Cloud Seeding Example (Dec. 2)

Case Study
A study claimed that the mean rainfall from "unseeded" clouds is 156.09 inches with SD 279.27.  26 clouds were randomly selected and were "seeded".  The amount of rainfall produced was:
[1]  129.6   31.4 2745.6  489.1  430.0  302.8  119.0    4.1   92.4   17.5
[11]  200.7  274.7    7.7 1656.0  978.0  198.6  703.4 1697.8  334.1  118.3
[21]  255.0  115.3  242.5   32.7   40.6

Does cloud-seeding work?

Discussion
Cloud-seeing "works" if seeded clouds produce more rain than non-seeded clouds.  You can see that some of the seeded clouds produced considerably less than the average of unseeded clouds, and others produced more.  So to determine if seeding is worthwhile, we need to see how the "typical" seeded cloud behaves.

Conceptually, we imagine that the population here consists of the rainfall from all unseeded clouds.  If seeding does nothing, then our random sample of 26 seeded clouds should look as if it came from this population.  This means that we change the question from "does seeding work?"  to "does the sample of seeded clouds look like a sample of unseeded clouds would look?"

A histogram would be useful here, but for the sake of brevity, let's look at the average and SD of this sample:
AVG: 448.676  SD = 663.2934

The average seems to be much larger than the "mean" of 156.09.  However, this difference could be due just to chance.  What's a 95% CI for the mean, assuming this sample came from the population of unseeded clouds?

We don't know the distribution, so at best we'll have an approximate confidence interval.  For the moment, let's assume the SD is known to be 279.27.  Then a 95% CI is
448.676 +- 1.96 * (279.27)/sqrt(26)
448.676 +- 107.3479
341.3281 to 556.0239

So we're approximately  95% confident that the true mean lies in this range.  But this range doesn't even include what we know to be the mean (156).  So either we got a bad confidence interval (5%) chance, or we were too approximate, or this population did NOT come from the
UNSEEDED population.
 

Significance Test Examples from Class

Example:  If a soda filling machine is working, it should dispense a mean of 12.00 oz.  The amount it dispenses actually follows a normal distribution.  A random sample of 4 cans gives these amounts:
12.26, 11.91, 12.15, 11.78
Does the machine work?
CHOOSE A SIGNIFICANCE LEVEL OF 5%.

H0: mean is 12 oz.
Ha: mean is not 12 oz.
Sigma is unknown, so we use a t-test:
(xbar - 12)/s/sqrt(4)
xbar is 12.025, and s is 0.2192
So our observed test statistic is:  .2281

What's the probability of getting something as extreme or more extreme than .2281?
2*P(T > .2281)

To do this, we need to know the degrees of freedom: n - 1 = 3.
Reading across this row, we don't see anything equal to .2281, but it's
LESS THAN .765, WHICH MEANS THAT P(T > .2281) MUST BE BIGGER THAN 25%.  WE DON'T CARE EXACTLY HOW MUCH BIGGER, BECAUSE AS LONG AS THE P-VALUE IS > 5% (OUR CHOSEN SIGNIFICANCE LEVEL) WE DON'T REJECT THE NULL HYPOTHESIS.