ssrs export to pdf barcode font Estimation Theory in Software

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CHAPTER 6 Estimation Theory
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the derivative. In this way we find 'f (x1, u) 1 f (x1, u) 'u c 'f (xn, u) 1 f (xn, u) 'u 0
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The solution of this equation, for u in terms of the xk, is known as the maximum likelihood estimator of u. The method is capable of generalization. In case there are several parameters, we take the partial derivatives with respect to each parameter, set them equal to zero, and solve the resulting equations simultaneously.
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Unbiased and efficient estimates 6.1. Give examples of estimators (or estimates) which are (a) unbiased and efficient, (b) unbiased and inefficient, (c) biased and inefficient.
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Assume that the population is normal. Then n S 2 are two such examples. n 1 (b) The sample median and the sample statistic 1 (Q1 Q3), where Q1 and Q3 are the lower and upper sample 2 quartiles, are two such examples. Both statistics are unbiased estimates of the population mean, since the mean of their sampling distributions can be shown to be the population mean. However, they are both # inefficient compared with X. # (a) The sample mean X and the modified sample variance S 2
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(c) The sample standard deviation S, the modified standard deviation S, the mean deviation, and the semiinterquartile range are four such examples for evaluating the population standard deviation, s.
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6.2. A sample of five measurements of the diameter of a sphere were recorded by a scientist as 6.33, 6.37, 6.36, 6.32, and 6.37 cm. Determine unbiased and efficient estimates of (a) the true mean, (b) the true variance. Assume that the measured diameter is normally distributed.
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(a) An unbiased and efficient estimate of the true mean (i.e., the population mean) is x # ax n 6.33 6.37 6.36 5 6.32 6.37 6.35 cm
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(b) An unbiased and efficient estimate of the true variance (i.e., the population variance) is
n n (6.33 1
a (x n
x)2 # 1 (6.37 6.35)2 (6.36 5 6.35)2 1 (6.32 6.35)2 (6.37 6.35)2
6.35)2
0.00055 cm2 Note that ^ s 20.00055 neither unbiased nor efficient. 0.023 is an estimate of the true standard deviation, but this estimate is
6.3. Suppose that the heights of 100 male students at XYZ University represent a random sample of the heights of all 1546 male students at the university. Determine unbiased and efficient estimates of (a) the true mean, (b) the true variance.
(a) From Problem 5.33: Unbiased and efficient estimate of true mean height (b) From Problem 5.38: Unbiased and efficient estimate of true variance
67.45 inch
n n 1
100 (8.5275) 99
CHAPTER 6 Estimation Theory
Therefore, ^ s !8.6136 and ^2 or between s and ^. s s
2.93. Note that since n is large there is essentially no difference between s2
6.4. Give an unbiased and inefficient estimate of the true (mean) diameter of the sphere of Problem 6.2.
The median is one example of an unbiased and inefficient estimate of the population mean. For the five measurements arranged in order of magnitude, the median is 6.36 cm.
Confidence interval estimates for means (large samples) 6.5. Find (a) 95%, (b) 99% confidence intervals for estimating the mean height of the XYZ University students in Problem 6.3.
# (a) The 95% confidence limits are X 1.96s> !n. Using x 67.45 inches and ^ 2.93 inches as an estimate of s (see Problem 6.3), the confidence limits s are 67.45 1.96(2.93> !100), or 67.45 0.57, inches. Then the 95% confidence interval for the population mean m is 66.88 to 68.02 inches, which can be denoted by 66.88 m 68.02. We can therefore say that the probability that the population mean height lies between 66.88 and 68.02 inches is about 95%, or 0.95. In symbols we write P(66.88 m 68.02) 0.95. This is equivalent to saying that we are 95% confident that the population mean (or true mean) lies between 66.88 and 68.02 inches. # (b) The 99% confidence limits are X
2.58s> !n. For the given sample, s !n 67.45 2.58 2.93 !100 67.45 0.76 inches
Therefore, the 99% confidence interval for the population mean m is 66.69 to 68.21 inches, which can be m 68.21. denoted by 66.69 In obtaining the above confidence intervals, we assumed that the population was infinite or so large that we could consider conditions to be the same as sampling with replacement. For finite populations where sampling is s N n s without replacement, we should use in place of . However, we can consider the factor !n A N 1 !n 1546 100 N n 0.967 as essentially 1.0, so that it need not be used. If it is used, the above confidence A 1546 1 AN 1 limits become 67.45 0.56 and 67.45 0.73 inches, respectively.
6.6. Measurements of the diameters of a random sample of 200 ball bearings made by a certain machine during one week showed a mean of 0.824 inch and a standard deviation of 0.042 inch. Find (a) 95%, (b) 99% confidence limits for the mean diameter of all the ball bearings.
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