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STA301 – Statistics and Probability
Lecture No 37:
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Hypothesis-Testing (continuation of basic concepts)
Hypothesis-Testing regarding  (based on Z-statistic)
In the last lecture, we commenced the discussion of the concept of Hypothesis-Testing.
We introduced the concepts of the Null and Alternative hypotheses as well as the concepts of Type-I and
Type-II error.
We now continue the discussion of the basic concepts of hypothesis-testing:
TEST-STATISTIC:
A statistic (i.e. a function of the sample data not containing any parameters), which provides a basis for
testing a null hypothesis, is called a test-statistic. Every test-statistic has a probability distribution (i.e. sampling
distribution) which gives the probability that our test-statistic will assume a value greater than or equal to a specified
value OR a value less than or equal to a specified value when the null hypothesis is true.
ACCEPTANCE AND REJECTION REGIONS:
All possible values which a test-statistic may assume can be divided into two mutually exclusive groups: one
group consisting of values which appear to be consistent with the null hypothesis (i.e. values which appear to support
the null hypothesis), and the other having values which lead to the rejection of the null hypothesis.
The first group is called the acceptance region and the second set of values is known as the rejection region for a test.
The rejection region is also called the critical region.
The value(s) that separates the critical region from the acceptance region, is called the critical value(s):
Critical
Value
Critical
Region
0
Acceptance
Region
Critical
Value
Z
Critical
Region
The critical value which can be in the same units as the parameter or in the standardized units, is to be decided by the
experimenter. The most frequently used values of , the significance level, are 0.05 and 0.01, i.e. 5 percent and 1
percent. By  = 5%, we mean that there are about 5 chances in 100 of incorrectly rejecting a true null hypothesis.
RELATIONSHIP BETWEEN THE LEVEL OF SIGNIFICANCE
AND THE CRITICAL REGION:
The level of significance acts as a basis for determining the CRITICAL REGION of the test.
For example, if we are testing H0:  = 45 against H1:   45, our test statistic is the standard normal variable Z, and the
level of significance is 5%, then the critical values are Z = +1.96 Corresponding to a level of significance of 5%, we
have:
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2.5%
-1.96
Critical
Region
2.5%
0
1.96
Acceptance
Region
Z
Critical
Region
ONE-TAILED AND TWO-TAILED TESTS:
A test, for which the entire rejection region lies in only one of the two tails – either in the right tail or in the
left tail – of the sampling distribution of the test-statistic, is called a one-tailed test or one-sided test. A one-tailed test is
used when the alternative hypothesis H1 is formulated in the following form:
H1 :  > 0
or
H1 :  < 0
For example, if we are interested in testing a hypothesis regarding the population mean, if n is large, and we are
conducting a one-tailed test, then our alternative hypothesis will be stated as
H1 :  > 0
or
H1 :  < 0
In this case, the rejection region consists of either all z-values which are greater than + z or less than – z (where  is
the level of significance):
If
H0 :  > 0
H1 :  < 0
Then (in case of large n):

REJECTION –z
0
Z
REGION
REJECT H0 if z < –z
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If
H0 :  < 0
H1 :  > 0
Then (in case of large n):

0
z REJECTION
Z
REGION
REJECT H0 if z > z/2
If, on the other hand, the rejection region is divided equally between the two tails of the sampling distribution of the
test-statistic, the test is referred to as a two-tailed test or two-sided test.
In this case, the alternative hypothesis H1 is set up as:
H1 :   0
meaning thereby
H1 :  < 0 or  > 0
If
H0 :  = 0
H1 :   0
Then (in case of large n):
/2
REJECTION
REGION
/2
–z/2
0
z/2
REJECTION
REGION
REJECT H0 if z < –z/2 or z > z/2
The location of critical region can be determined only after the alternative hypothesis H1 has been stated. It is
important to note that the one-tailed and the two-tailed tests differ only in location of the critical region, not in the size.
We illustrate the concept and methodology of hypothesis-testing with the help of an example:
EXAMPLE:
A steel company manufactures and assembles desks and other office equipment at several plants in a
particular country. The weekly production of the desks of Model A at Plant-I has a mean of 200 and a standard
deviation of 16.
Recently, due to market expansion, new production methods have been introduced and new employees hired.
The vice president of manufacturing would like to investigate whether there has been a change in the weekly
production of the desks of Model A. To put it another way, is the mean number of desks produced at Plant-I different
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from 200 at the 0.05 significance level? The mean number of desks produced last year (50 weeks, because the plant
was shut down 2 weeks for vacation) is 203.5.
On the basis of the above result, should the vice president conclude that the there has been a change in the
weekly production of the desks of Model A.
SOLUTION:
We use the statistical hypothesis-testing procedure to investigate whether the production rate has changed
from 200 per month.
Step-1:
Formulation of the Null and Alternative Hypotheses:
The null hypothesis is “The population mean is 200.”
The alternative hypothesis is ‘The mean is different from 200” or “The mean is not 200.”
These two hypotheses are written as follows:
H0 :  = 200
H1 : µ  200
Note:
This is a two-tailed test because the alternative hypothesis does not state a direction. In other words, it does
not state whether the mean production is greater than 200 or less than 200. The vice president only wants to find out
whether the production rate is different from 200.
Step-2:
Decision Regarding the Level of Significance (i.e. the Probability of Committing Type-I Error):
Here, the level of significance is 0.05.
This is , the probability of committing a Type-I error (i.e. the risk of rejecting a true null hypothesis).
Step-3:
Test Statistic (that statistic that will enable us to test our hypothesis):
The test statistic for a large sample mean is
z
X 

n
Transforming the production data to standard units (z values) permits the use of the area table of the standard normal
distribution.
Step-4:
Calculations:
In this problem, we have n = 50,
X = 203.5, and  = 16.
Hence, the computed value of z comes out to be:
z
X 

n

203.5  200
16
 1.55
50
Step-5:
Critical Region (that portion of the X-axis which compels us to reject the null hypothesis):Since this is a two-tailed test,
half of 0.05, or 0.025, is in each tail.
The area where H0 is not rejected, located between the two critical values, is therefore 0.95.
Applying the inverse use of the Area Table, we find that, corresponding to  = 0.05, the critical values are 1.96 and 1.96, as shown below:
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0.5000
 0.05

 0.025
2
2
-1.96
-1.96
Region of
rejection
0.5000
0.4750
 0.05

 0.025
2
2
0.4750
0
H0 is not rejected
+1.96
Scale to z
Region of
rejection
Critical
Value
Critical
Value
Decision Rule for the 0.05 Significance Level
The decision rule is, therefore:
Reject the null hypothesis and accept the alternative hypothesis if the computed value of z is not between –
1.96 and +1.96.
Do not reject the null hypothesis if z falls between –1.96 and + 1.96.
Step-6:
Conclusion:
The computed value of z i.e. 1.55 lies between -1.96 and + 1.96, as shown below:
Computed
value of z
- 1.96
Reject H0
0
Do not reject H0
1.55
1.96
z scale
Reject H0
Because 1.55 lies between -1.96 and + 1.96, therefore, it does not fall in the rejection region, and hence H0 is not
rejected.
In other words, we conclude that the population mean is not different from 200. So, we would report to
the vice president of manufacturing that the sample evidence does not show that the production rate at Plant-I has
changed from 200 per week.
The difference of 3.5 units between the historical weekly production rate and the production rate of
last year can reasonably be attributed to chance.
The above example pertained to a two-tailed test.
Let us now consider a few examples of one-tailed tests:
EXAMPLE:
A random sample of 100 workers with children in day care show a mean day-care cost of Rs.2650 and
a standard deviation of Rs.500. Verify the department’s claim that the mean exceeds Rs.2500 at the 0.05 level with this
information.
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SOLUTION:
In this problem, we regard the department’s claim, that the mean exceeds Rs.2500, as H1,
and regard the negation of this claim as H0.
Thus, we have
i)
H0 :  < 2500
H1 :  > 2500 (exceeds 2500)
(Important Note: We should always regard that hypothesis as the null hypothesis which contains the equal sign.)
ii)
We are given the significance level at  = 0.05.
iii)
The test-statistic, under H0 is
Z
X  0
S
n
,
which is approximately normal as n = 100 is large enough to make use of the central limit theorem.
iv)
The rejection region is
Z > Z0.05 = 1.645
0.05
0
v)
Z
Z0.05 REJECTION
=1.645 REGION
Computing the value of Z from sample information, we find
z
2650  2500 150

3
50
500 100
vi) Conclusion:
Since the calculated value z = 3 is greater than 1.645, hence it falls in the rejection region, and, therefore, we
reject H0, and may conclude that the department’s claim is supported by the sample evidence.
An Interesting and Important Point:
For  = 0.01, Z = 2.33.
As our computed value of Z i.e. 3 is even greater than 2.33, the computed value of X is highly significant.
(With only 1% chance of being wrong, the department’s claim was correct).
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