Page 427 - DMTH404_STATISTICS

Unit 31: Hypothesis Testing

Remarks: Notes

Alternatively, a null hypothesis can be tested by computing critical sample mean X for a
C
given standard error and the level of significance.
(i) Let H :  =  against H :   
0 0 a 0
s
If a = 0.05, then X =  ± 1.96
C
0
n
s s
If  – 1.96 < X <  + 1.96 , we accept H .
0 0
n n
(ii) Let H :  £  against H :  >  (Right tailed test)
0 0 a 0
s
If a = 0.05 then X =  + 1.645
C 0
n
If X > X , we reject H .
C
0
In the above example,
H :  ³ 200 against  < 200 (Left tailed test)
0
10
 X = 200 – 1.645 ´ = 197.67
C
50
It is given that X = 198. Since X > X , we accept H at 5% level of significance.
C
0
31.2.2 Test of Hypothesis Concerning Population Mean (s being
unknown)

When s is not known, we use its estimate computed from the given sample. Here, the nature of
the sampling distribution of X would depend upon sample size n. There are the following two
possibilities:
(i) If parent population is normal and n < 30 (popularly known as small sample case), use
å (X - X ) 2
i
t - test. The unbiased estimate of s in this case is given by s = .
n - 1
Also, like normal test, the hypothesis may be one or two tailed.

(ii) If n ³ 30 (large sample case), use standard normal test. The unbiased estimate of s in this
å (X - X ) 2
i
case can be taken as S = , since the difference between n and n - 1 is
n
negligible for large values of n. Note that the parent population may or may not be
normal in this case.

Example 14: The yield of alfalfa from six test plots is 2.75, 5.25, 4.50, 2.50, 4.25 and 3.25
tonnes per hectare. Test at 5% level of significance whether this supports the contention that true
average yield for this kind of alfalfa is 3.50 tonnes per hectare.

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