x
2
4
5
6
y
7 11 13 20
STRATEGY:
Put the data in L1 & L2.
Use prgm
PREDICT
in group CH10Vn
.
Enter the default alpha (.05).
(Note: Same data as 13, 15, 16, 17 and 18)
12)
The paired data below consists of test scores and hours of preparation for 5 randomly selected students.
The equation of the regression line is
y
ˆ
= 44.8447 + 3.52427
x
. Find the explained variation.
x
5
2
9
6
10
y
64 48 72 73 80
Note: Find the explained variation =
Σ
(
y
ˆ

y
)
2
when
y
ˆ
= 44.8447 + 3.52427
x
.
STRATEGY: Put the data in L1 & L2.
Use prgm
VARIATN
in group CH10Vn
.
press 1
(Note: Same data as 12, 15, 16, 17 and 18)
13)
The paired data below consists of test scores and hours of preparation for 5 randomly selected students.
The equation of the regression line is y^ = 44.8447 + 3.52427x. Find the unexplained variation.
x Hours of preparation
5
2
9
6
10
y Test score
64 48 72 73 80
STRATEGY: Put the data in L1 & L2.
Use prgm
VARIATN
in group CH10Vn
.
press 1
Use program PREDICT
Use program VARIATN
Answer:
511.724 the explained variation
E U = Explained Variation
U V = Unexplained Variation
T V = Total Variation
E U = Explained + Unexplained (used to check T V)
Answer:
87.476 the unexplained variation
E U = Explained Variation
U V = Unexplained Variation
T V = Total Variation
E U = Explained + Unexplained (used to check T V)
Use program VARIATN
Answer: .8873
14)
Use the computer display to answer the question.
A collection of paired data consists of the number of
years that students have studied Spanish and their scores on a Spanish language proficiency test. A computer
program was used to obtain the least squares linear regression line and the computer output is shown below. Along
with the paired sample data, the program was also given an x value of 2 (years of study) to be used for predicting
test score.
The regression equation is
Score = 31.55 + 10.90 Years.
Predictor
Coef
StDev T
P
Constant
31.55
6.360
4.96
0.000
Years
10.90
1.744
6.25
0.000
S = 5.651
RSq = 83.0%
RSq (Adj) = 82.7%
Predicted values
Fit
StDev Fit
95.0% CI
95.0% PI
53.35
3.168
(42.72, 63.98)
(31.61, 75.09)
Use the information in the display to find the value of the linear correlation coefficient
r
. Determine whether there is
significant linear correlation between years of study and test scores. Use a significance level of 0.05. There are 10
pairs of data.
ANSWER: (No calculator program to do thisdo by reasoning)
Use the Coef column. The constant is 31.55 and the
x
coefficient is 10.90.
x
y
90
.
10
3155
ˆ
+
=
.
Now find
r
and test it.
R
Sq = 83.0%,
R
=
83
= .911,
n
= 5,
α
= .05. Look up the critical value in A6 or use
prgmA6PEARSN.
The critical value is .632.
The test stat 
r 
=  .911  > .632 the critical so Reject Ho.
There is
significant linear correlation.
Conclusions: r = 0.91; There is significant linear correlation.
Note: Pval = 0 so we have another reason to reject
Ho.
((Note: Same data as 12, 13, 16 and 17)
15)
Find the total variation for the paired data.
The paired data below consists of test scores and hours of
preparation for 5 randomly selected students. The equation of the regression line is y^ = 44.8447 + 3.52427x. Find
the total variation.
x Hours of preparation
5
2
9
6
10
y Test score
64 48 72 73 80
(Note: Same data as 12 and 13)
STRATEGY: Put the data in L1 & L2.
Use prgm
VARIATN
in group CH10Vn
.
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 Spring '10
 evans
 Regression Analysis, significant linear correlation