Question

1. A farmer was interested in a relationship between the amount of fertilizer (x) and the number of bushels (y) of soybeans produced. The farmer conducted an experiment and obtained the following data.

Hundreds of pounds per acre (x)	Bushels per acre (y)
1.0	25
2.5	32
3.0	35
3.0	32
3.4	35
4.0	39
4.0	41
4.5	40

1. Draw a scatter plot. Do the sample data appear to indicate a linear relationship between the amount of fertilizer and the number of bushels?

b. Do the data appear to be positively or negatively correlated?

3. Find and graph (on the scatter diagram in part a) the line of the best fit for estimating the bushels per acre given hundreds of pounds of fertilizer.

4. Estimate the number of bushels given 3.5 hundred pounds of fertilizer.

Answer 1

Sol:

with plot function in R obtain a scatterplot

Rcode:

df2 =read.table(header = TRUE, text ="
fertilzer_peracre bushels_produced
1.0 25
2.5 32
3.0 35
3.0 32
3.4 35
4.0 39
4.0 41
4.5 40

"
)
df2

linreg <- lm(bushels_produced~fertilzer_peracre,data=df2)
summary(linreg)
plot(x= df2$fertilzer_peracre,y=df2$bushels_produced,
main="Scatterplot of bushels_produced vs fertilzer_peracre",
xlab="fertilzer_peracre",ylab="bushels_produced")

Intrpetation:

From scatterplot we observe there exists a strong linear realtionship between  amount of fertilizer (x) and the number of bushels (y) of soybeans produced.

With lm fucntion in R fit a linear regression of  the number of bushels (y) of soybeans produced and amount of fertilizer (x)

Rcode:

linreg <- lm(bushels_produced~fertilzer_peracre,data=df2)
summary(linreg)
plot(x= df2$fertilzer_peracre,y=df2$bushels_produced,
main="Scatterplot of bushels_produced vs fertilzer_peracre",
xlab="fertilzer_peracre",ylab="bushels_produced")
abline(coef(linreg)[1:2],col='red')

## rounded coefficients for better output
cf <- round(coef(linreg), 4)

check to avoid having plus followed by minus for negative coefficients
eq <- paste0("bushels_produced = ", cf[1],
ifelse(sign(cf[2])==1, " + ", " - "), abs(cf[2]), "fertilzer_peracre "
)

## printing of the equation
mtext(eq, 3, line=0)

Solution-b:

we have

bushels_produced= 20.094177 + 4.655377  *fertilzer_peracre

for x=fertilzer_peracre =3.5

bushels_produced= 20.094177 + 4.655377  *3.5

=36.388

the number of bushels given 3.5 hundred pounds of fertilizer.=36.388