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CHAPTER 25 Mathematical Statistics Changes Regression and a short introduction to correlation have been combined in the same section (Sec. 25.9). SECTION 25.1. Introduction. Random Sampling, page 1044 Purpose. To explain the role of (random) samples from populations. Main Content, Important Concepts Population Sample Random numbers, random number generator Sample mean: x w ; see (1) Sample variance s 2 ; see (2) Comments on Content Sample mean and sample variance are the two most important parameters of a sample. x w measures the central location of the sample values and s 2 their spread (their variability). Small s 2 may indicate high quality of production, high accuracy of measurement, etc. Note well that x w and s 2 will generally vary from sample to sample taken from the same population, whose mean m and variance s 2 are unique, of course. This is an important conceptual distinction that should be mentioned explicitly to the students. SECTION 25.2. Point Estimation of Parameters, page 1046 Purpose. As a first statistical task we discuss methods for obtaining approximate values of unknown population parameters from samples; this is called estimation of parameters. Main Content, Important Concepts Point estimate, interval estimate Method of moments Maximum likelihood method SOLUTIONS TO PROBLEM SET 25.2, page 1048 2. Put 5 0 in Example 1 and proceed with the second equation in (8), as in the example, to get the estimate | 2 5 o n i 5 1 x i 2 . 4. The likelihood function is (we can drop the binomial factors) < 5 p k 1 (1 2 p ) n 2 k 1 ••• p k m (1 2 p ) n 2 k m 5 p k 1 1 ••• 1 k m (1 2 p ) nm 2 ( k 1 1 ••• 1 k m ) . 1 } n 392 im25.qxd 9/21/05 2:06 PM Page 392

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The logarithm is ln < 5 ( k 1 1 ••• 1 k m ) ln p 1 [ nm 2 ( k 1 1 1 k m ) ] ln (1 2 p ). Equating the derivative with respect to p to zero, we get, with a factor 2 1 from the chain rule, ( k 1 1 1 k m ) 5 [ nm 2 ( k 1 1 1 k m )] . Multiplication by p (1 2 p ) gives ( k 1 1 •• 1 k m )(1 2 p ) 5 [ nm 2 ( k 1 1 1 k m ) ] p . By simplification, k 1 1 k m 5 nmp . The result is p ˆ 5 o m i 5 1 k i . 6. We obtain < 5 ƒ( x 1 )ƒ( x 2 ) ƒ( x n ) 5 p (1 2 p ) x 1 2 1 p (1 2 p ) x 2 2 1 p (1 2 p ) x n 2 1 5 p n (1 2 p ) x 1 1 ••• 1 x n 2 n . The logarithm is ln < 5 n ln p 1 ( o n m 5 1 x m 2 n ) ln (1 2 p ). Differentiation with respect to p gives, with a factor 2 1 from the chain rule, 52 o n m 5 1 . Equating this derivative to zero gives n (1 2 p ˆ ) 5 p ˆ ( o n m 5 1 x m 2 n ) . Thus p ˆ 5 1/ x w . 8. p ˆ 5 2/(7 1 6) 5 2/13, by Prob. 6. 10. < 5 1( b 2 a ) n is maximum if b 2 a is as small as possible, that is, a equal to the smallest sample value and b equal to the largest. 12. m 5 1/ u , ˆ 5 x w 14. ˆ 5 1/ x w 5 2, F ( x ) 5 1 2 e 2 2 x if x ^ 0 and 0 otherwise. A graph shows that the step function F ˜ ( x ) (the sample distribution function) approximates F ( x ) reasonably well. (For goodness of fit, see Sec. 25.7.) SECTION 25.3. Confidence Intervals, page 1049 Purpose. To obtain interval estimates (“confidence intervals”) for unknown population parameters for the normal distribution and other distributions.
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ch25 - im25.qxd 2:06 PM Page 392 CHAPTER 25 Mathematical...

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