Exp1 Error Analysis Case1


Case 1

[ Graphing Techniques | Error Analysis | eg Case 2]

You measure the values of resistance along a wire at intervals in x, with distances xi determined very accurately and values of resistance Ri each with an error so that the error flags would be (+) and (-) .04 ohms.

The data are given below along with useful sums. The values of R are from the equation

R = b + m x, for the values of b and m determined in the least squares fit.

 

#

xi

Ri

x2i

xiRi

R

 

m

ohms

   

ohms

1

10

1.32

100

13.2

1.31

2

20

2.30

400

46.0

2.27

3

30

3.20

900

96.0

3.22

4

40

4.19

1600

167.6

4.18

5

50

5.08

2500

254.0

5.13

6

60

6.09

3600

265.4

6.08

7

70

7.01

4900

490.7

7.04

8

80

7.99

6400

639.2

7.99

9

90

8.96

8100

806.4

8.95

10

100

9.92

10000

992.0

9.90

SUMS

550

56.06

38500

3870

 

 

From the sums in the table,

so

then b is given by

so

b = (38500 x 56.06 - 550 x 3870)/82500 = 0.361

 

where the common error on each measurement has canceled in this case. Similarly m is found from:

so

m=(10 x 3870 - 550 x 56.06)/82500 =0.0954

 

where again the common measurement error cancels. Now we can determine the uncertainty in these fitted values.For our case of common measurement errors:

giving the result that: m=.0954 ±.0004. The error in b can be calculated from

so that b=0.36 ±.03 Ohms.

 

The final result is: R = 0.36 + (0.0954 x) ohms with an error in R given by

A plot of the data with fitted line:

 

From the appearance of the plot, namely that the fitted line passes within all the error bars,we can infer that the measurement error is probably overestimated.

 

[ Graphing Techniques | Error Analysis | Case 1 | Case 2]