HYPOTHESIS TESTING
OBJECTIVES:
At the end of this module, you will be able to
1. Define hypothesis testing in inferential statistics
2. Discuss the steps in the process of hypothesis testing
2.1 stating the null and alternative hypothesis
2.2 specifying the significance level
2.3 computing the value of relevant test statistic
2.4 determining the critical value of the test statistic
2.5 making decision to reject or not reject the null hypothesis
2.6 interpreting the results of the test
3. apply the method of hypothesis testing using
3.1 Z-test
3.2 t – test
INTRODUCTION
In the field of Research, we are interested in making conclusions about the population of
interest. Most often, it is not possible to study the entire population and, therefore, we study instead
a sample which is taken from this population. For instance, we might be interested in finding out the
performance of elementary teachers in the region. Since it is impossible to take all the teachers, we
can study a representative sample and on the basis of our findings we can make decisions about the
population.
TEST OF HYPOTHESIS
A statistical hypothesis is a guess made concerning the population. The hypothesis may or
may not, in fact, be true. There are two types of statistical hypothesis, the null hypothesis Ho and the
alternative Ha. A null hypothesis is a statement that a parameter, say the population mean is equal
to a certain value, for a one sample case. For two samples, the null hypothesis is a statement of no
difference between two population parameters. The alternative hypothesis is opposite to the null
hypothesis. The null hypothesis is always paired with a corresponding alternative hypothesis. The
examples are given below:
Ho
Ha1
Ha2
Ha3
:
:
:
:
µ = 50
µ ≠ 50
µ > 50
µ > 50
(Non-directional or two -tailed test)
(Directional or one- tailed test)
(Directional or one- tailed test)
Ho
Ha1
Ha2
Ha3
:
:
:
:
µ1 = µ2
µ1 ≠µ2
µ1 >µ2
µ1 <µ2
(Non-directional or two -tailed test)
(Directional or one- tailed test)
(Directional or one- tailed test)
ACCEPTING OR REJECTING THE NULL HYPOTHESIS
If the computed value of the test statistic is greater than a positive critical value (value taken
from the table of normal curve) or the computed value is less than a negative critical value (value
taken from the table of normal curve), then the null hypothesis is rejected, in favor of the alternative.
LEVEL OF SIGNIFICANCE AND TYPE OF ERRORS.
A Type I error is made if a null hypothesis is rejected when it should NOT be rejected, that is,
the null hypothesis is TRUE; while a Type II error is committed when the null hypothesis is not
rejected when it SHOULD BE rejected, that is the null hypothesis is FALSE. The level of significance is
the probability of making a Type I error and is symbolized by α, Alpha, with a common value from 1%
to 5%.
ILLUSTRATIVE EXAMPLES.
Two problems are given below to illustrate the concepts discussed above. In the first
example, a Z-test is used since the sample size is 30 or more (large sample). In the second example, a
t-test is used because the sample size is small, that is, less than 30.
Example 1. (Large sample, n ≥ 30)
The mean lifetime of a sample of 100 fluorescent bulbs produced by a company is
computed to be 1570 hours with a standard deviation of 120 hours. If it is claimed that the
mean lifetime of all bulbs produced by the company is 1600hours, test the hypothesis that µ
= 1600 hours against the alternative hypothesis that µ ≠ 1600 using a significance level of
0.05.
Ho
Ha
α
:
:
:
µ = 1600 hours
µ ≠1600 hours (Non-directional or two-tailed test
0.05
Z =
Mean - µ
------------ ( Test statistic for comparing the sample mean
s / √n
with the population mean)
Z =
1570 - 1600
----------------- =
120 / √ 100
- 2.50 (computed Z-value)
From the Table of Normal Curve, Z = - 1.96.
Since the computed value of Z (= -2.50) is less than the critical Z-value (= - 1.96), the null
hypothesis is rejected in favor of the alternative hypothesis at 5% level of significance.
Conclusion.
The mean lifetime of the population of fluorescent bulbs produced by the company is not
equal to 1600 hours.
The t – test of significance.
A large sample consists of 30 or more members. As the sample size becomes smaller, the
distribution becomes peaked and the tails of the curve becoming somewhat higher than those of the
normal curve. For a sample size of say n = 20, Z may turn out to be significant at 5% level when in fact
it is significant at 8%. Thus using a Z-test for this sample of 20 will result into an erroneous
conclusion.
Degrees of freedom. In a broadest sense, degrees of freedom pertain to the number of
quantities that are allowed to vary around a parameter. For example, if the mean if 10 scores in a
distribution is known, only 9 of the scores are free to vary in value but not the tenth score. Here,
there are 9 degrees of freedom. The degree of freedom, df, is found by subtracting from the sample
size n, the number of fixed parameters.
Example 2. (Small sample n < 30)
A test of breaking strength of six ropes manufactured by a company showed a mean breaking
strength of 7750 pounds(lb) and a standard deviation of 145 lb whereas the manufacturer claimed a
mean breaking strength of 8000 lb. Can we support the manufacturer’s claim at 0.01 level of
significance.
Ho
:
µ = 8000 lb
Ha
:
µ < 8000 lb
α
:
0.01
t =
Mean - µ
-----------s / √n-1
t =
7750 - 8000
----------------- =
145 / √ 6 - 1
(Directional or one-tailed test)
( Test statistic for comparing the sample mean
with the population mean)
- 3.86 (computed t-value)
From the t – table with one-tailed test at 0.01 level of significance and df = 5
t = - 3.36 (negative critical t-value)
Since the computed t-value (-3.86) is less than the critical t-value (-3.36), we reject Ho in
favor of the Ha at 1% level of significance and 5 degrees of freedom. Thus, it may be concluded that
the manufacturer’s claim seems to be unjustified.
Exercise 1 ( Module 1)
1. On an examination given to students at a large number of different schools, the mean
grade was 74.5 and the standard deviation was 8.0. At one school where 200 students
took the examination, the mean grade was 75.9. Can we say that the mean grade in this
particular school is significantly higher than 74.5? Use a one-tailed test at 5% level of
significance.
2. In a Biology experiment, a certain type of seed has always grown to a mean height of 8.5
inches. A sample of 26 seeds grown under new conditions has a mean height of 8.8
inches and a standard deviation of 1 inch. At 5% level of significance, test the hypothesis
that the new conditions do not grow better plants.
3. One hundred sixth-graders were tested and found to have a mean IQ of 105 with a
standard deviation of 16. Does this mean that these pupils have significantly higher IQ
than 100? Use 1% level of significance.