Z Test
Z Test
Z Test
T- Test
Santos, Freddie
Corpuz, Grace
Molina, Jappy
Z- test= Grouped
Data (30 or more)
T- test= Ungrouped
Data (less than 30)
TEST OF MEANS
(One Sample) aims
to find out if a
population
characteristic, as
indicated by the
mean has changed.
T- Test
Given :
Population Mean. =60 %
Sample mean X = 57%
Sample standard deviation ,
s = 7%
Sample size, n = 15
Level of significance = 5%
2. At 5% level of significance, a
professor wants to know if her
introductory statistics class
has a good grasp of basic math.
Six students are chosen at
random from the class and
given a math proficiency test.
The professor wants the class
to be able to score above 70 on
the test. The six students get
scores of 62, 92, 75, 68, 83, and
Z-Test
At 1% level of significance, do
you think that the mining firm
will decide to buy from the
said manufacturer? Random
samples of 100 flashlight
batteries are tested. It is found
that the average life of the
sample batteries is 38 hours
with the standard deviation of
5
hours.
Given:
Population mean, =35
hours
Sample mean, X=38
hours
Sample size, n = 100
Level of significance = 1%
4- Step
Solution
1. Null Hypothesis
Ho: =35 (That the average life
of the manufacturers batteries
is
not
different
from
the
established population standard
of 35 hours.)
Alternative Hypothesis
Ha: >35 (that the average life
of the manufacturers batteries
is greater than the established
standard of 35 hours)
2.Tabular Value
Zt = 2.33
Based on 1%
level of
significance and
one tailed test
4. Conclusion
The computed Z score of 6 is
greater than the tabular value
of 2.33. Therefore, Reject Ho
(accept Ha)
The average life of the
manufacturers
batteries
is
greater than the established
average of 35 hours.
TEST OF DIFFERENCE
OF MEANS aims to
determine if the
same characteristic
between two
populations is
significantly
different.
Variety A
Variety B
Average
yield per
hectare, X
250
cavans
240 cavans
Standard
deviation, s
20
cavans
15 cavans
At 1% level of significance, is
there a significant difference in
the yield of the two palay
Solution
Given:
Variety
A
Sample
X1 =
mean
250
Sample N1= 40
size
hectare
s
Variety B
X2 = 240
N2= 30
hectares
4. Conclusion
4-STEP SOLUTION
1.State the null and alternative
hypothesis.
2.Get the tabular value.
3.Solve for the computed value.
4.Compare the computed value
(zc) from the tabular value (zt).
An experiment is conducted to
determine whether intensive tutoring
(covering a great deal of material in a
fixed amount of time) is more
effective than paced tutoring
(covering less material in the same
amount of time). Two randomly
chosen groups are tutored separately
and then administered proficiency
tests. Use a significance level of 0.05.
Let 1= population mean for the
intensive tutoring group
2= population mean for the paced
tutoring group.