- Level: University Degree
- Subject: Mathematical and Computer Sciences
- Word count: 1265
Instrument Calibration.
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Introduction
Instrument Calibration
Introduction
A new device that measures the concentration of fluorescence requires calibration. The device is calibrated using the calibration line, method of least square estimation can be used to estimate the parameters α and β and the validity of the estimates using measurement readings from observed data.
The following are the measurements of fluorescence y, of a substance, A, in known concentrations x in μg/m3, using the new device.
Fluorescence (y) | 1.0 | 8.0 | 16.0 | 24.0 | 32.0 | 38.0 |
Concentration (x) | 0.0 | 2.0 | 4.0 | 6.0 | 8.0 | 10.0 |
Description of Data
The fluorescence (y) of a substance, A and its concentration (x) has a linear relationship, where concentration (x) increases as fluorescence (y) of the substance increases.
A linear model will be fitted to the observed data. The model is as follows:
Straight-line regression
- Fluorescence (y) is the dependent variable
- Concentration (x) is called the regressor variable.
1, 2, … , n
here =
e = Error
ei is the error in the ith observation yi.
Assumption on the errors
- {ei} are mutually independent
- E(ei) = 0
- Var(ei) = σ2
We assume ei’s are normal i.e. ei ~ N(0, σ2) 1, 2, … , n
Regression by Formula
Middle
16.05
23.62
31.19
38.76
Total
0.095
-0.476
-0.048
0.381
0.810
-0.762
0.000
For the model to be a good fit to the data we require the sum of the errors equal zero.
∴
∴
Variance-covariance matrix
∴
The above variance-covariance matrix gives , and
Design matrix
Below is the full fitted model for the data.
Analysis of Variance
Source | Sum of squares | df | Mean (SS) | F ratio |
Due to regression | ||||
Error SS (residual SS) | ||||
Total |
The ratiois distributed as F with (1, n –2 ) df.
Here we reject the null hypothesis that β = 0 if the> Fα, 1, n – 2
Rejecting the null hypothesis implies that the variable x influences the variable y. That is the concentration increases as the fluorescence of substance A increases.
Analysis of Variance
Source DF SS MS F P
Regression 1 1003.21 1003.21 2478.53 0.000
Error 4 1.62 0.40
Total 5 1004.83
From the ANOVA table we see that the P-value is 0.000 < 0.01, which is significant at the 1% significance level. We can conclude there is overwhelming evidence to reject the null hypothesis in favour of the alternative hypothesis, i.e. the concentration depends on fluorescence of the substance.
Goodness of fit in regression
Having found the best straight line, the next question is how well it describes the data.
Conclusion
When y = 17.5
hence
Having now calculated the value of at a given , we can calculate the associated error.
Given some the predictor is just , the error is given by
Where
hence
= (17.35, 17.65) with 90% confidence.
Conclusion
The regression model fitted for the data is of the form:
where is the dependent variable
is the intercept
is the slope of the regression coefficient
is the independent variable
is the error term.
Where
The regression equation is
The equation will specify the average magnitude of the expected change in Y given a change in X.
Limitations of Regression
An assumption is made with the regression line that it is where it should be.
You cannot assume that the regression line is valid outside the range of the data.
- You can interpolate, but you cannot extrapolate.
In the model the unknown parameter α and β where calculated. From the 90% confidence interval we know we have a good estimate for the gradient β of the model however more data is required before anything very definite can be said about the intercept of the model the parameter α.
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