Parametric Test
(The Z – test)
Table Contents
1.1.
Introduction
1.2.
Introduction to Hypothesis Testing
1.3.
Levels of Significance (a)
1.4.
Critical Values
1.5.
Z – Test
Activity 1.1
Quiz 5.2
Learning Outcomes
1. Manifest knowledge and skills on the principles and concepts of z – test.
2. Solve problems involving z – test.
Overview
This chapter discusses on the introduction to parametric test and z – test. It also
discussed the introduction to hypothesis testing as well as the determination of level of
significance and critical values. Example problems in this chapter are discussed in step
by step manner. In addition, this chapter provides solution with the aid of computer, i.e.,
Microsoft excel.
At the end of this chapter, provides self – test in order to assess the reader and
students.
1.1. Introduction
A parametric test is a statistical test that specifies certain conditions about the distribution
of responses in the population from which the research sample was drawn. Most statisticians
perceive that parametric statistical tests are more powerful than nonparametric tests. The “power”
of a statistical test refers to the probability of rejecting the null hypothesis when it is actually false.
Many applications of statistics in educational research involve comparison of two population
means. This is done with the use of the t – test and Z – test.
1.2. Introduction to Hypothesis Testing
In any research, hypothesis is essential. Hypothesis refers to the tentative explanation for
a phenomenon, used as a basis for further investigation. Apparently a research question is
sometimes stated as a hypothesis, which provides the general scope of the investigation and work
to delineate the statement of the specific problem and the variables one wishes to investigate.
There are two types of hypothesis, i.e., null hypothesis (H0) and alternative hypothesis (H1).
The former is used for any hypothesis set up primarily to see whether it can be rejected or nullified.
The latter is an affirmation of the existence of difference or relationship between variables. The
alternative hypothesis has two kinds, i.e., directional hypothesis and non - directional hypothesis.
In directional hypothesis, the directions, i.e., higher, lower, more or less, of the outcome of the
study are specified so that the researcher will determine the relationship of the variables. It is often
referred as a research as a research hypothesis. It also uses one tailed test. The non – directional
does not make specific prediction about the direction of the outcome of the study will take. This
kind of hypothesis uses a two – tailed test. Example 1.1 shows some application of constructing a
hypothesis for a given problem. As shown in example 1.1, the directional alternative hypothesis
provides a better view on the directions of the outcome whether it is higher or lower. Basically,
the two kinds of alternative hypothesis are significant in all aspect of research study.
Example Problem 1.1
Title: The Mathematical Ability of the Male and Female Grade 8 Students of the
Pasig Catholic College.
H0:
There is no significant difference in the mathematical ability of the male
and female grade 8 students of the Pasig Catholic College.
H1 (Directional): The mathematical ability of the female grade 8 students of the Pasig
Catholic College is higher than the mathematical ability of the
male students.
H1 (Non – Directional): There is significant difference in the mathematical ability of the
male and female grade 8 students of the Pasig Catholic College.
Example Problem 1.2
Title: The Drinking Habits and Diabetes Cases among the Class D and Class E
Residents of Pasig City.
H0:
There is no significant relationship between the drinking habits and
diabetes cases among the class D and class E residents of Pasig City.
H1 (Directional): The class D and class E residents of Pasig City who drink more
frequently are more prone to suffer diabetes.
H1 (Non – Directional): There is significant relationship between the drinking habits and
diabetes cases among the class D and class E residents of Pasig City.
1.3. Levels of Significance (a)
We have already learned the basic concept of hypothesis. Recall that hypothesis testing is
important in an educational research. Researchers must learn how to accept or reject hypothesis
which can only be done if they have the knowledge on the level of significance. The level of
significance is designated by lower case Greek alphabet alpha (a) and it refers to the probability
to commit a type I error.
In an educational research, there are two kinds of errors, i.e., type I and type II. The former
is used when the null hypothesis which is true, is rejected. The latter is used when the null
hypothesis is false and is accepted. These errors are common and may be identified once the
hypothesis is tested by using statistical measurement and level of significance.
The most commonly used significance levels in research are 0.05 and 0.01. The former
implies that the probability of committing a type I error by chance is 5 out of 100 while the later
means that the likelihood of making a type I error by chance is 1 out of 100. The authors of this
book recommend the use of the 0.01 level of significance rather than 0.05. However, the decision
on the value of a depends on the researchers’ choice. Evidently, the researchers’ know the best
value of level of significance to be used in order to obtain a good result.
1.4. Critical Values
The critical value is associated to the level of significance. It is the critical region of the
test distribution. The critical region, i.e., rejection region, is just one part of the test distribution.
The other part of the test distribution is the acceptance region. The acceptance region of the test
distribution coincides with the level of significance.
Figure 1.1
The difference on the critical region and acceptance region of
the one – tailed test and two tailed test
Figure 1.1a and b show the difference on the critical region and acceptance region of one –
tailed test and two – tailed test. As shown in the figure 1.1 a and b, the one – tailed test exhibit
only one side of critical region while the two – tailed test shows two sides of critical region.
The critical values, e.g., Z, are determined by the use of t – distribution table, see appendix
A. Since the level of significance is related to critical value, 0.05 and 0.01 values of the level
significance are important for the purpose of reference in the t – distribution table.
1.5. Z – test
Many research problems in the undergraduate and graduate studies involve comparison of
two population means. The statistical tool for this will be t – test or Z – test. The Z – test is a
statistical tool for larger sample. Apparently the mathematical formulas for calculating the T – test
and the Z – test are similar, differing only in the determination of the standard error terms in their
denominator. Equation 1.1 and 1.2 show the mathematical formula for one – sample mean and two
– sample mean Z – test respectively.
The following conditions should be met before the Z – test is used:
1. Population standard deviation is known.
2. Sample size is greater than or equal to 30 even if the population standard deviation is not
known.
Equation 5.1
Equation 5.2
Z = (ẋ - │┘) √n/∂
Z = ẋ2 - ẋ2/√S1^2/n1 + S2^2/n2
Where: Z = Z – test
Where: Z = Z – test Z = Z - test
ẋ = sample mean
ẋ2 = sample mean of one group
│┘ = population mean
ẋ2 = sample mean of the other group
∂ = population standard deviation
S1 = sample variance of one group
N = sample size
S2 = sample variance of the other group
n1 = sample size of one group (n1 ≥ 30)
n2 = sample size of the other group n2 ≥30)
Example Problem 1.3
The achievement test results in Biology revealed that the 500 second year high school
students had a mean performance of 112.50 and standard deviation of 24. Would you agree in
this claim if a random sample of 50 second year students showed an average performance of
105.50 and standard deviation of 20.50? Use 5% level of significance to test its significance.
Given:
ẋ (sample mean) = 105.50
n (sample size) = 50
│┘(population mean) = 112.50
∂ (population standard deviation) = 24
Solution:
Step 1: Write the problem,
Problem: Is the Biology achievement mean performance of the second year high school
students 112.50?
Step 2: Determine the variable,
Variable: Achievement test performance of the second year high school students
Step 3: Determine the instrument,
Instrument: Biology achievement test
Step 4: Determine the null hypothesis,
Hypothesis: The mean performance of the second year high school students in the
Biology achievement test is 112.50
Step 5: Determine the alternative hypothesis,
Alternative Hypothesis: The mean performance of the second year high school
students in the Biology achievement test is not 112.50
Step 6: Determine the critical value,
Critical Value: The alternative hypothesis is non – directional; hence the two – tailed
test shall be used. At 5% level of significance, the critical value is 1.96, see appendix A.
Step 7: Determine the test statistic,
Test Statistic: One sample mean test, population standard deviation is known.
Step 8: Solve the Z – test value using equation 5.1,
Z = (ẋ - │┘) √n/∂
= (105.50 – 112.50) √50/24
= (-7) (7.07)/24
= -49.49/24
= -2.06
We can ignore the negative sign since the are of consideration is the students
distribution, instead consider the absolute value of Z thus,
Z = │-2.06│
Z = 2.06
Step 9: Write the data in tabular presentation,
Table 1
Comparison of the Sample Mean and the Population
Mean of the Biology Achievement Test Results
of the second Year High School Students
Number
of
Students
Mean
Standard
Deviatio
n
Compute
d
Z Value
2.06
Critica
l
Z
Value,
0.05
1.96
Decision
Interpretation
Reject
H0
Significant
Populatio
500
112.50
24
n
Sample
50
105.50
20.50
Since the computed value of 2.06 is greater than the critical value of 1.96, reject the null
hypothesis. Hence, at 5% level of significance, the mean performance of the second year high
school students in Biology achievement test is not 112.50.
Example Problem 1.4
A professor gave a final examination to his 400 students. He noted that 40 of the female
students had a mean performance of 122 and a variance of 60 while 35 of the male students had a
mean performance of 126 and a variance of 74. Assuming that the students were randomly selected
and the distribution is normal, is there a reason to believe that there is significant difference
between the examination mean performance of the male and female students. Use 0.01 level to
tests its significance.
Given:
ẋ1 (sample mean of one group) = 122 ẋ2 (sample mean of the other group) =126
S12 (mean variance of one group) = 60 S22 (mean variance of the other group) =74
n1 (sample size of one group) = 40 n2 (sample size of the other group) =35
Solution:
Step 1: Write the problem,
Problem: How significant is the difference between the examination performance of
the male and female students?
Step 2: Determine the variable,
Variable: Final examination performances of the male and female students.
Step 3: Determine the instrument,
Instrument: Final examination
Step 4: Determine the null hypothesis,
Hypothesis: There is no significant difference between the examination mean
performances of the male and female students.
Step 5: Determine the alternative hypothesis,
Alternative Hypothesis: There is significant difference between the examination
mean performances of the male and female students.
Step 6: Determine the critical value,
Critical Value: The alternative hypothesis is non – directional; hence the two – tailed
test shall be used. At 1% level of significance, the critical value is 2.576, see appendix A.
Step 7: Determine the test statistic,
Test Statistic: Two sample mean tests or separate – variance t model for large sample
sizes (n1 ≥ 30).
Step 8: Solve the Z – test value using equation 5.2,
Z = ẋ2 - ẋ2/√S1^2/n1 + S2^2/n2 = 122 – 126/√ = -4/√1.5 + 2.1 = -4/√2.61 = -4/1.9 = -2.105
Step 9: Write the data in tabular presentation,
Table 1
Comparison of the Sample Mean and the Population Mean of the Biology
Achievement Test Results of the second Year High School Students
Female
Male
Number
of
Students
Mean
40
35
122
126
Standard Computed Critical Decision Interpretation
Deviation Z Value
Z
Value,
0.01
Not
2.105
2.576 Accept
Significant
H0
60
74
Since the computed value of 2.105 is less than critical value of 2.576, accept the null
hypothesis. Therefore, there is no significant difference between the final examination
performances of the male and female students at 0.01 level.