Question

1. The following sample observations were randomly selected: X: 4 3 6 128 Y: 4 6 5 7a7129 Determiner and r?. Determine the regression equation. Determine Sand Sy. Draw the scatter diagram. At the .05 significance level, is the correlation in the population greater than zero? (Say what the hypotheses are, what the test statistic is, what the critical value(s) is (are), what your decision is regarding the null hypothesis.)

Answer 1

CORRELATION DATA

( X)	( Y)	X^2	Y^2	X*Y
4	4	16	16	16
5	6	25	36	30
3	5	9	25	15
6	7	36	49	42
10	7	100	49	70

calculation procedure for correlation
sum of (x) = 28
sum of (y) = 29
sum of (x^2) = 186
sum of (y^2) = 175
sum of (x*y) = 173
to calculate value of r( x,y) = co-variance ( x,y ) / sd (x) * sd (y)
co-variance ( x,y ) = [ sum (x*y - N *(sum (x/N) * (sum (y/N) ]/n-1
= 173 - [ 5 * (28/5) * (29/5) ]/5- 1
= 2.12
and now to calculate r( x,y) = 2.12/ (SQRT(1/5*173-(1/5*28)^2) ) * ( SQRT(1/5*173-(1/5*29)^2)
=2.12 / (2.4166*1.1662)
=0.7522
value of correlation is =0.7522
co-efficient of determination = r^2 = 0.5659
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REGRESSION EQUATION

X	Y	(Xi - Mean)^2	(Yi - Mean)^2	(Xi-Mean)*(Yi-Mean)
4	4	2.56	3.24	2.88
5	6	0.36	0.04	-0.12
3	5	6.76	0.64	2.08
6	7	0.16	1.44	0.48
10	7	19.36	1.44	5.28

calculation procedure for regression
mean of X = sum ( X / n ) = 5.6
mean of Y = sum ( Y / n ) = 5.8
sum ( (Xi - Mean)^2 ) = 29.2
sum ( (Yi - Mean)^2 ) = 6.8
sum ( (Xi-Mean)*(Yi-Mean) ) = 10.6
b1 = sum ( (Xi-Mean)*(Yi-Mean) ) / sum ( (Xi - Mean)^2 )
= 10.6 / 29.2
= 0.363
bo = sum ( Y / n ) - b1 * sum ( X / n )
bo = 5.8 - 0.363*5.6 = 3.767
value of regression equation is, Y = bo + b1 X
Y'=3.767+0.363* X          
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SCATTER DIAGRAM

onion V Y y = 0.363 x + 3.767 R2 = 0.566 w in HHHHHHHHHHHHHHHHHHHHHHHHHHHHHH 24 6 8 10 12

HYPOTHESIS STEPS

Given that,
value of r =0.7522
number (n)=5
null, Ho: row(ρ) =0
alternate, H1: row(ρ)>0
level of significance, alpha = 0.05
from standard normal table,right tailed t alpha/2 =2.353
since our test is right-tailed
reject Ho, if to > 2.353
we use test statistic (t) = r / sqrt(1-r^2/(n-2))
to=0.7522/(sqrt( ( 1-0.7522^2 )/(5-2) )
to =1.977
|to | =1.977
critical value
the value of |t alpha| at los 0.05% is 2.353
we got |to| =1.977 & | t alpha | =2.353
make decision
hence value of |to | < | t alpha | and here we do not reject Ho
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null, Ho: row(ρ) <0
alternate, H1: row(ρ)>=0
test statistic: 1.977
critical value: 2.353
decision: do not reject Ho
correlation in the data does n't greater than zero