Page 434 - DMTH404_STATISTICS

Statistics

Notes Solution.

We have to test H :  =  against H :    .
0 1 2 a 1 2
Note that H implies that students have not benefited by the extra coaching.
0
Let X and X denote the marks in 1st and 2nd tests respectively.
1 2
Calculation of d and s
d
Student No . : 1 2 3 4 5 6 7 8 9 10 11

d i : - 4 - 4 - 4 - 6 0 - 8 4 - 2 2 - 6 6
d 2 : 40 44 40 40 46 40 34 48 38 44 36
i
From the above table, we can write å d = - 22 and å d = 244
2
i
i
- 22 244 11 4
-
´
Thus, d = = - 2 and s = = 4.47
d
11 10
- 2 11
´
Further, t = = 1.48 . The value of t at 5% level of significance and 10 d.f. is 2.228.
cal
4.47
Therefore, the sample information provides no evidence that students have benefited by extra
coaching.
Example 22: A random sample of heights of 20 students gave a mean of 68 inches with
S.D. of 3 inches. Test the hypothesis that mean height in population is 70 inches under the
assumption that the heights are normally distributed. Also construct a 95% confidence interval
for the population mean.
Solution.
We have to test H :  = 70 against H :   70.
0 1 a 1
It is given that n = 20, X =68 and S = 3.

n 20
The unbiased estimate of s.d. is s S= = 3 = 3.08 .
n - 1 19
s 3.08
 . .S E = = = 0.688.
X
n 20
Alternatively, we can directly write

s n 1 S 3
S . . = = S ´ = = = 0.688.
E
X
n n - 1 n n - 1 19
-
68 70 ´ 19
Thus, t = = 2.906
cal
3
This value is greater than 2.093, the value of t from tables at 5% level of significance and 19 d.f.
Thus, H is rejected.
0

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