ISM_chapter7 - Chapter 7 Sampling Distributions and the...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
143 Chapter 7: Sampling Distributions and the Central Limit Theorem 7.1 a. c. Answers vary. d. The histogram exhibits a mound shape. The sample mean should be close to 3.5 = μ e. The standard deviation should be close to σ / 3 = 1.708/ 3 = .9860. f. Very similar pictures. 7.2 a. P ( Y = 2) = P ( W = 6) = p (4, 1, 1) + p (1, 4, 1) + p (1, 1, 4) + p (3, 2, 1) + p (3, 1, 2) = p (2, 3, 1) + p (2, 1, 3) + p (1, 3, 2)+ p (1, 2, 3) + p (2, 2, 2) = 216 10 . b. Answers vary, but the relative frequency should be fairly close. c. The relative frequency should be even closer than what was observed in part b. 7.3 a. The histogram should be similar in shape, but this histogram has a smaller spread. b. Answers vary. c. The normal curve should approximate the histogram fairly well. 7.4 a. The histogram has a right–skewed shape. It appears to follow p ( y ) = y /21, y = 1, …, 6. b. From the Stat Report window, μ = 2.667, σ = 1.491. c. Answers vary. d. i. It has a right–skewed shape. ii. The mean is larger, but the std. dev. is smaller. e. i. sample mean = 2.667, sample std. dev = 1.491/ 12 = .4304. ii. The histogram is closely mound shaped. iii. Very close indeed. 7.5 a. Answers vary. b. Answers vary, but the means are probably not equal. c. The sample mean values cluster around the population mean. d. The theoretical standard deviation for the sample mean is 6.03/ 5 = 2.6967. e. The histogram has a mound shape. f. Yes. 7.6 The larger the sample size, the smaller the spread of the histogram. The normal curves approximate the histograms equally well. 7.7 a. b. Answers vary. c. The mean should be close to the population variance d. The sampling distribution is not mound–shaped for this case. e. The theoretical density should fit well. f. Yes, because the chi–square density is right–skewed. 7.8 a. σ 2 = (6.03) 2 = 36.3609. b. The two histograms have similar shapes, but the histogram generated from the smaller sample size exhibits a greater spread. The means are similar (and close to the value found in part a). The theoretical density should fit well in both cases. c. The histogram generated with n = 50 exhibits a mound shape. Here, the theoretical density is chi–square with ν = 50 – 1 = 49 degrees of freedom (a large value).
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
144 Chapter 7: Sampling Distributions and the Central Limit Theorem Instructor’s Solutions Manual 7.9 a. P (| Y μ | .3) = P (–1.2 Z 1.2) = .7698. b. P (| Y μ | .3) = P (–.3 n Z .3 n ) = 1 – 2 P ( Z > .3 n ). For n = 25, 36, 69, and 64, the probabilities are (respectively) .8664, .9284, .9642, and .9836. c. The probabilities increase with n , which is intuitive since the variance of Y decreases with n . d. Yes, these results are consistent since the probability was less than .95 for values of n less than 43. 7.10 a. P (| Y μ | .3) = P (–.15 n Z .15 n ) = 1 – 2 P ( Z > .15 n ). For n = 9, the probability is .3472 (a smaller value). b. For n = 25: P (| Y μ | .3) = 1 – 2 P ( Z > .75) = .5468 For n = 36: P (| Y μ | .3) = 1 – 2 P ( Z > .9) = .6318 For n = 49: P (| Y μ | .3) = 1 – 2 P ( Z > 1.05) = .7062 For n = 64: P (| Y μ | .3) = 1 – 2 P ( Z > 1.2) = .7698 c. The probabilities increase with n .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 15

ISM_chapter7 - Chapter 7 Sampling Distributions and the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online