Page 426 - DMTH404_STATISTICS

Statistics

Notes Remarks:

1. If the manufacturer claims that the average mileage is more than 40 kms/litre rather than
equal to 40 kms/litre, we have to use a one tailed test. Now we shall test H :  = 40 against
0
X 
-
H :  > 40 and z would be defined as z = . Since this value is also equal to 4.47
a cal cal
s / n
and lies in the critical region, we reject at 5% level of significance. This implies that the
claim of the manufacturer may be taken as correct.
2. In one tailed tests the alternative hypothesis is expressed as a strict inequality and the null
hypothesis as a weak inequality or simply equality.
3. The decision rule can also be specified in terms of prob or p-value of the observed sample
result. The p-value is the smallest level of significance at which the null hypothesis can be
rejected. We define p-value

æ X  ö
-
= 2P z ³ ÷ , for a two tailed test,
ç
è s / n ø
æ X  ö
-
= P z ³ , when H : m > m and
ç è s / n ø ÷ a 0
-
æ X  ö
= P z £ , when H : m < m
ç ÷ a 0
è s / n ø
The decision rule is : If p-value < , reject H .
0
In the above example p -value is approximately equal to zero when H is either   40 or
a
 > 40, therefore H is rejected. However, if H is taken as  < 40, the p -value is almost
0 a
equal to unity and consequently H would be accepted.
0
4. As per the central limit theorem, even if the parent population is not normal, the sampling
distribution of z will be approximately normal when n > 30.

Example 13: A filling machine at a soft drink factory is designed to fill bottles of 200 ml
with a standard deviation of 10 ml. A sample of 50 bottles was selected at random from the filled
bottles and the volume of soft drink was computed to be 198 ml per bottle. Test the hypothesis
that the mean volume of soft drink per bottle is not less than 200 ml.
Solution.
Here n > 30, therefore, the sampling distribution of mean volume of soft drink per bottle will be
normal.
We have to test H :  ³ 200 against H :  < 200.
0 a
It is given that X = 198 and s = 10.

-
X  198 200
-
Thus, the test static is z = 0 = = - 1.41
cal
s / n 10/ 50
Since this value is greater than - 1.645, z lies in the acceptance region. Hence, there is no
cal
evidence against H at 5% level of significance.
0

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