What are the criteria for using a z-zone test statistic versus a
t-test statistic in determining the decision rule in your hypothesis
testing?  Explain with one real world application for either test.

A search on "z-zone test" for statistics yields no hits. I assume here
you mean tests based on the standard z-score, or z-tests.

The z-test is used when one has a sample of data X_1, X_2, ... , X_n
drawn from a normal distribution with known variance but unknown mean
one wants to determine whether, for some value M_0, the mean of the
distribution is larger than M_0 .

Similarly, one can ask, is the mean of the distribution smaller than
M_0; or is the mean of the distribution equal to M_0.

An applet and definition of z-test is at:
http://www.stat.sc.edu/~ogden/javahtml/power/power.html .

The t-test is used to compare the means of two groups. A definition
and description of the t-test is given at
http://trochim.human.cornell.edu/kb/stat_t.htm .

According to the site at
http://www.physics.csbsju.edu/stats/t-test.html , "Student", whose
real name was W.S.Gossett [1876-1937] developed his "Student's t-test"
to solve problems stemming from his employment in a brewery (he
published under a pseudonym because his employer wanted to protect his
trade secrets, see:
http://intrepid.mcs.kent.edu/~blewis/stat/tTest.html for this and
other information about the test).

t-tests are often used to determine whether some way of doing things
is worth trying. For example, a drug might be given to a thousand
patients and a placebo to a thousand patients, and the outcomes
compared. The t-test can answer the question: are the outcomes
significantly better among those patients who took the drug than among
those who took the placebo? (E.g. see:
http://web.umr.edu/~rhall/virtualstat/t/texamp.html )

Gossett might have had a problem like this:

Suppose there are two ways to label beer bottles: a manual system and
a new-fangled electrical one. The electrical one is more expensive: is
it worth using? Suppose the daily productions of the manual system the
electrical ones have standard deviations of 100 and 150
respectively. Suppose we sample over 50 days the outputs from the two
system and find that their sample means are 2500 and 2300. Is it worth
upgrading the beer bottling plant to use electricity?

LINKS
-----
z-test: http://www.stat.sc.edu/~ogden/javahtml/power/power.html .

t-test: http://trochim.human.cornell.edu/kb/stat_t.htm .

http://www.physics.csbsju.edu/stats/t-test.html ,
http://intrepid.mcs.kent.edu/~blewis/stat/tTest.html ,
http://web.umr.edu/~rhall/virtualstat/t/texamp.html 

SEARCH STRATEGY
--------------
t-test
z-test

---

Clarification of Answer by rbnn-ga on 19 Sep 2002 11:36 PDT

I want to mention that there is another interpretation of the word
"t-test"
that makes a little more sense in the context of the question "when
does one use a t-test rather than a z-test". Usually the phrase
"t-test" is used to compare populations, but the t-distribution
itself is used even for testing the mean of a single population when
the variance of this population is not known.

The z-test may be used to test means from a population when the
variance is known. However, when the variance is unknown then we want
to use the sample variance, and this requires the
t-distribution. Generally you want to use the t-distribution for
smaller-size samples, since for large samples the t-distribution
approaches the normal distribution anyway.

Here is an example. Suppose we have a machine that fills up beer
bottles, but we are worried that it is not dispensing the right amount
of beer on average.

We take 5 random samples and find that the number of grams of beer in
the five samples is:

101.8, 101.9, 102.0, 102.1, 102.2 grams

We want to test the null hypothesis that the the machine does dispense
100 grams of beer on average.

H0: mu = 100
H1: mu ~= 100

Specify alpha=0.05. We want the t-value for the two-tailed rejection
region for 4 degrees of freedom (since there are 5 data points, there
are 4 degrees of freedom).

I used the t-distibution applet at:
http://www.econtools.com/jevons/java/Graphics2D/tDist.html to compute
this. I find that the rejection region is |t|>2.78 .

Now I compute the mean of the data:

Mean = 102.0  [by inspection]

and the sample variance

(.2^2+ .1^2 +0^2 + .1^2 + .2^2)/ (5-1) = .025

The t-statistic has the value: 

(Mean-100)/sqrt(sampleVariance/n)

2/sqrt(.025/5) = 28.28

Since this exceeds the critical value of 2.78, we reject the null
hypothesis.

rbnn-ga appears to have really put forth effort in answering this question.  

Thanks