IT233: Applied Statistics TIHE 2005
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Example 2: A sample of size 16 is to be taken from a population having a normal distribution with = 6.25. Find a number such that
Solution: Given: = 6.25
= 16 d. f. () = = 15
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= 0.025
= 0.025
= 0.025
0.025
0 2.4
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From Chi-Square Table we get
2.4 = 27.488
= 11.453
Example 3: For the Example 2, find a constants and such that
Solution:
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= 0.90
= 0.90
0.05
0.05
0.90
0 2.4 2.4
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From Chi-Square Table we find and as follows:
2.4 = 7.261 2.4 = 24.996
= 3.025 = 10.415
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Theorem: If and are the variances of independent random samples of size and taken from normal populations with variances and , respectively, then the statistic
has an - distribution with and d. f.
Applications: The sampling distribution of could be used:
· To find the probabilities about
· To find the confidence interval for
· To test the hypothesis
Example 1: (Page 229, No.9) If and represent the variances of independent random samples of size = 25 and = 31, taken from normal populations with variances = 10 and = 15, respectively, find .
Solution: We know,
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1
0.05
0 1.89
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Looking F-table for = 24 and = 30 at = 0.05 we found = 0.05
= 0.05
Example 2: In the above example, find a value such that
= 0.01
Solution:
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1
= 0.01
= 0.01
= 0.01
= 24, = 30
0.01
0 1.5
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Looking F-table for = 24 and = 30 at = 0.01 we get
1.5 = 2.47
= 1.647
NOTE: Since F-distribution is also used for finding confidence interval and hypothesis testing, sometimes we need to find lower tail as well as upper tail values of F.
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0.95
Lower-tail value: Its cuts off a large area to the right.
0.05
Upper-tail value: Its cuts off a small area to the right.
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The F-table gives only upper-tail values. The lower-tail values are computed as
Hence,
Similarly,
Example: Find (i) with = 19 and = 20
(ii) with = 28 and = 12
Solution:
(i)
= 0.4639
(ii)
= 0.3448
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