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# ch 14 - Chapter 14 Solutions 14.1(a The standard deviation...

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Chapter 14 Solutions 14.1. (a) The standard deviation of x is σ/ 1000 . = 1 . 8974. (b) m = 2 ( 1 . 8974 ) . = 3 . 8 (or “ ± 3 . 8”). (c) The confidence intervals drawn may vary, of course, but they should be 2 m = 7 . 6 units wide. (d) 95%. 281.9 283.8 285.7 278.1 276.2 274.3 280 14.2. (a) 51% ± 3% = 48% to 54%. (b) “95% confidence” means that this interval was found using a procedure that produces correct results (i.e., includes the true population proportion) 95% of the time. 14.3. Shown below are sample output screens for (a) 10 and (b) 1000 SRSs. In 99.4% of all repetitions of part (a), students should see between 5 and 10 hits (that is, at least 5 of the 10 SRSs capture the true mean µ ). Out of 1000 80% confidence intervals, nearly all students will observe between 76% and 84% capturing the mean. 14.4. Search Table A for 0.0125 (half of the 2.5% that is not included in a 97.5% confidence interval). This area corresponds to z = 2 . 24. Software gives z = 2 . 2414. z * z * 0 Standard Normal curve Probability = 0.0125 Probability = 0.0125 14.5. State: What is the concentration of the active ingredient in this batch? Formulate: We will estimate the true concentration in this batch of the product—the mean µ of the distribution of all measurements—by giving a 95% confidence interval. Solve: We assume that the three measurements can be considered an SRS of all such measurements. We have been told that repeated measurements follow a Normal distribution with standard deviation 0.0068 g / l. The mean of the sample is x = 0 . 8403 + 0 . 8363 + 0 . 8447 3 = 0 . 8404 3 g / l . 169

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170 Chapter 14 Confidence Intervals: The Basics For 95% confidence, the critical value is z = 1 . 960, so the 95% confidence interval for µ is x ± z σ n = 0 . 8404 3 ± 1 . 960 0 . 0068 3 = 0 . 8404 3 ± 0 . 00770 = 0 . 8327 to 0 . 8481 g / l .
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ch 14 - Chapter 14 Solutions 14.1(a The standard deviation...

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