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Answers to Practice Test #2

    1. The Social Readjustment Rating Scale (SRRS) rates a person based on the number of changes that have occurred recently in his or her life. Some people claim that life change is related to decreased immunity and therefore an increased susceptibility to illness. The data below depicts the score on the SRRS in life change units (LCU) and the number of illness a particular individual has had for the past year.

Patient

X = LCU

X2

Y = Illness

Y2

X·Y

M. B.

192

 36864

6

 36 1152

T. G.

267

 71289

7

 49 1869

A. H.

200

 40000

4

 16 800

D. O.

235

 55225

4

 16 940

M. G.

220

 48400

8

 64 1760

J. E.

289

 83521

9

 81 2601

Totals

 1403 335299 38 262 9122

Please use this data to:

(a) Plot a scatterplot of this data

test2 scatterplot.gif (3927 bytes)

(b) Compute the correlation coefficient between LCU and number of illnesses

 

 

 

(c) Compute the degrees of freedom and find the critical value for this correlation coefficient

    df = 6 - 2 = 4             C.V. = .811

(d) Compute the coefficient of determination and explain what this coefficient of determination means

    r2 = .6022 = .362

    .362 is the proportion of the variability of illness that can be explained by the variability in life change units and visa versa.

(e) Tell your conclusions about the relationship between LCU and illness.

    Because the computed value of r is less than the critical value then we fail to reject the hypothesis that r is greater than 0.           The correlation coefficient is not significant.

2. The Social Readjustment Rating Scale (SRRS) rates a person based on the number of changes that have occurred recently in their life. Some people claim that life change is related to decreased immunity and therefore an increases susceptibility to illness. The data below depicts the score on the SRRS in life change units (LCU) and the number of illness a particular individual has had for the past year.

LCU (X) Illness (Y) Correlation (r)
Mean = 180 Mean = 4 r =.80
Standard Deviation = 25 Standard Deviaiton = 1.5  

First we must compute the standard error of the estimate for X and for Y

    t2 error.gif (1385 bytes)

a. How many illnesses do you predict will occur in the next year for a person who has an LCU of 160?

   

 

b. How many illnesses do you predict will occur in the next year for a person who has an LCU of 185?

   

 

c. What do you predict are the number of LCU’s for a person who has had 7 illnesses in the past year?

   

 

d. What do you predict are the number of LCU’s for a person who has had 3 illnesses in the past year?

   

(REMEMBER EACH PREDICTION MUST BE ± SOME ERROR)

Copyright © 2004 by Mark W. Vernoy