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# Instrument Calibration.

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Introduction

## 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 8 16 24 32 38 Concentration (x) 0 2 4 6 8 10

## 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

1. Fluorescence (y) is the dependent variable
2. 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

1. {ei} are mutually independent
2. E(ei) = 0
3. 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 ratio is 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 and make the subject of the formula.   When y = 17.5

## 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.

• ## 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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