northern hemisphere southern hemisphere clouds26966092 craters15283156 ratio1.7641.930 are craters...

Northern Hemisphere

Southern Hemisphere

Clouds 2696 6092

Craters 1528 3156

Ratio 1.764 1.930

Are craters the cause? Are these two ratios equal?

This is NOT a matter of opinion! This is what error analysis is for.

Normal (Gaussian) Distribution

Error in a single measurement = e = x – X

σ = standard deviation of the distribution

= uncertainty in a single measurement

= variance

σm = standard error of the mean

= uncertainty in mean value after n measurements

Error in ith measurement = ei = xi - X

Error in the mean = E

Thus

Experimental results presented as: m = (1.10 +/- 0.04) × 102 kg

Sometimes you’ll see P = (4.30 +/- 0.04sys +/- 0.05stat) × 10-2 W

Systematic uncertainty – two ways to take into account:

• Given by device documentation

• Half of last digit on display

(voltmeter at left: +/- 0.005 V)

Can also have a systematic shift (such as a mis-zeroed measuring device) – trickier to handle

Statistical uncertainty – quantity changes from one measurement to next

Example: What is my bowling average?

160 169 212 191 178 190 153 144

The problem is that I don’t know the true mean, so how do I find the standard deviation?

Residual of the ith measurement

Standard deviation of the sample = s:

And using , we find that

For my bowling scores,

Mean = (160+169+212+191+178+190+153+144)/8 = 174.6

s = sqrt[(160-174.6)2 + (169-174.6)2 + … + (144-174.6)2]/8 = 21.1

σm = sqrt(1/7) * 21.1 = 8.0

Thus my bowling average is 175 +/- 8, or even better, (1.75 +/- 0.08) × 102

Propagation of errors: For a function f(x,y,z),

Some common examples:

f = x + y OR f = x – y (σf)2 = (σx)2 + (σy)2

f = xy OR f = x/y (σf/f)2 = (σx/x)2 + (σy/y)2

f = xn σf/f = n(σx/x)

f = ln x σf = σx/x

f = ex σf/f = σx

A random process is one in which an observed quantity will be different from moment to moment, such as:

• The number of photons emitted by a laser

• Decays from a radioactive source

Useful rule:

If a measurement of a random process is N, the standard error is √N.

Northern Hemisphere

Southern Hemisphere

Clouds 2696 6092

Craters 1528 3156

Ratio 1.764 1.930

Clouds (North) = 2696 +/- √2696 = 2696 +/- 52

Clouds (South) = 6092 +/- √6092 = 6092 +/- 78

Ratio = Clouds / Craters, so σR = σCl / Craters

Ratio (North) = (2696 +/- 52) / 1528 = 1.764 +/- 0.034

Ratio (South) = (6092 +/- 78) / 3156 = 1.930 +/- 0.025

Difference = South – North, so (σDiff)2 = (σS)2 + (σN)2

The difference is 0.17 +/- 0.04, which is not consistent with zero to within three standard deviations. Thus the two ratios are not equivalent, and cratering can not be the sole reason for the discrepancy in cloud formation.

Student’s T-test: a tool for determining the equivalency of two numbers, taking into account sample size and confidence level.

Procedure:

1. Find the variance in the difference, (σDiff)2 = (σ1)2 + (σ2)2

2. Calculate the t value: t = (x1 – x2)/σDiff

3. Find the number of degrees of freedom, n1 + n2 – 2

4. Choose a confidence level (p = 0.05 means 95% probability)

5. Look up on a t-table; if your t value exceeds the number in the table, the two means are not equivalent.

Example: Earlier we used 8 scores to find my bowling average = 175 +/- 8. Let’s say Joe also bowled 8 games, for an average of 150 +/- 6. Is it 95% probable that I am a better bowler?

σDiff = √(62 + 82) = 10

t value = (175 – 150) / 10 = 2.5

# of degrees of freedom = (8 + 8 – 2) = 14

For p = 0.05 and 14 degrees of freedom, the table reads 2.14 < 2.50. Thus the two averages are not equivalent, so I am clearly the superior bowler.

To fit a function to your data, test a range of possibilities for each variable in the function, and choose those which minimize

Σ[xi – f(x)]2.

This is called least-squares fitting.

0

0.5

1

1.5

2

2.5

3

0 20 40 60 80 100 120 140

Tim e (m inutes)

Pro

be

sig

nal

(m

V)

6/7/2006

6/8/2006

6/9/2006

You can also perform a weighted fit by applying a coefficient to each term which scales inversely with the size of the error bar (the larger the error bar, the smaller the coefficient, and the less effect the point has on the fit):

Σai[xi – f(x)]2.

Useful references:

• Practical Physics, 4th ed. by G. L. Squires

• http://davidmlane.com/hyperstat/

• Data analysis functions in Excel, Matlab

• Your tutors or Sheets, for other software applications

• Your mentor!

The MOST IMPORTANT thing to draw from this session:

Much of science is comparing experiment and theory, or quantity A to quantity B. Proper error analysis gives you a systematic way to determine if two numbers agree or disagree, and makes your paper much more convincing.