Correlation and Regression Analysis

Surname 1
Student Name
Professor
Course
Date
Question 1
a)
x y xy x^2
420 2.8 1176 176400
610 3.6 2196 372100
625 3.75 2343.75 390625
500 3 1500 250000
400 2.5 1000 160000
450 2.7 1215 202500
550 3.5 1925 302500
650 3.9 2535 422500
480 2.95 1416 230400
565 3.3 1864.5 319225
∑ x =¿ ¿
5250
∑ y=¿ ¿
32
∑ xy =¿ ¿17171.251
7171.25
∑ x2=¿ ¿282
6250
x = 525
y= 3.2
sx = 88.19171
sy = 0.477842

Surname 2
Cov(x, y) = ∑ xy−n x y
n−1 = 17171.25−(10∗525∗3.2)
10−1 = 41.25
r = Cov(x , y)
sx sx
= 41.25
88.19171∗0.477842 = 0.97884
b)
r = 0.97884.
There exists a 97.88% correlation between x and y. Since 97.88% is very close to 1, then
x and y are strongly and positively correlated.
c)
Let the linear model take the form: y = a+ bx
Where:
a is the y-intercept
b is the slope/gradient and a correlation coefficient of price
b= n∑ xy−∑ x ∑ y
n ∑ ( x2 ) − (∑ x )2 = 10 (17171.25 )−(5250∗32)
10 ( 2826250 )−52502 =0.00530
a=∑ y−b ∑ x
n = 32−0.00530(5250)
10 = 0.4175
Thus,
y = 0.4175+0.00530x
d)

Surname 3
350 400 450 500 550 600 650 700
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
A scatterplot
x
y
Question 2
a)
Hypothesis statement
H0 : b =0
H1 : b ≠0
The observed p-value= 0.6019. The observed p-value (0.6019) > 0.05 thus we fail to
reject the null hypothesis. Therefore, no enough evidence to claim that the coefficient of
the price is different from zero at a 5 % significance level.
b)
R2= 0.01104;
The regression model explains 1.104% variation of the independent variable in a
dependent variable. The fitted regression model does not explain the variation of
independent variable to the dependent variable sufficiently.
c)