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[DOC File]Exam 3 Practice Questions
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49. A The 95% confidence interval for the population mean is. In this particular problem, we have. Using df = n – 1 = 4, we look up (in a table) that t.025 = 2.776. Then our confidence interval is. 50. C Assumptions for the confidence interval for the mean are as follows: data is quantitative, random sample, data comes from a normal distribution.
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12) The width of a confidence interval estimate for a mean will be . A) narrower for a sample of size 200 than for a sample of 100 B) wider for 80% confidence than for 99% confidence. C) wider when the variance is 0.4 than when it is 0.9. D independent of sample size. E) independent of the confidence level.
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9. Find the z scores (critical values) that corresponds to a level of confidence of 98.22. 10. Use the given degree of confidence and sample data to construct a confidence interval . for the population mean. A random sample of 86 light bulbs had a mean life of 545 . hours with a standard deviation of 29 hours. Construct a 90 percent confidence
96% confidence interval z score

[DOC File]AP Statistics LEOCT
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On a final exam in Algebra, the mean score was 23 with a standard deviation of 9. What is the z-score for a student who scored 18 on the exam? A. 28. B. 9. C. -5. D. 0.56. E. -0.56. Suppose that the results of a recent poll in a particular state said that 63% of respondents were …
96% confidence interval z

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a. below a positive Z score b. above a negative Z score c. between two positive Z scores d. above a positive Z score ____ 6. The area between the mean and a Z score of +1.50 is 43.32%. This score is less than ____ of the scores in the distribution. a. 43.32% b. 6.68% c. 3.32% d. 93.32% ____ 7. The mean score on a final chemistry exam was 75 ...
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Apr 08, 2010 · Construct the 90% confidence interval for occupational prestige for respondents with a bachelor’s degree (N=750). State in words the meaning of the result. The z value for 90% confidence is 1.645. The interval goes from: mean-z*sd/sqrt(N) to mean+z*sd/sqrt(N) 52.74-1.645*12.675/sqrt(750) to 52.74+1.645*12.675/sqrt(750) 51.98 to 53.50
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b) The confidence interval should be . c) is the estimate for (. So the approximate confidence interval is . d) The T distribution approximates the normal distribution as n increases, but at. small values of n, the T distribution is wider, meaning that a 95% confidence. interval would be wider, too.
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[DOC File]CHAPTER 8—INTERVAL ESTIMATION
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The 95% confidence interval for the average examination score of the population of the examinations is a. 76.00 to 84.00 b. 77.40 to 86.60 c. 83.00 to 85.00 d. 68.00 to 100.00 ANS: C PTS: 1 TOP: Interval …
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68 confidence interval z score