Lecture 5

Random Errors in

Chemical Analysis -II

A measurement x1, with

m and s

m

I am making a second measurement x2

of the same analyte.

Random!

m

It has the same m We do not know it!

It has the same s

As an estimate of my unknown m

I will use average 2

21 xxx

sssss 22 22

2

2

121

22

2

2

21 ssss

av

…………

If I am making N measurements of the same value

N

xxxx N

...21

Nav

ss

Case 3: We know:

Standard deviation sReal value ?

Take N measurements; calculate average as

N

xxxx N

...21

Nx

sm

2 with 95% probability

Coinfidence interval (CI)

Case 4 We know nothing:

Mean - ? Standard deviation -?

Take N measurements; calculate average as

N

xxxx N

...21

Calculate standard deviation as

1

)(...)()( 22

2

2

1

N

xxxxxxs N

Confidence interval?

I do not know s !

Case 4 We know nothing:

Mean - ? Standard deviation -?

Take N measurements; calculate average as

N

xxxx N

...21

Calculate standard deviation as

1

)(...)()( 22

2

2

1

N

xxxxxxs N

N

stx 95m with 95% probability

Coinfidence interval (CI)

6 8 10 12 14 16

m=12.34

s=0.50

N=2

6 8 10 12 14 16N=3

6 8 10 12 14 16

6 8 10 12 14 16

N=4

N=6

11.8 12.2 12.6

m=12.34

s=0.50

N=6

11.8 12.2 12.6

11.8 12.2 12.6

N=15

N=60

I titrate an unknown solution using a class A

50 mL burette. My results show standard deviation

of 0.05 mL. How many mesurements I need

in order to get confidence interval of ±0.01 mL?

44.02

05.07.12

N=2

08.04

05.018.3

N=4

038.09

05.03.2

N=9

02.020

05.01.2

N=20

009.0120

05.098.1

N=120

Ubur = 0.05 mL

22

randomburtotal uuu

44.005.044.0 22 totalu

095.005.008.0 22 totalu

06.005.0038.0 22 totalu

054.005.002.0 22 totalu

05.005.0009.0 22 totalu

A very simple rule:

CIs overlap – the same

CIs do not overlap - different