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P z2 c 2n zc P(1 A z2 c n n P) z2 c 4n2
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(b) Use the formula derived in (a) to obtain the 99.73% confidence limits of Problem 6.13. (c) Show that for large n the formula in (a) reduces to P zc !P(1 P)>n, as used in Problem 6.13.
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(a) The sample proportion P in standard units is P sP p P !p(1 p p)>n zc, where zc determines the level of
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The largest and smallest values of this standardized variable are confidence. At these extreme values we must therefore have P Squaring both sides, P2 2pP p2 z2 c p(1 n p) p5 zc p(1 A n p)
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Multiplying both sides by n and simplifying, we find
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0 bp 4 An c 0, whose solution
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If a n z2, b A 2nP z2 B and c c c for p is given by the quadratic formula as p b 2nP
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nP2, this equation becomes ap2 z2 B 2 c
2nP z2 2 A 2nP c 2b2 4ac 2a 2 An z2 zc 24nP(1 P) z2 c c 2 An z2 B c
z2 B c
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CHAPTER 6 Estimation Theory
Dividing the numerator and denominator by 2n, this becomes P p z2 c 2n zc P(1 A z2 c n 100 in the formula derived in (a), we n P) z2 c 4n2
(b) For 99.73% confidence limits, zc 3. Then using P 0.55 and n find p 0.40 and 0.69, agreeing with Problem 6.13(c).
(c) If n is large, then z2 >2n, z2 >4n2, and z2 >n are all negligibly small and can essentially be replaced by zero, so c c c that the required result is obtained.
6.28. Is it possible to obtain a 95% confidence interval for the population standard deviation whose expected width is smaller than that found in Problem 6.22(a)
The 95% confidence limits for the population standard deviation as found in Problem 6.22(a) were obtained by choosing critical values of x2 such that the area in each tail was 2.5%. It is possible to find other 95% confidence limits by choosing critical values of x2 for which the sum of the areas in the tails is 5%, or 0.05, but such that the areas in each tail are not equal. In Table 6-2 several such critical values have been obtained and the corresponding 95% confidence intervals shown. Table 6-2 Critical Values x0.01 x0.02 x0.03 x0.04 12.44, x0.96 12.64, x0.97 12.76, x0.98 12.85, x0.99 15.32 15.42 15.54 15.73 95% Confidence Interval 92.3 to 113.7 91.7 to 111.9 91.0 to 110.8 89.9 to 110.0 Width 21.4 20.2 19.8 20.1
From this table it is seen that a 95% interval of width only 19.8 is 91.0 to 110.8. An interval with even smaller width can be found by continuing the same method of approach, using critical values such as x0.031 and x0.981, x0.032 and x0.982, etc. In general, however, the decrease in the interval that is thereby obtained is usually negligible and is not worth the labor involved.
SUPPLEMENTARY PROBLEMS
Unbiased and efficient estimates
6.29. Measurements of a sample of weights were determined as 8.3, 10.6, 9.7, 8.8, 10.2, and 9.4 lb, respectively. Determine unbiased and efficient estimates of (a) the population mean, and (b) the population variance. (c) Compare the sample standard deviation with the estimated population standard deviation. 6.30. A sample of 10 television tubes produced by a company showed a mean lifetime of 1200 hours and a standard deviation of 100 hours. Estimate (a) the mean, (b) the standard deviation of the population of all television tubes produced by this company. 6.31. (a) Work Problem 6.30 if the same results are obtained for 30, 50, and 100 television tubes, (b) What can you conclude about the relation between sample standard deviations and estimates of population standard deviations for different sample sizes
CHAPTER 6 Estimation Theory
Confidence interval estimates for means (large samples)
6.32. The mean and standard deviation of the maximum loads supported by 60 cables (see Problem 5.98) are 11.09 tons and 0.73 tons, respectively. Find (a) 95%, (b) 99% confidence limits for the mean of the maximum loads of all cables produced by the company. 6.33. The mean and standard deviation of the diameters of a sample of 250 rivet heads manufactured by a company are 0.72642 inch and 0.00058 inch, respectively (see Problem 5.99). Find (a) 99%, (b) 98%, (c) 95%, (d) 90% confidence limits for the mean diameter of all the rivet heads manufactured by the company. 6.34. Find (a) the 50% confidence limits, (b) the probable error for the mean diameter in Problem 6.33. 6.35. If the standard deviation of the lifetimes of television tubes is estimated as 100 hours, how large a sample must we take in order to be (a) 95%, (b) 90%, (c) 99%, (d) 99.73% confident that the error in the estimated mean lifetime will not exceed 20 hours. 6.36. What are the sample sizes in Problem 6.35 if the error in the estimated mean lifetime must not exceed 10 hours
Confidence interval estimates for means (small samples)
6.37. A sample of 12 measurements of the breaking strengths of cotton threads gave a mean of 7.38 oz and a standard deviation of 1.24 oz. Find (a) 95%, (b) 99% confidence limits for the actual mean breaking strength. 6.38. Work Problem 6.37 assuming that the methods of large sampling theory are applicable, and compare the results obtained. 6.39. Five measurements of the reaction time of an individual to certain stimuli were recorded as 0.28, 0.30, 0.27, 0.33, 0.31 second. Find (a) 95%, (b) 99% confidence limits for the actual mean reaction time.
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