APRIL 1995 EXAM: QUESTION 4

The following data come from a paper "Changes in growth hormone status related to body weight in cattle" with x = body weight (kg) and y = metabolic clearance rate.

   x:  110  110  110  230  230  230  360  360  360  360  505  505  505  505	
   y:  235  198  173  174  149  124  115  130  102   95  122  112   98   96	

(a) Fit a straight line to the data by least squares. Plot the data and the fitted line on a graph. Can metabolic clearance rate be predicted as a linear function of body weight? Present your analysis in an ANOVA table with F-tests for non-linearity and for the slope of the regression line. State your assumptions and your conclusions.

(b) What metabolic clearance rate would you predict for a body weight of 300 kg?

(c) Give a 95% confidence interval for s² = Var( y | x ).

MY SOLUTION

(a) The analysis below was done in Systat, but you should have no difficulty duplicating it on your calculator, even under examination conditions.

Begin by computing the following sums, sums of squares, and sum of products:

n = 14,  S x = 4480,  S y = 1923
S x2 = 1733500,  S xy = 544730,  S y2 = 288013
Lxx = 299900,  Lxy = -70630,  Lyy = 23875.2143
 

Fit a straight line and compute the simple linear regression ANOVA:

 
DEP VAR:CLEARANC      N:      14  MULTIPLE R: 0.835  SQUARED MULTIPLE R: 0.697
 
ADJUSTED SQUARED MULTIPLE R: 0.671    STANDARD ERROR OF ESTIMATE:      24.5646
 
VARIABLE      COEFFICIENT    STD ERROR     STD COEF TOLERANCE    T   P(2 TAIL)
CONSTANT         212.7209      15.7841       0.0000     .      13.4769   0.0000
WEIGHT            -0.2355       0.0449      -0.8347    1.0000  -5.2504   0.0002
 
                       ANALYSIS OF VARIANCE
SOURCE       SUM-OF-SQUARES   DF  MEAN-SQUARE     F-RATIO       P
REGRESSION      16634.2011     1   16634.2011     27.5666      0.0002
RESIDUAL         7241.0132    12     603.4178
 

There are 4 distinct level of Body Weight in the data set. Treat Body Weight as a factor at 4 levels and analyse as a one-way ANOVA:

LEVELS ENCOUNTERED DURING PROCESSING ARE:
WT_GRP
       1.0000       2.0000       3.0000       4.0000
 
DEP VAR:CLEARANC      N:      14  MULTIPLE R: 0.904  SQUARED MULTIPLE R: 0.817
 
                       ANALYSIS OF VARIANCE
SOURCE       SUM-OF-SQUARES   DF  MEAN-SQUARE     F-RATIO       P
WT_GRP          19514.2143    3    6504.7381     14.9157      0.0005
ERROR            4361.0000   10     436.1000
 

Combine the results from the two ANOVA tables above to get a Regression Analysis with a Pure Error term and a test for Lack of Fit:

(This step was done manually; some statistics packages can't do an analysis like this with Pure Error.)

                       ANALYSIS OF VARIANCE
SOURCE       SUM-OF-SQUARES   DF  MEAN-SQUARE     F-RATIO       P
REGRESSION      16634.2011     1   16634.2011     38.1431      0.0001
LACK OF FIT      2880.0132     2    1440.0066      3.3020      0.0792
ERROR            4361.0000    10     436.1000

Conclusions:

The lack of fit F-test is not significant at the 5% level; hence, over the range of body weights studied, linear regression is an adequate model for Metabolic Clearance. The regression is significant, and accounts for 90% of the variation.

(b) From the fitted line: E( y | x = 300 ) = 142.07

(c) s² = MSPE = 436.1 on 10 df.

However, the lack of fit test was not significant, so if you prefer you could use the residual mean square from the regression ANOVA to estimate the conditional variance and get 2 more df:

s² = MSR = 603.4178 on 12 df.

95% Confidence Interval for the conditional variance s²:

Using s² = MSPE gives (436.1/(20.48/10), 436.1/(3.25/10)) = (212.9, 1324)

Using s² = MSR gives (603.4178/(23.34/12), 603.4178/(4.40/12) = (310.2, 1646)

Assumptions:

• Independence (can't test)
• Normality (can't test, but at least note that there are no outliers in the data)
• Homoscedasticity (not enough data to test formally, but, looking at the plot, the scatter of points is about the same at each of the four levels of Body Weight, so the assumption is reasonable).