In: Statistics and Probability
The following data were obtained in a study of the relationship between diastolic blood pressure (Y) and age (X) for boys 5 to 13 years old.
X |
5 |
8 |
11 |
7 |
13 |
12 |
12 |
6 |
Y |
63 |
67 |
74 |
64 |
75 |
69 |
90 |
60 |
Fit a simple linear regression model to the data and plot the residuals against the fitted values.
Omit case 7 and refit the model. Plot the residuals versus the fitted values and compare to what you got in (a).
Using the fitted model from (b), obtain a 99% prediction interval for a new Y observation at X = 12. Does observation y7 fall outside this prediction interval? What is the significance of this?
ANS::
as per given data
a) Regression analysis performed in Minitab:
Regression Analysis: y versus x
The regression equation is:
y = 48.7 + 2.33*x
Predictor Coef SE
Coef T P
Constant 48.667
7.887
6.17 0.001
x
2.3333 0.8135
2.87 0.028
S = 6.68331 R-Sq = 57.8%
R-Sq(adj) = 50.8%
Plot of Ei vs Xi:
From the residual plot we can see that the residuals are not close to zero or the average value of the residuals will not be zero.
b) After omitting the 7th case we again fit the regression equation.
Regression Analysis: y versus x
The regression equation is:
y = 53.1 + 1.62*x
Predictor
Coef SE Coef
T P
Constant 53.068 3.214
16.51 0.000
x
1.6214 0.3448
4.70 0.005
S = 2.64538 R-Sq = 81.6%
R-Sq(adj) = 77.9%
Thus as we can see that the R-square value in this case is 81.6% which means that the fitted regression equation is more or less ok. It is obviously much better than that obtained in part a. Thus the 7th observation being an outlier was influencing the regression equation and was making it a bad fit.
c)
Predicted Values for New Observations:
New Obs Fit SE Fit
99%
CI 99%
PI
1 72.52
1.47 (66.58, 78.47)
(60.31, 84.74)
Values of Predictors for New Observations:
New Obs x
1 12.0
Thus the 99% P.I is : (60.31, 84.74)
We can see that the observation Y7=90 is falling outside the 99% P.I. Thus which means that Y7 is an outlier.
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