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IT233: Applied Statistics TIHE 2005 Example 2: A sample of size 16 is to be taken from a population having a normal distribution with 2 = 6.25. Find a number C such that P(S 2 C) 0.025 Solution: 2 = 6.25 Given: n = 16 d. f. ( ) = n 1 = 15 P(S 2 C) 0.025 (n 1) S 2 P (n 1)C = 0.025 2 2 15 C P 2 = 0.025 6.25 P 2 2.4C = 0.025 0.025 0 2.4 C From Chi-Square Table we get 2.4 C = 27.488 C = 11.453 Example 3: For the Example 2, find a constants C1 and C2 such that P(C1 S 2 C2 ) 0.90 1 Solution: P(C1 S 2 C2 ) 0.90 0.05 15C2 15C 1 2 P = 0.90 6.25 6.25 P 2.4C 2 2.4C = 0.90 1 2 0.05 0.90 0 2.4 C1 2.4 C2 From Chi-Square Table we find C1 and C2 as follows: 2.4 C1 = 7.261 2.4 C2 = 24.996 C1 = 3.025 C2 = 10.415 S12 Sampling Distribution of the Ratio of Two Sample Variances - 2 S2 Theorem: If S 2 and S 2 are the variances of independent random samples of 1 2 size n1 and n2 taken from normal populations with variances 2 and 2 , 1 respectively, then the statistic S12 12 22 S12 F S22 22 12 S22 has an F - distribution with 1 n1 1 and 2 n2 1 d. f. Applications: The sampling distribution of S12 S22 To find the probabilities about 2 S12 S22 could be used: 2 12 To find the confidence interval for To test the hypothesis H 0 : 12 22 22 Example 1: (Page 229, No.9) If S 2 and S 2 represent the variances of 1 2 independent random samples of size n1 = 25 and n2 = 31, taken from normal populations with variances 2 = 10 and 2 = 15, respectively, find 1 2 P S 2 1 2 S 1.26 . 2 Solution: We know, F 22 S12 12 S22 Fn 1, n 1 1 2 2 S2 2 S2 P 1 1.26 P 2 1 1.26 2 S2 2 S2 12 2 1 2 15 P F 1.26 10 0.05 P F 1.89 0 1.89 Looking F-table for 1 = 24 and 2 = 30 at = 0.05 we found P F 1.89 = 0.05 S2 1 P 1.26 = 0.05 2 S 2 3 Example 2: In the above example, find a value C such that S2 1 P C = 0.01 2 S 2 Solution: S2 1 P C = 0.01 S2 2 2 S2 22 2 1 P C = 0.01 2 2 S2 1 1 2 1 = 24, 2 = 30 0.01 P F 1.5C = 0.01 0 1.5 C Looking F-table for 1 = 24 and 2 = 30 at = 0.01 we get NOTE: 1.5 C = 2.47 C = 1.647 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. 0.95 0.05 f 0.95 1, 2 f0.05 1, 2 Upper-tail value: Its cuts off a small area to the right. Lower-tail value: Its cuts off a large area to the right. 4 The F-table gives only upper-tail values. The lower-tail values are computed as f1 1, 2 f 0.95 1, 2 Hence, f 0.975 1, 2 Similarly, Example: 19, 20 2 1 2 f 0.99 with = 28 and = 12 1 2.1555 f 0.99 28,12 1 = 0.4639 (ii) f 0.05 2 , 1 f 0.025 2 , 1 1 f 0.05 20,19 0.95 1 1 Solution: f f 2 , 1 f 0.95 with = 19 and = 20 Find (i) (ii) (i) 1 1 f 0.01 12, 28 1 2.90 = 0.3448 5