ssrs export to pdf barcode font Work Problem 6.19 using small or exact sampling theory. in Software

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6.22. Work Problem 6.19 using small or exact sampling theory.
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(a) 95% confidence limits are given by S !n>x0.975 and S !n>x0.025. For n 200 1 x2 0.975 x2 0.025 199 degrees of freedom, we find as in Problem 4.41, page 136, 1 (z 2 0.975 1 (z 2 0.025 !2(199) !2(199) 1)2 1)2 1 (1.96 2 1 ( 1.96 2 19.92)2 19.92)2 239 161
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from which x0.975 15.5 and x0.025 12.7. Then the 95% confidence limits are 100 !200>15.5 91.2 and 100 !200>12.7 111.3 hours respectively. Therefore, we can be 95% confident that the population standard deviation will lie between 91.2 and 111.3 hours. This should be compared with the result of Problem 6.19(a). (b) 99% confidence limits are given by S !n>x0.995 and S !n>x0.005. For n 200 1 199 degrees of freedom, 1 (z 2 0.995 1 (z 2 0.005 !2(199) !2(199) 1)2 1)2 1 (2.58 2 1 ( 2.58 2 19.92)2 19.92)2 253 150
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x2 0.995 x2 0.005
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from which x0.995 15.9 and x0.005 12.2. Then the 99% confidence limits are 100 !200>15.9 88.9 and 100 !200>12.2 115.9 hours respectively. Therefore, we can be 99% confident that the population standard deviation will lie between 88.9 and 115.9 hours. This should be compared with the result of Problem 6.19(b).
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Confidence intervals for variance ratios 6.23. Two samples of sizes 16 and 10, respectively, are drawn at random from two normal populations. If their variances are found to be 24 and 18, respectively, find (a) 98%, (b) 90% confidence limits for the ratio of the variances.
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CHAPTER 6 Estimation Theory
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16, n 10, s2 1 20, s2 2
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(a) We have m
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18 so that m m n n 1 1 s2 1
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s2 2
From Problem 4.47(b), page 139, we have F0.99 4.96 for n1 16 1 15 and n2 10 1 9 degrees of freedom. Also, from Problem 4.47(d), we have for n1 15 and n2 9 degrees of freedom F0.01 1 > 3.89 so that 1 > F0.01 3.89. Then using (17), page 198, we find for the 98% confidence interval 1 25.2 4.96 20.0 0.283 s2 1 s2 2 s2 1 s2 2 (3.89) 25.2 20.0
10 (18) 9
16 (24) 15
(b) As in (a) we find from Appendix F that F0.95 interval is
3.01 and F0.05 s2 1 s2 2
1 > 2.59. Therefore, the 90% confidence 25.2 20.0
25.2 1 3.01 20.0
s2 1 s2 2
(2.59)
Note that the 90% confidence interval is much smaller than the 98% confidence interval, as we would of course expect.
6.24. Find the (a) 98%, (b) 90% confidence limits for the ratio of the standard deviations in Problem 6.23.
By taking square roots of the inequalities in Problem 6.23, we find for the 98% and 90% confidence limits (a) (b) 0.53 0.65 s1 s2 s1 s2 2.21 1.81
Maximum likelihood estimates 6.25. Suppose that n observations, X1, c, Xn, are made from a normally distributed population of which the mean is unknown and the variance is known. Find the maximum likelihood estimate of the mean.
Since we have (1) Therefore, (2) ln L n ln (2ps2) 2 1 (x 2s2 a k m)2 L f (x1, m) c f (xn, m) (2ps2)
n>2e a(xk m)2>2s2
f (xk, m)
1 22ps2
(xk m)2>2s2
Taking the partial derivative with respect to m yields (3) Setting 'L>'m 0 gives 1 'L L 'm 1 (x s2 a k m)
CHAPTER 6 Estimation Theory
(4) or (5)
a (xk
i.e.
a xk
a xk n
Therefore, the maximum likelihood estimate is the sample mean.
6.26. If in Problem 6.25 the mean is known but the variance is unknown, find the maximum likelihood estimate of the variance.
If we write f (xk , s2) instead of f (xk, m), everything done in Problem 6.25 through equation (2) still applies. Then, taking the partial derivative with respect to s2, we have 1 'L L 's2 Setting 'L>'s2 0, we find s2 a (xk n m)2 n 2s2 1 (x 2(s2)2 a k m)2
Miscellaneous problems 6.27. (a) If P is the observed proportion of successes in a sample of size n, show that the confidence limits for estimating the population proportion of successes p at the level of confidence determined by zc are given by
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