
What does the symbol mean?
n
∑ Xi
i =1
Please provide an example.
Suppose X is variable.

The following sample observations were randomly selected.
X: 4, 5, 3, 6, 10
Y: 4, 6, 5, 7, 7
Obsevation
Number
1
2
3
4
5
Total
X
Y
4
5
3
6
10
4
6
5
7
7
𝑋2
28
29
Number of observations 𝑛 = 5
5
∑ 𝑋𝑖 = 28
𝑖=1
5
∑ 𝑌𝑖 = 29
𝑖=1
5
∑ 𝑋𝑖 2 = 186
𝑖=1
5
∑ 𝑌𝑖 2 = 175
𝑖=1
5
∑ 𝑋𝑖 𝑌𝑖 = 173
𝑖=1
Coefficient of correlation
𝑌2
16
25
9
36
100
186
XY
16
36
25
49
49
175
16
30
15
42
70
173
𝑟=
∑ 𝑋𝑌 − (∑ 𝑋)(∑ 𝑌)
√(∑ 𝑋 2 − (∑ 𝑋)2 /𝑛)(∑ 𝑌 2 − (∑ 𝑌)2 /𝑁)
=
173 − (28)(29)/5
√(186 − 282 /5)(175 − 292 /5)
173 − 812/5
=
√(186 − 784/5)(175 − 841/5)
173 − 162.4
=
√(186 − 156.8)(175 − 168.2)
=
10.6
√(29.2)(6.8)
=
10.6
√198.56
=
10.6
= 0.7522
14.0911
The correlation coefficient is 0.7522. The value is positive. So the variables are
positively or directly correlated. An increase in one variable is accompanied an
increase in the other variable and vice versa.
The coefficient of determination is 𝑟 2 = 0.75222 = 0.5659
56.59% of the variation in Y is explained by the variation in X.
Determine the coefficient of correlation and the coefficient of determination. Interpret the results.

The following regression equation was computed from a sample of 20 observations:
Y’ = 15 - 5X
SSE was found to be 100 and SST total 400.
Determine the standard error of estimate.
Determine the coefficient of determination.
Determine the coefficient of correlation. (Caution: Watch the sign!)
This question is due on day 5.
𝑦 = 𝛽0 + 𝛽1 𝑥
𝑦̂ = 𝛽̂0 + 𝛽̂𝑥 = 15 − 5𝑥
𝛽̂0 = 15
𝛽̂1 = −5
Standard error of estimate
𝑆𝑆𝐸
100
100
𝑆𝑦𝑥 = √
=√
=√
= √5.5556 = 2.357
𝑛−2
20 − 2
18
Coefficient of determination =
𝑆𝑆𝑇−𝑆𝑆𝐸
𝑆𝑆𝑇
=
400−100
400
300
= 400 = 0.75
Coefficient of correlation is negative, since 𝛽̂1 is negative.
The magnitude is √𝑟 2 = √0.75 = 0.866
The coefficient of correlation is 𝑟 = −0.866