ch14 - Chapter 14 Solutions . 14.1. The standard deviation...

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Chapter 14 Solutions 14.1. (a) The standard deviation is σ/ 840 . = 2 . 0702. (b) The missing number is 2 840 . = 4 . 1404. (c) The 95% confdence interval is x ± 2 840 . = 268 to 276. 14.2. 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% confdence intervals, nearly all students will observe between 76% and 84% capturing the mean. 14.3. Search Table A For 0.0125 (halF oF the 2.5% that is not included in a 97.5% confdence 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 14.4. State: What is the true conductivity oF this liquid? Plan: We will estimate the true conductivity µ , the mean oF all measurements oF its conductivity, by giving a 90% confdence interval. Solve: The statement oF the problem in the text suggests that the conditions For inFerence should be satisfed. The mean oF the sample is x = 4 . 988 3 microsiemens per centimeter ( µ S / cm). ±or 90% confdence, the critical value is z = 1 . 645. A 90% confdence interval For µ is thereFore x ± z µ σ n = 4 . 988 3 ± 1 . 645 µ 0 . 2 6 = 4 . 988 3 ± 0 . 1343 = 4 . 8540 to 5 . 1226 µ S / cm . Conclude: We are 90% confdent that the true conductivity is between 4.8540 and 5.1226 µ S / cm. 169
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170 Chapter 14 Introduction to Inference 14.5. (a) The two low scores (72 and 74) are both possible outliers, but there are no other apparent deviations from Normality. (b) State: What is the mean IQ µ of all seventh-grade girls in this school district? Plan: We will estimate µ by giving a 99% conFdence interval. Solve: The problem states that these girls are an SRS of the popula- tion. In part (a), we saw that the scores appear to come from a Normal distribution. With x . = 105 . 84, our 99% conFdence interval for µ is 105 . 84 ± 2 . 576 µ 15 31 . = 105 . 84 ± 6 . 94 = 98 . 90 to 112 . 78 IQ points. Conclude: We are 99% conFdent that the mean IQ of seventh-grade girls in this district is between 98.90 and 112.78. 7 24 7 8 8 69 9 13 9 68 10 023334 10 578 11 11222444 11 89 12 0 12 8 13 02 14.6. (a) If µ = 115, the distribution is approximately Normal with mean µ = 115 and standard deviation σ/ 25 = 6. (b) The actual result lies out toward the high tail of the curve, while 118.6 is fairly close to the middle. If µ = 115, observing a value similar to 118.6 would not be too surprising, but 125.8 is less likely, and it therefore provides some evidence that µ> 115. 121 127 133 109 103 97 115 125.8 118.6 14.7. (a) If the claim is true, the sampling distribution of x is Normal with mean 5 µ S / cm and standard deviation 6 . = 0 . 0816 µ S / cm. (b) 4.98 is less than 0.25 standard deviations below the presumed mean, while 4.7 is about 3.67 standard deviations below. If µ = 5, observing a value similar to 4.98 would not be too surprising, but 4.7 is less likely, and it therefore provides evidence that µ is different from 5. (SpeciFcally, it suggests that µ< 5.) 5.08 5.16 5.24 4.92 4.84 4.76 5 4.7 4.98 14.8. H 0 : µ = 115 vs. H a : 115. (Because the teacher suspects that older students have a higher mean, we use a one-sided alternative.) 14.9. H 0 : µ = 5 vs. H a : µ 6= 5. (We are concerned about deviation from 5 in either direction, so we use a two-sided alternative.) 14.10. H 0 : µ = 50 vs.
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This note was uploaded on 11/23/2010 for the course STAT 2325151 taught by Professor T during the Spring '10 term at Waters College.

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ch14 - Chapter 14 Solutions . 14.1. The standard deviation...

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