Chpt07 - Chapter 7 Estimation and Statistical Intervals...

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Chapter 7 Estimation and Statistical Intervals Section 7.1 1. A single randomly selected item forms a sample of size n = 1. If x denotes the value of this item, then the sample average is also equal to x. Since the sample average, based on any sample size, is an unbiased estimator or the population mean, then the length x is indeed an unbiased estimator of the population mean. Of course, this estimator is not as precise as the sample average based on larger sample sizes (i.e., it has a larger standard error), but it is still unbiased. 2. (a) An unbiased point estimate of μ , the average amount of gas used by all houses in the area is: therms x 6 . 120 10 206 , 1 = = (b) An unbiased point estimate of π , the proportion of all homes that use over 100 therms is: 80 . 10 8 = = p 3. (a) Because the population is known to be normal and its standard deviation is known to be σ = 5, the sampling distribution of x is also normal. Standardizing, P( μ -1 x μ + 1) = P( n σ μμ ) 1 ( z n + ) 1 ( ) = P(- n z n ) = P(-.63 z .63) = .7357 - .2643 = .4714. (b) In each case, as in part (a), the desired probability is P(- n z n ): For n = 50, P(- n z n ) = P(-1.41 z 1.41) = .9207 - .0793 = .8414. For n = 100, P(- n z n ) = P(-2.00 z 2.00) = .9772 - .0228 = .9544. For n = 1000, P(- n z n ) = P(-6.32 z 6.32) 1.0000 - .0000 = 1.0000. In other words, as n increases, it becomes more and more likely that the sample mean will fall within ± 1 standard deviation from the population mean.
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Chapter 7 Estimation and Statistical Intervals 2 (c) As the following graph shows, the probability that the sample mean lies within ± 1 unit of μ increases as the sample size n increases: 1000 500 0 1.0 0.9 0.8 0.7 0.6 0.5 Sample Size, n Proability 4. (a) Recall: ( ) n p p ππ σπ μ = = 1 and Even though π is unknown, we can still set an upper bound on . p σ W h e n is maximized. So, p , 5 . = n p 2 1 = . In our case, 15811 . 10 = = p n () 4714 . 63 . 63 . 15811 . 10 . 15811 . 10 . 10 . 10 . So, = < < = < < = < < z P z P p P π (b) When 07071 . 50 2 1 , 50 = = = p n So, < < = < < 07071 . 10 . 07071 . 10 . 10 . 10 . z P p P 8414 . 41 . 1 41 . 1 = < < = z P When 05 . 100 2 1 , 100 = = = p n So, < < = < < 05 . 10 . 05 . 10 . 10 . 10 . z P p P 9544 . 2 2 = < < = z P
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Chapter 7 Estimation and Statistical Intervals 3 When 015811 . 000 , 1 2 1 , 000 , 1 = = = p n σ So, () < < = < < 015811 . 10 . 015811 . 10 . 10 . 10 . z P p P π 1 33 . 6 32 . 6 < < = z P (c) 0 0.2 0.4 0.6 0.8 1 0 200 400 600 800 1000 n probability Based on the graph, the probability that the sample proportion, p, lies within 10% of π increases as the sample size, n, increases. However, as you can see, the increase is not linear. 5. (a) As in exercise 3, we have P( μ -1 x μ + 1) = P( n μ ) 1 ( z n + ) 1 ( ) = P(- n z n ). In this exercise, σ = 2. To make this probability equal to 90%, first use Table I (pages 534- 535) to find the (approximate) value z = 1.645 for which P(-1.645 z 1.645) .90. Then equate 1.645 and n and solve for n: n = 1.645, so n =1.645( σ ) = 1.645(2) and, squaring, n = (4)(1.645) 2 = 10.82. Rounding off to the nearest integer value, n 11 will guarantee that there is at
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This note was uploaded on 02/23/2009 for the course STAT 350 taught by Professor Staff during the Fall '08 term at Purdue University-West Lafayette.

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Chpt07 - Chapter 7 Estimation and Statistical Intervals...

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