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Unit 31: Hypothesis Testing
Remarks: Notes
1. Out of the two types of errors, the type I error is considered to be more serious.
Consequently, the probability of type I error is fixed at a low value (often 0.05 or lower).
Thus, when the computed value of a statistic falls in the critical region, implying thereby
that the probability of H being true is low or equivalently the probability of H being
0 0
false is high, we reject H . However, if the computed value of statistics lies in the acceptance
0
region, it would not be appropriate to say that the probability of H being true is very
0
high because the probability of accepting a false H (the value of b) may also be high. Thus,
0
accepting H only implies that the sample information does not provide any evidence of
0
H being false. Because of this nature of the tests of hypothesis, the conclusion "accept H "
0 0
is often replaced by "do not reject H " or "there is no evidence against H on the basis of
0 0
available sample information", etc.
2. The tests of hypothesis are also known as the Tests of Significance. We know that if the
sample result is highly unlikely, H is rejected because the sample result is significantly
0
different from the hypothesised value. Alternatively, it implies that the observed difference
between the computed and the hypothesised value is not attributable due to chance or
fluctuations of sampling.
31.2 Tests of Hypothesis Concerning Mean
These tests can be divided into two broad categories depending upon whether s, the population
standard deviation, is known or not.
31.2.1 Test of Hypothesis Concerning Population Mean (s being known)
This test is applicable when the random sample X , X , ...... X is drawn from a normal population.
1 2 n
We can write
H : = (specified) against H : (two tailed test)
0 0 a 0
-
X
The test statistic ~ N (0,1 . Let the value of this statistic calculated from sample be denoted
)
s / n
X
-
as z = . The decision rule would be :
cal
s / n
Reject H at 5%(say) level of significance if z > 1.96. Otherwise, there is no evidence against
0 cal
H at 5% level of significance.
0
Example 12: A manufacturer claims that the average mileage of scooters of his company
is 40 kms/litre. A random sample of 20 scooters of the company showed an average mileage of
42 kms/litre. Test the claim of the manufacturer on the assumption that the mileage of scooter
is normally distributed with a standard deviation of 2 kms/litre.
Solution.
Here, we have to test H : = 40 against H : 40.
0 a
X 42 40
-
-
z = = = 4.47.
cal
s / n 2/ 20
Since z > 1.96, is rejected at 5% level of significance.
cal
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