Question

A plant distills liquid air to produce oxygen, nitrogen and argon. the percentage of impurity in the oxygen is thought to be linearly related to the amount of impurities in the air as measured by the pollution count. in parts per million (ppm). A sample of plant operating data is show below. (a) state the hypothesis of a linear relationship between oxygen purity and input air purity and fit a linear regression model to the data. (b) Test significance of the regression. (c) Plot and analyze the residuals from the sample data and comment on model adequacy. Evaluate all three assumptions regrading the residuals. Explain how to do all of this using Minitab.

Count(ppm) Purity (%)

1.10 93.3
1.45 92.0
1.36 92.4
1.59 91.7
1.09 94.0
0.75 94.6
1.20 93.6
0.99 93.1
0.83 93.2
1.22 92.9
1.47 92.2
1.81 91.3
2.03 90.1
1.76 91.6
1.68 91.9

Answer 1

Sol:

Regression Analysis: purity versus pollution count

Analysis of Variance

Source             DF Adj SS   Adj MS F-Value P-Value

Regression          1 16.491 16.4908    90.13    0.000

pollution count   1 16.491 16.4908    90.13    0.000

Error              13   2.379   0.1830

Total              14 18.869

Model Summary

       S    R-sq R-sq(adj) R-sq(pred)

0.427745 87.39%     86.42%      81.77%

Coefficients

Term               Coef SE Coef T-Value P-Value   VIF

Constant         96.455    0.428   225.24    0.000

pollution count -2.901    0.306    -9.49    0.000 1.00

Regression Equation

purity = 96.455 - 2.901 pollution count

Fits and Diagnostics for Unusual Observations

Obs purity     Fit   Resid Std Resid

9 93.200 94.047 -0.847      -2.22 R

R Large residual

a).

Regression Equation

purity = 96.455 - 2.901 pollution count

b).

calculated F=90.13,

P=0.00 which is < 0.05 level.

The model is significant.

c).

FITS1	RESI1
93.26351	0.03649
92.24817	-0.24817
92.50926	-0.10926
91.84204	-0.14204
93.32153	0.678471
94.27884	0.321155
92.97341	0.626586
93.58262	-0.48262
94.04677	-0.84677
92.91539	-0.01539
92.19016	0.009844
91.20383	0.09617
90.56562	-0.46562
91.37789	0.222112
91.58095	0.319045

d).

R square =87.39%

proportion of variability explained = 0.8739.