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					         Physics 380: Intermediate Laboratory
              Coin Counting Laboratory
Experiment: Obtain a set of 20 pennies. Toss the coins and record the number of heads
that you obtain. Repeat this for 30 tosses total, recording the results in the order you take
them.

Make a table of the coin-tossing data obtained from all students in the class (one column
for each student). Do all the appropriate statistical analysis for your coins. Examples are
shown below.

Here are my results:

12,12, 9, 15, 12, 8, 10, 8, 7, 9, 9, 10, 12, 10, 8, 9, 13, 7, 9, 9, 11, 10, 10, 11, 12, 9, 11, 12,
8, 12

For the first 5 tosses, the mean and standard deviation are 12 and 2.1.

We can compute the standard error in the mean: s<x>= 2.1/root5 = 1.1
and hence can get uncertainties at different levels of confidence:
       At the 50% level with a sample of 5, 0.7
       At the 95% level, 2.6

Thus at the 95% level we can say that the data supports the hypothesis of the average
equals ten since the average lies between 12 +/- 2.6. At 68% or 50% levels we cannot
claim agreement – this is a common result for such a small sample.

For all 30 tosses, the mean and standard deviation are 10.1 and 1.9.

Notice that the sample standard deviation is only changed by a small amount. The
standard error in the mean is 0.35 and this is an improvement over the sample of 5 tosses.

At the 50% level the confidence is 0.24
At the 95% level the confidence is 0.70.

And again at the 95% level we support the hypothesis that the mean is 10 since it lies
within 10.1 +/- 0.70, now we also agree with the hypothesis at the 50% level!

As a lab group, we will pool all our data to get an even smaller standard error in the
mean, but we will probably find that the sample standard deviation is similar to the above
results.

You should find the mean, standard deviation, standard error in the mean, and the level of
confidence at 50% and 95% levels for (a) the first 5 tosses (b) all 30 of your tosses, and
(c) the aggregate data from the class. Comment on the agreement between the means in
each sample and the expected population mean of 10.

What else can we do with the data? We can plot the histogram of results and see if the
distribution looks like a Gaussian curve. One approach is to see if the appropriate number
of points lie within s, 2s, etc.

Between one standard deviation from the mean on either side, we expect to find 68% of
the results, for our sample of 30 this means 68%(30) = 20.4 results. We look between
10.1 +/- 1.9 (from 8.2 to 12.0). I counted all values of 8, 9, 10, 11, and 12 and found there
are 22 values or 73% of the total. Likewise between two standard deviations I expect
95% or 28.5 values. I find that between 10.1 +/- 2 (that is from 6.3 to 13.9) there are 29
values or 97%. I counted tosses resulting in 6 to 14 heads. Both measurements agree well
with theory.

You should determine the number of tosses which result in values within s and 2s of the
mean. Do this for (a) your sample of 30 tosses and (b) for the aggregate data from the
class.

I can predict how many readings I expect for each bin of the histogram and compare with
what I actually got. A histogram is discrete while Gaussian curve is continuous. So to
compare I must digitize the Gaussian. I do this by integrating all values of the curve
between 10.5 and 11.5 and equating this to the expected value for the bin marked 11.

The Gaussian is usually written in terms of a variable z as Aexp(-z2/2). Tables generally
give the integral from zero to a positive value of z. For negative values, we rely on the
symmetry of the curve. Thus to find the value for bin 11, we first evaluate z at the limits
of the bin to get z1 = 0.211 and z2 = 0.737 and then consult the tables to get the areas
which we put into the following:

= NORMDIST(11.5,10.1,1.9,TRUE)-NORMDIST(10.5,10.1,1.9,TRUE)

To measure the goodness of fit we can use a chi-squared (X2) test. We define X2 in terms
of the predicted and measured values. In our case we define X2 as:

X2 = (number predicted – number measured)2 /number predicted

= (0.2)2/0.2 + (0.6)2/0.6 + (0.3)2/1.7 + …. + (0.8)2/0.2 + (0.1)2/0.1 = 10.2

The number of degrees of freedom we have is equal to the number of terms in our sum
minus 2 – this accounts for the fact we have used the numbers to get mean and standard
deviation already.

DOF = Nbins – 2 = 10
How good is the fit? We can calculate the reduced X2 as

X2reduced = X2/DOF = 1.02 for our numbers

The reduced X2 should be closed to 1.00, since it is we claim to have a good fit between
the Gaussian prediction and our data. Refer to any statistics text for a full treatment of X2.

For your sample of 30 tosses and the sample mean and standard deviations, predict
what you should observe from a Gaussian curve. Plot a histogram showing
predictions and observations. Compute a X2 for your histogram. Comment on the
goodness of fit.

				
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