Question
Consider the following bivariate data set where $p, q \in \mathbb{Z}^+$.
The regression line of $y$ on $x$ has equation $y = 2.1875x + 0.6875$.
The regression line passes through the mean point $(\overline{x}, \overline{y})$.
(a) Given that $\overline{x} = 7$, verify that $\overline{y} = 16$.
(b) Given that $q – p = 3$, find the value of $p$ and the value of $q$
▶️Answer/Explanation
Detail Solution
part(a)
From the given data
$$\overline {x}= \frac{5+6+6+8+10}{5}=\frac{35}{5}=7$$
$$\overline{x}=7$$
Substitute $\overline{x} = 7$ into the regression line equation.
The regression line is given by $y = 2.1875x + 0.6875$
Substituting $\overline{x} = 7$
$$\overline{y} = 2.1875(7) + 0.6875$$
Calculate $\overline{y}$
$$\overline{y} = 2.1875 \times 7 + 0.6875 = 15.3125 + 0.6875 = 16$$
Thus, $$\overline{y} = 16$$ is verified.
part(b)
The mean
is the sum of the
-values divided by the number of data points (5):
$$\overline{y}=\frac{9+13+p+q+21}{5}$$
$\text{since} \overline{y}=16$ $$16=\frac{43+p+q}{5}$$
$$43+p+q=80$$
$$p+q=80-43$$
$$p+q=37 …………(1)$$
from the question ,given that $q-p=3 …………..(2)$
from the first and second equation
$ (1)+(2) $
$$p+q +q-p=37+3$$
$$2q=40$$
$$q=20$$
from first equation $$ p+q=37$$
$$p=37-q$$
$$p=37-20$$
$$p=17$$
$ q=20, p=17$
————Markscheme—————–
(a) EITHER
$\overline{y} = 2.1875 \times 7 + 0.6875$
OR
$\overline{y} = 15.3125 + 0.6875$
THEN
$\overline{y} = 16$
(b) attempts to use $16 = \frac{\sum y}{n}$ to form a linear equation in p and q
$16 = \frac{9+13+p+q+21}{5} (80 = p+q+43 \Rightarrow p+q=37)$
attempts to solve two linear equations simultaneously for p and q (one of which is q = p+3)
$16 = \frac{9+13+p+p+3+21}{5} (80 = 2p+46)$
p = 17 and q = 20