Practical Statistics
Z-Tests
There are six statistics that will
answer 90% of all questions!
1.
2.
3.
4.
5.
6.
Descriptive
Chi-square
Z-tests
Comparison of Means
Correlation
Regression
Z-test are for proportions.
This test is so easy….
That it is not even given in some stat
programs like SPSS…..
Z-test are for proportions.
What is the probability that out of 25
customers, 22 would like the service
when the usual percent is that 70%
are satisfied?
Z-test are for proportions.
They come in two types:
1. A sample proportion against a hypothesis.
Z-test are for proportions.
They come in two types:
1. A sample proportion against a hypothesis.
2. Two samples compared to each other.
Z-test are for proportions.
The standard error for proportions is:
SE 
pq
n
Where p = freq/total and q = 1 - p
Z-test are for proportions.
Hence:
pt  p
Z
pq
n
Where p is the hypothesized value, and pt is
the proportion found in a sample of size n.
Z-test are for proportions.
Suppose that XYZ Company believed that 20%
of their customers bought 80% of their product.
A sample of 200 customers found that 25% bought
80% of the product. Was the company correct in
their estimate?
Z-test are for proportions.
The test statistic looks like this:
Z
.25 .20
0.05

 177
.
(.20 X .80) 0.0283
200
Z-test are for proportions.
Z
.25 .20
0.05

 177
.
(.20 X .80) 0.0283
200
Since the test was “two-tailed,” the critical
value of Z would be 1.96. Therefore, we would
conclude that there is not enough evidence to
over-ride the assumption that 20% of the
customers bought 80% of the product.
Z-test are for proportions.
Z
.25 .20
0.05

 177
.
(.20 X .80) 0.0283
200
http://faculty.vassar.edu/lowry/tabs.html#z
P = 0.077
Z-test are for proportions.
They come in two types:
1. A sample proportion against a hypothesis.
2. Two samples compared to each other.
Z-test are for proportions.
The test for this case looks like this:
( pt 1  pt 2 )  ( p1  p2 )
Z
p1q1
p2 q2
(
) (
)
n1
n2
Z-test are for proportions.
Usually, the test assumes that the two groups
Are equal, or:
( pt 1  pt 2 )  0
Z
1 1
pq (  )
n1 n2
There is a problem here.
What is the value of:
?
( pt 1  pt 2 )  0
Z
1 1
pq (  )
n1 n2
p is the value of the population
proportion, but we usually don’t know that value,
so p is estimated by the weighted average
of the two groups….
p1n1  p2 n2
n1  n2
Suppose that a new product was test marketed in
the United States and in Japan. The company
hypothesizes that both countries response to the
product will be the same.
80% of a sample of 500 said they would buy the
product again in the U.S., while 75% of a sample
of 200 in Japan said they would buy the product
again.
Test the hypothesize…..
H0: p1  p2
The test would be:
(.80 .75)
Z
p?
But what is p?
Since p = 0.80 in the U.S., and 0.75 in Japan,
the weighted average is used for p.
So:
P = ((.8 x 500)+(.75x200))/700 = 0.786
The test would be:
Z
(.80 .75)
1
1
(.785)(.215)(

)
500 200
Z = .05/.0343 = 1.45 The critical value is 1.96;
The probability of z = 1.45 is P = 0.147.
The U.S. and Japanese customers are the same.
Questions:
1. Jason works in a building with four floors served by an elevator.
He has to use the elevator several times a day to carry heavy material
from one floor to the next. He states, only partly in jest, that the
elevator hates him because it is never on the floor he is on, and he
always has to wait for it to come. To test his hypothesis, he kept
track of where the elevator was for one month when he used it.
He had 60 recorded elevator uses and 9 times the elevator was at
his floor when he needed it. Test Jason’s hypothesis.
2. Suppose Jason always needs to start at the second floor so, along
with the info above, he also kept track of which floor the elevator was
actually on when he needed it. He found: first floor: 20; second floor:
9; third floor, 12; and fourth floor: 19.
Would the answer to #1 be changed if Jason worked on a different floor?
Questions:
1. Women buy 20% more BicTac than men. A new company produces
a new version of BicTac supposedly targeted at men. A MR survey of
200 users found that 100 customers of the new product were women.
Has the new product made any inroad into the male market?
2. An administrator believes that the number of men and women
in an MBA program is the same, in a sample of 500, how many more
women than men must be found to reject the administrator’s
hypothesis?
Service Encounter and Personality
Recall that 60% of shoppers normally are women,
but this sample found that 55% were women. Is
the sample bad?
Z = (.55 - .60)/SQ((.6 X .4)/271) = -1.68
p = 0.093