Statistics Homework Solutions 120

Statistics Homework Solutions 120 - 120 CHAPTER 5 (c) Let X...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 120 CHAPTER 5 (c) Let X be the number of defective components in a lot of 200. ¥ ¤ ¡ ¡ ¡ 20 5 . ¡ ¥¨ ¤ ¡ To find a 95% confidence interval for r, express the z-score of P X upper and lower confidence limits for p. ¡ ¤ p . Now find a 95% confidence interval for z by substituting ¥ £¥ The z-score of 20.5 is 20 5 200 p 200 p 1 the upper and lower confidence limits for p. 20 5 as a function of p and substitute the ¥¨ PX ¤ Using the continuity correction, r  Let r represent the proportion of lots that are returned. Bin 200 p , so X is approxi¥ Let p be the population proportion of components that are defective. Then X mately normally distributed with mean µX 200 p and σX 200 p 1 p . ¤ ¡ ¨ ¡ ¤ From part (a), the 95% confidence interval for p is (0.052873, 0.10554). Extra precision is used for this confidence interval to get good precision in the final answer. 3 14. ¡ z ¨ ¡ 3 14 is 1 ¡ 0 14 ¨ ¡ ¨ 0 5557. The area to the right of z ¨ 0 10554, we are 95% confident that 0 9992 ¨ ¨ p ¡ ¡ ¡ ¡ ¨ 0 4443 ¡ ¨ ¨ 0 14 is 1 ¡ ¨ ¡ ¡ ¡ The area to the right of z 0 0008. 0 14. ¡ Since we are 95% confident that 0 052873 ¡ 3 14. Substituting 0.10554 for p yields z ¨ Substituting 0.052873 for p yields z ¨ ¡ r 0 5557. ¨ ¡ Therefore we are 95% confident that 0 0008 ¨ The confidence interval is (0.0008, 0.5557). 4 , or (0.490, 0.727). 1 96. ¨ 0 93478, z 025 0 93478 ¨ ¤ ¨ ¨ ¨ ¡¥ ¡ ¡ ¨ ¢ ¤ ©¥ £ ¢ ¤ ¡ ¡ ¡ ¨ Since the upper limit is greater than 1, replace it with 1. The confidence interval is (0.863, 1). 2 58. ¤ ¥ £ 0 73913 42 4 , or (0.572, 0.906). ¢ ¢ ¨ 2 58 0 73913 1 ¤ ¨ ¤ ©¥ £ ¢ ¤ ¡ ¡ ¡ The confidence interval is 0 73913 0 73913, z 005 ¥ 4 ¡¥ 42 ¨ 2 32 ¡ 42, p ˜ ¡ 32, n ¨ (c) X ¨ ¨ 13. (a) Let n be the required sample size. p ˜ ¤ ¥ £ ¡ p1 ˜ n 4. ¢ ¤ 1 96 ¥ ¨ ¡ Then n satisfies the equation 0 05 ¨ Since there is no preliminary estimate of p available, replace p with 0.5. ˜ ˜ 0 21154, z 025 ¨ ¡ ¡¥ 4 ¢ 1 96 0 21154 1 1 96. 0 21154 ¨ ¤ ¨ ¨ 100 ¡ ¤ ©¥ £ 2 ¨ ¢ 20 ¤ ¥ £ ¤ ¡ 100, p ˜ 100 ¢ ¡ 20, n ¡ The confidence interval is 0 21154 4 , or (0.133, 0.290). ¥ (b) X 381. ¡ Solving for n yields n ¤ ¥ £ 0 93478 1 42 ¢ 1 96 4, ¥ ¡ ¡ The expression for the confidence interval yields 0 93478 or (0.863, 1.006). ¨ 4 42 ¡ 42 ¨ 2 ¤ ¤ 41 ¨ ¢ 42, p ˜ ¤ ¥ £ ¤ ¥ £ 41, n 0 60870 ¢ 1 645 0 60870 1 (b) X 1 645. ¥ ¢ The confidence interval is 0 60870 0 60870, z 05 ¡¥ 4 ¨ 42 2 ¡ 26 ¡ 42, p ˜ ¨ 26, n 11. (a) X ¨ ...
View Full Document

Ask a homework question - tutors are online