Question

Find the regression​ equation, letting the diameter be the predictor​ (x) variable. Find the best predicted circumference of a marble marble with a diameter of 1.4 cm. How does the result compare to the actual circumference of 4.4 ​cm? Use a significance level of 0.05.

_   Diameter   Circumference
Baseball   7.4   23.2
Basketball   24.3   76.3
Golf   4.2   13.2
Soccer   21.8   68.5
Tennis   6.9   21.7
Ping-Pong   4.0   12.6
Volleyball   21.2   66.6

Answer 1

Sol:

in R studio create a table with the given data and asisgn to df1

with lm function fit a model of Diameter on circumference

Diameter--y

circumference-x

df1 =read.table(header = TRUE, text ="
Diameter Circumference
Baseball 7.4 23.2
Basketball 24.3 76.3
Golf 4.2 13.2
Soccer 21.8 68.5
Tennis 6.9 21.7
Ping-Pong 4.0 12.6
Volleyball 21.2 66.6
"
)
df1

plot(Circumference~Diameter,pch=16,data=df1)

linmod <- lm(Circumference~Diameter,data=df1)
coefficients(linmod)

Output:

(Intercept) Diameter
0.01281831 3.14042619

Circumference= 0.01281831 +3.14042619 *Diameter

For diamteter=1.4

predicted circumference

= 0.01281831 +3.14042619 *1.4

= 4.409415

predicted circumference=4.409415

Residual=observed circumference-predicted circumference

=4.4-4.409415

=-0.009415

predicted value is more than obseved value