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Question to be answered

Does anyone know how to solve the attached problem in two lines of code? I believe an as.matrix would work to create a matrix, X, and then use X %*% X, t(X), and solve(X) to get the answer. However, it does not seem to be working. Any answers will help, thanks.

nghauran
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Zach
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    Please show us what you've tried so far. It would be helpful for you to produce a [reproducible example](https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example) – Kevin Arseneau Oct 04 '17 at 23:42

2 Answers2

1

I would recommend using read.csv instead of read.table

It would be useful for you to go over the difference of the two functions in this thread: read.csv vs. read.table 

df <- read.csv("http://pengstats.macssa.com/download/rcc/lmdata.csv")
model1 <- lm(y ~ x1 + x2, data = df)
coefficients(model1) # get the coefficients of your regression model1
summary(model1) # get the summary of model1
TechWizard
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Based on the answer of @kon_u, here is an example to do it by hands:

df <- read.csv("http://pengstats.macssa.com/download/rcc/lmdata.csv")
model1 <- lm(y ~ x1 + x2, data = df)
coefficients(model1) # get the coefficients of your regression model1
summary(model1) # get the summary of model1 

### Based on the formula
X <- cbind(1, df$x1, df$x2) # the column of 1 is to consider the intercept
Y <- df$y
bhat <- solve(t(X) %*% X) %*% t(X) %*% Y # coefficients 
bhat # Note that we got the same coefficients with the lm function
nghauran
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