    Fred> calculation thus:

    Fred>   yp - E <= y <= yp + E

    Fred>   Where E = tc*sC *sqrt(1 + 1/n + (x-xBar)^2/SSx) yp
    Fred> is the predicted value from the regression line tc is
    Fred> the value from Student's t distribution for a
    Fred> confidence level, c, using n-2 degrees of freedom, sC
    Fred> is the standard error of estimate SSx is
    Fred> Sum(x^2)-[Sum(x)]^2/n n is the number of data pairs.

    Fred> So that even though the model, predicted value,
    Fred> standard error all agree, R gives a much smaller
    Fred> confidence interval than the book does.

    Fred> Thanks for any advice/help.

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