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Unit 31: Hypothesis Testing
Notes
X - X - k
2
1
or Z = under H .
cal 2 2 0
s s
1 2
+
n n
1 2
In a similar way, we can write the expressions for t under different situations.
cal
Example 20: A sample of 100 electric bulbs of 'Philips' gave a mean life of 1500 hours
with a standard deviation of 60 hours. Another sample of 100 electric bulbs of "HMT" gave a
mean life of 1615 hours with a standard deviation of 80 hours. Can we conclude that the mean
life of 'HMT' bulbs is greater then that of 'Philips' bulbs by 100 hours?
Let X = 1615, S = 80, n = 100, X = 1500, S = 60, n = 100.
1 1 1 2 2 2
We can write
H : £ + 100 against H : > + 100
0 1 2 a 1 2
-
-
1615 1500 100
Z = = 1.5
cal 2 2
80 60
+
100 100
Since Z < 1.645, we accept H at 5% and say that the difference in mean life of 'HMT' bulbs and
cal 0
that of 'Philips' bulbs is less than or equal to 100 hours.
31.2.4 Paired t - Test
This test is used in situations where there is a pairing of observations (X , X ), like marks
1i 2i
obtained by students of a class in two subjects, performance of the patients before and after the
administration of a drug, etc. We define d = X - X , the difference in the observations for the i
i 1i 2i
th item.
d -
2
å d i å ( i d ) 2 å d - nd 2
i
s =
Then, we compute d = and d =
n n - 1 n - 1
As before, we can test H : = against H : (two tailed test) or H : £ (or ³) against
0 1 2 a 1 2 0 1 2
H : > (or <) (one tailed test).
a 1 2
d d n
t
The test statistic t = = ~ -distribution with (n - 1) d.f.
s d / n s d
Example 21: Eleven students of B.Com. (Hons) were given a test in economic analysis.
They were imparted a month's special coaching and a second test was held at the end of it. The
result were as follows :
Student No . : 1 2 3 4 5 6 7 8 9 10 11
Marks 1 Test : 36 40 36 34 46 32 38 46 40 38 42
st
in
Marks 2nd Test : 40 44 40 40 46 40 34 48 38 44 36
in
Do the marks give an evidence that the students have benefited by extra coaching?
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