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Unformatted text preview: Chapter 7 Estimates and Sample Sizes 72 Estimating a Population Proportion 1. The confidence level was not stated. The most common level of confidence is 95%, and sometimes that level is carelessly assumed without actually being stated. 2. The margin of error is the maximum likely difference between the point estimate for a parameter and its true value. 3. By including a statement of the maximum likely error, a confidence interval provides information about the accuracy of an estimate. 4. No. A voluntary response sample is not necessarily representative of the population. 5. For 99% confidence, α = 1–0.99 = 0.01 and α /2 = 0.01/2 = 0.005. For the upper 0.005, A = 0.9950 and z = 2.575. z α /2 = z 0.005 = 2.575 6. For 99.5% confidence, α = 1–0.995 = 0.005 and α /2 = 0.005/2 = 0.0025. For the upper 0.0025, A = 0.9975 and z = 2.81. z α /2 = z 0.0025 = 2.81 7. For α = 0.10, α /2 = 0.10/2 = 0.05. For the upper 0.05, A = 0.9500 and z = 1.645. z α /2 = z 0.05 = 1.645 8. For α = 0.02, α /2 = 0.02/2 = 0.01. For the upper 0.01, A = 0.9900 [0.9901] and z = 2.33. z α /2 = z 0.01 = 2.33 9. Let L = the lower confidence limit; U = the upper confidence limit. ˆp = (L+U)/2 = (0.200+0.500)/2 = 0.700/2 = 0.350 E = (U–L)/2 = (0.500–0.200)/2 = 0.300/2 = 0.150 The interval can be expressed as 0.350 ± 0.150. 10. Let L = the lower confidence limit; U = the upper confidence limit. ˆp = (L+U)/2 = (0.720+0.780)/2 = 1.500/2 = 0.750 E = (U–L)/2 = (0.780–0.720)/2 = 0.060/2 = 0.030 The interval can be expressed as 0.750 ± 0.030. 11. Let L = the lower confidence limit; U = the upper confidence limit. ˆp = (L+U)/2 = (0.437+0.529)/2 = 0.966/2 = 0.483 E = (U–L)/2 = (0.529–0.437)/2 = 0.092/2 = 0.046 The interval can be expressed as 0.483 ± 0.046. 12. Given that ˆp = 0.222 and E = 0.044, L = ˆp – E = 0.222 – 0.044 = 0.178 U = ˆp +E = 0.222 + 0.044 = 0.266 The interval can be expressed as 0.178 < p < 0.266. 198 CHAPTER 7 Estimates and Sample Sizes 13. Let L = the lower confidence limit; U = the upper confidence limit. ˆp = (L+U)/2 = (0.320+0.420)/2 = 0.740/2 = 0.370 E = (U–L)/2 = (0.420–0.320)/2 = 0.100/2 = 0.050 14. Let L = the lower confidence limit; U = the upper confidence limit. ˆp = (L+U)/2 = (0.772+0.776)/2 = 1.548/2 = 0.774 E = (U–L)/2 = (0.776–0.772)/2 = 0.004/2 = 0.002 15. Let L = the lower confidence limit; U = the upper confidence limit. ˆp = (L+U)/2 = (0.433+0.527)/2 = 0.960/2 = 0.480 E = (U–L)/2 = (0.527–0.433)/2 = 0.094/2 = 0.047 16. Let L = the lower confidence limit; U = the upper confidence limit. ˆp = (L+U)/2 = (0.102+0.236)/2 = 0.338/2 = 0.169 E = (U–L)/2 = (0.236–0.102)/2 = 0.134/2 = 0.067 IMPORTANT NOTE: When calculating E = α /2 ˆ ˆ z pq/n do not round off in the middle of the problem. This, and the subsequent calculations of U = ˆp + E and L = ˆp – E may accomplished conveniently on most calculators having a memory as follows....
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 Spring '11
 Dr.Kalluri
 Normal Distribution, σx

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