Chapter 7--Problem 10, A population has a mean of u = 100 and astandard deviatio

Question

Chapter 7--Problem 10, A population has a mean of u = 100 and astandard deviation of a = 20. Find the z-scores corresponding toeach of the following sample means obtained from this population.A. M = 102 for a sample of a n = 4 scores. B. M = 102 for a sampleof n = 100 scores. C. M = 95 for a sample of n = 16 scores D. M =95 for a sample of n = 25 scores

Chapter 7--Problem 18, For a normal-shaped distribution with u =50 and a = 8; A. What proportion of the scores have values between46 and 54? B. For samples of n = 4, what proportion of the samplesmeans have values between 46 and 54? C. For sample of n = 16, whatproportion of the sample mean have values between 46 and 54?

Explanation / Answer

Standard Error for a sample mean = population stdev / sqrt(n) z = (X - ) / SE A) z(102) = (102 - 100) / (20/sqrt(4)) = 2 / 10 = .2 B) z(102) = (102 - 100) / (20/sqrt(100)) = 2 / 2 = 1 C) z(95) = (95 - 100) / (20/sqrt(16)) = -5 / 5 = -1 D) z(95) = (95 - 100) / (20/sqrt(25)) = -5 / 4 = -1.25 Note: As the sample size increases in both instances, the samevariation from the mean becomes more significant. This isalways the case. A) z(46) = (46 - 50) / 8 = -.5 --> p = .3085 z(54) = (54 - 50) / 8 = .5 --> p = .6915 .6915 - .3085 = .383 = 38.3% B) z(46) = (46 - 50) / (8/sqrt(4)) = -1 --> p = .1587 z(54) = (54 - 50) / (8/sqrt(4)) = 1 --> p = .8413 .8413 - .1587 = .6824 = 68.24% Note: This is a common range and is a part of the 68-95-99.7 rule(which is the area between ±1, 2, and 3 standard deviationsrespectively). C) z(46) = (46 - 50) / (8/sqrt(16)) = -2 --> p = .0228 z(54) = (54 - 50) / (8/sqrt(16)) = 2 --> p = .9772 .9772 - .0228 = .9544 = 95.54%

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