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Bluman 5th_Chapter 7 HW Soln

# Bluman 5th_Chapter 7 HW Soln - Math 227 Elementary...

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Math 227 – Elementary Statistics: A Brief Version, 5/e Bluman Ch 7.1 pg. 364 #11, 13, 15, 17, 19, 21, 23, 25 11. Reading Scores : A sample of the reading scores of 35 fifth-graders has a mean of 82. The standard deviation of the sample is 15. Note: All values we calculated were rounded. a) Find the best point estimate of the mean. X Is the best point estimate for μ , therefore, the best point estimate of the population mean is 82 μ = b) Find the 95% confidence interval of the mean reading scores of all the fifth-graders. n=35, Confidence interval (C.I.) = 0.95, 82, 15 x s = = / 2 0.05 0.05 0.025 2 2 1 0.025 0.975 ( ) 1.96 Z α α α = = = - = = / 2 / 2 [ ] [ ] s s X Z X Z n n α α μ - < < + 15 15 82 [1.96 ] 82 [1.96 ] 35 35 μ - < < + 77 87 μ < < c) Find the 99% confidence interval of the mean reading scores of all fifth graders. 0.95 /2 1.96 Z α = 0.025 0.975 0.95

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/2 0.99 ( ) 0.495 2 Z α = = To find go to table E and look for area = .4950, the corresponding z value for this area is 2.58. /2 /2 [ ] [ ] X Z X Z n n α α σ σ μ - < < + 82 [2.58 (2.5355)] 82 [2.58 (2.5355)] μ - < < - 75 89 μ < < d) Which interval is larger? Why? The 99% confidence interval is larger because the confidence level is larger. 13. Workers’ Distractions A recent study showed that the modern working person experiences an average of 2.1 hours per day of distractions (phone calls, e-mails, impromptu visits, etc.). A random sample of 50 workers for a large corporation found that these workers were distracted an average of 1.8 hours per day and the population standard deviation was 20 minutes. Estimate the true mean population distraction time with 90% confidence, and compare your answer to the results of the study. n=50, Confidence interval (C.I.) = 0.90, 1.8hr, 20min 0.33 hr x s = = / 2 0.10 0.10 0.05 2 2 1 0.05 0.95 ( ) 1.65 Z α α α = = = - = = / 2 / 2 [ ] [ ] s s X Z X Z n n α α μ - < < + 0.33 0.33 1.8 [1.65 ] 1.8 [1.65 ] 50 50 μ - < < + /2 Z α /2 2.58 Z α = 0.90 / 2 1.65 Z α = 0.05 0.95
1.8 0.08 1.8 0.08 μ - < < + 1.72 1.88 μ < < The estimate is lower. 15. Actuary Exams A survey of 35 individuals who passed the seven exams and obtained the rank of Fellow in the actuarial field finds the average salary to be \$150,000. If the standard deviation for the population is \$15,000, construct a 95% confidence interval for all Fellows. n=35, Confidence interval (C.I.) = 0.95, \$150,000, \$15,000 x s = = / 2 0.05 0.05 0.025 2 2 1 0.025 0.975 ( ) 1.96 Z α α α = = = - = = /2 /2 [ ] [ ] X Z X Z n n α α σ σ μ - < < + 15,000 15,000 150,000 [1.96 ] 150,000 [1.96 ] 35 35 μ - < < + 150,000 4969.51 150,000 4969.51 \$145,030 \$154,970 μ μ - < < + < < 17. Television viewing a study of 415 kindergarten students showed that they have seen on average 5000 hours of television. If the sample standard deviation is 900, find the 95% confidence level of the mean for all students. If a parent claimed that his children watched 4000 hours, would the claim be believable? n=415, Confidence interval (C.I.) = 0.95, 5,000, 900 x σ = = / 2 0.05 0.05 0.025 2 2 1 0.025 0.975 ( ) 1.96 Z α α α = = = - = = /2 /2 ( ) ( ) X Z X Z n n α α σ σ μ - < < + 0.95 /2 1.96 Z α = 0.025 0.975 0.95 0.95 /2 1.96 Z α = 0.025 0.975 0.95

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900 900 5000 (1.96 ) 5000 (1.96 ) 415 415 4913 5087 μ μ - < < + < < Since 4000 hours is not within a 95% confidence interval, we can say that the claim is not believable. 19. Hospital Noise Levels Noise levels at various areas urban hospitals were measured in decibels. The mean of the noise levels in 84 corridors was 61.2 decibels, and the standard deviation was 7.9. Find the 95% confidence interval of the true mean.
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