Time after injection (hours) Temperature (0F)
24 102.8
32 104.5
48 106.5
56 107.0
Test the hypothesis that the correlation in significant at the 0.05 level.
| ( X) | ( Y) | X^2 | Y^2 | X*Y |
| 24 | 102.8 | 576 | 10567.8 | 2467.2 |
| 32 | 104.5 | 1024 | 10920.3 | 3344 |
| 48 | 106.5 | 2304 | 11342.3 | 5112 |
| 56 | 107 | 3136 | 11449 | 5992 |
calculation procedure for correlation
sum of (x) = ∑x = 160
sum of (y) = ∑y = 420.8
sum of (x^2)= ∑x^2 = 7040
sum of (y^2)= ∑y^2 = 44279.34
sum of (x*y)= ∑x*y = 16915.2
to caluclate value of r( x,y) = covariance ( x,y ) / sd (x) * sd
(y)
covariance ( x,y ) = [ ∑x*y - N *(∑x/N) * (∑y/N) ]/n-1
= 16915.2 - [ 4 * (160/4) * (420.8/4) ]/4- 1
= 20.8
and now to calculate r( x,y) = 20.8/ (SQRT(1/4*16915.2-(1/4*160)^2)
) * ( SQRT(1/4*16915.2-(1/4*420.8)^2)
=20.8 / (12.649*1.672)
=0.984
value of correlation is =0.984
coeffcient of determination = r^2 = 0.967
properties of correlation
1. If r = 1 Corrlation is called Perfect Positive
Corrlelation
2. If r = -1 Correlation is called Perfect Negative
Correlation
3. If r = 0 Correlation is called Zero Correlation
& with above we conclude that correlation ( r ) is = 0.9835>
0 ,perfect positive correlation
Given that,
value of r =0.984
number (n)=4
null, Ho: ρ =0
alternate, H1: ρ!=0
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =4.303
since our test is two-tailed
reject Ho, if to < -4.303 OR if to > 4.303
we use test statistic (t) = r / sqrt(1-r^2/(n-2))
to=0.984/(sqrt( ( 1-0.984^2 )/(4-2) )
to =7.81
|to | =7.81
critical value
the value of |t α| at los 0.05% is 4.303
we got |to| =7.81 & | t α | =4.303
make decision
hence value of | to | > | t α| and here we reject Ho
ANSWERS
---------------
null, Ho: ρ =0
alternate, H1: ρ!=0
test statistic: 7.81
critical value: -4.303 , 4.303
decision: reject Ho
we have enough evidence to support the claim that the correlation
in significant at the 0.05 level.
55 pts The following temperatures were recorded in a rabbit at various times after being inoculated...