confidence interval with R

[Snowy88]

I am trying to predict the amount that a male with average status, income and verbal score would spend along with an appropriate 95% CI.

I used my linear model with all my variables and sex is coded as male=0 and female=1 in data set.
I think I did something wrong because I get all 47 observations but there are 19 females and 28 males.
I set up the prediction as:
> predict(mdl, sex=0, interval='confidence', level=0.90)
fit lwr upr
1 -10.6507430 -21.4372267 0.1357407
2 -9.3711318 -21.9428731 3.2006095
3 -5.4630298 -15.0782882 4.1522286
4 24.7957487 12.5630143 37.0284831...
please help...

[Dason]

It's not clear what you're trying to do. Here is an example of building a model and using predict:

n <- 30
x1 <- rnorm(n)
x2 <- runif(n)
y <- 2 + x1 - .5*x2 + rnorm(n)
dat <- data.frame(y, x1, x2)

o <- lm(y ~ x1 + x2, data = dat)

# Gives predictions for every value in the data set
predict(o, interval = "confidence", level = .90)

# If we want new predictions we need to use the newdata
# argument

newdat <- data.frame(x1 = c(1, 2), x2 = c(.5, .7))
predict(o, newdata = newdat, interval = "confidence", level = .9)

[Snowy88]

I have a data set with 47 observation and 5 variables (sex, income, status, verbal score and spending) where sex is coded as 0=male and female for 1.
The questions was to predict the amount that a male with average income, status and verbal score would spend with a 95%CI.
i know i have to use my linear model to predict average, but my prediction set up is wrong. How do I predict for male when its coded as such in my data...that is where i get confused...

[Dason]

You would use the exact same setup as I gave in my previous post but you would plug in your variables and the values you want them to take into the newdata parameter. In your case sex=0 gives male so you would make sure that the sex portion of newdata had a 0...

[Snowy88]

sorry Dason but I like I said I am new to R and I still dont understand. I am trying to clarify the prediction and CI for the problem because I have to then repeat the prediction with maximal values and compare CI to determine which is wider. Thanks

[Snowy88]

kindly explain in more detail so i can try to complete the problem. Thank you

[Dason]

I guess it's still not clear to me what is giving you trouble. Have you worked through that example I gave you?

[Snowy88]

I need the prediction for male but i get all 47 observation instead of the prediction for male on average?

then when doing the newdata as you suggested > I used the max. values found in my summary for each variable ...Was that correct?
g2<-data.frame(status=75, income=15, verbal=10, sex=0)
> predict(g,g2, level=.90)
> predict(g,g2, interval='confidence', level=.90)
fit lwr upr
1 71.30794 47.07516 95.54072

[Dason]

So to get the predictions for the males where everything else is average put in the average values for the other predictors.

[Snowy88]

but I also need the "prediction" band for max values as well correct?
> predict(g,g2, interval='prediction', level=.90)
fit lwr upr
1 71.30794 26.10037 116.5155

[Dason]

Ok... I don't know what to do with this thread anymore. You aren't actually asking a question there. It's not clear what is giving you problems.

[hlsmith]

If it is still unclear, I believe Snowy88 wants 95% confidence interval around the sex estimate (beta coefficient).

[Dason]

confint(o)

will do that. But I don't think that's what they want. They talk about prediction and this doesn't get you a prediction.

[hlsmith]

I was assuming this from their posts yesterday. Another way not using R would be:

If sample size > 30 the following is typically used:
-22.1133-1.96(8.21111) and -22.1133+1.96(8.21111)
Or (95% CI: -38.20708, -6.01952)

[Snowy88]

I need the prediction on average for male with a 95%CI then repeat with maximal values using the variables status, income and verbal. After I need to determine which CI is wider and explain...Now If I did the predictions correctly shown above with the mean then the max values. But doesn't that give 2 prediction intervals and 2 confident intervals for both the average and max. values ie..Average
> g1<-data.frame(status=43, income=4.64, verbal=6.66, sex=0)
> predict(g,g1, interval='confidence', level=.90)
fit lwr upr
1 28.11506 19.7603 36.46983
> predict(g,g1, interval='prediction', level=.90)
fit lwr upr
1 28.11506 -10.95281 67.18293

How will I create a plot to show which CI are wider?