For the following assignment, use the Rank Correlation that was demonstrated in
ID: 3207871 • Letter: F
Question
For the following assignment, use the Rank Correlation that was demonstrated in Chapter 12 of the textbook (page 566). Utilizing Excel or SPSS: Use a rank correlation coefficient to test for a correlation between two variables. Use a significance level of =0.05. The new health care program in the United States makes provisions for capitation programs where health care insurers work with clinical facilities to perform risk analysis of patients to determine the cost of providing care. The following assignment might be used to assess how much a person smokes. When nicotine is absorbed by the body, cotinine is produced. A measurement of cotinine in the body is therefore a good indicator of how much a person smokes. The reported number of cigarettes smoked per day and the measured amounts of cotinine (in ng/ml) are provided. (The values are from randomly selected subjects in a National Health Examination Survey.) Is there a significant linear correlation? How would you measure the cotinine level in the body? Explain the result.
Refer to the "Rank Correlation Table." Rank Correlation Table x
x (cigarettes per day)
60
10
4
15
10
1
20
8
7
10
10
20
y(cotinine)
179
283
75.6
174
209
9.51
350
1.85
43.4
25.1
408
344
x (cigarettes per day)
60
10
4
15
10
1
20
8
7
10
10
20
y(cotinine)
179
283
75.6
174
209
9.51
350
1.85
43.4
25.1
408
344
Explanation / Answer
CORRELATION
r( X,Y) =Co V ( X,Y) / S.D (X) * S.D (y)
r( X,Y) = Sum(XY) / N- Mean of (X) * Mean of (Y) / Sqrt( X^2/n - ( Mean of X)^2 ) Sqrt( Y^2/n - ( Mean of Y)^2 )
Co v ( X, Y ) = 1 /12 (37111.51) - [ 1/12 *175 ] [ 1/12 *2102.46] = 537.553
S. D ( X ) = Sqrt( 1/12*5155-(1/12*175)^2) = 14.728
S .D (Y) = Sqrt( 1/12*601709.7926-(1/12*2102.46)^2) = 139.448
r(x,y) = 537.553 / 14.728*139.448 = 0.2617
If r = 0.2617> 0 ,Positive Correlation
HYPOTHESIS
Given that,
value of r =0.2617
number (n)=12
null, Ho: =0
alternate, H1: !=0
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.228
since our test is two-tailed
reject Ho, if to < -2.228 OR if to > 2.228
we use test statistic (t) = r / sqrt(1-r^2/(n-2))
to=0.2617/(sqrt( ( 1-0.2617^2 )/(12-2) )
to =0.86
|to | =0.86
critical value
the value of |t | at los 0.05% is 2.228
we got |to| =0.86 & | t | =2.228
make decision
hence value of |to | < | t | and here we do not reject Ho
ANSWERS
---------------
null, Ho: =0
alternate, H1: !=0
test statistic: 0.86
critical value: -2.228 , 2.228
decision: do not reject Ho
no significance relationship b/w them
( X) ( Y) X^2 Y^2 X*Y 60 179 3600 32041 10740 10 283 100 80089 2830 4 75.6 16 5715.36 302.4 15 174 225 30276 2610 10 209 100 43681 2090 1 9.51 1 90.4401 9.51 20 350 400 122500 7000 8 1.85 64 3.4225 14.8 7 43.4 49 1883.56 303.8 10 25.1 100 630.01 251 10 408 100 166464 4080 20 344 400 118336 6880 175 2102.46 5155 601709.79 37111.51 = TOTALS