Download Statistics Homework Solution: Calculating Percentages and Confidence Intervals - Prof. Abd and more Assignments Statistics in PDF only on Docsity! Statistics 100 Solution for Homework #3 *5.15 L/ (a) In the population, p = 176 and o = 30. For y = 186, - y-c~ 186-176 Z = 0 .= 30 z.33. From Table 3, the area below .33 is .6293. For y = 166, 166 - 176 z = y =T = -.33. From Table 3, the area below -.33 is .3707. Thus, the percentage with 166 S y I 186 is .6293 - -3707 = .2586, or 25.86%. (b) We are concerned with the sampling distribution of for n = 9. From Theorem 5.1, the mean of the sampling distribution of is = 1.1 = 176, the -&ndard deviation is I '0 30 "ir =G = g = l O , and the shape of the distribution is normal because the population distribution is normal (part 3a of Theorem 5.1). We need to find the shaded area in the figure. . . For V = 186. . * - Y - % 186-176 z=- - 0-z. - 10 = 1.00. From Table 3, the area below 1-00 is 3413. For ji = 166, - y - % 166-176 z=- - 10 = -1.00. o* A From Table 3, the area below -1.00 is .1587. 166 176 186 J - 1 -00 0 ' 1 .a0 . z Thus, the percentage with 166 I B I 186 is 3413 - .I587 = .6826, or 68.26%. (c) The probability of an event can be interpreted as the long-run relative frequency of occurrence of the event (Section 3.3). Thus, the question in p a (c) is just a rephrasing of the question in part (b). It follows from part @) that 5.18 p = 145; o = 22. d 155 - 145 (c) R{135 I P I 155) = .9312 (from patt (b)). (a) z = 22 = .45; Table 3 gives .6736. (d) n = 36; Oy = 2 ~ m = 3.67. 135 - 145 z = 7 = -.45; Table 3 gives .3264. 155 - 145 nm. ,6736 - .3264 = .3472 or 34.72% of the plants. = 3-67 = 2-72; Table 3 gives .9967. - - -. - 135 - 145 Z = (b) n = 16; a - = 2 2 1 m = 5.5. 3-67 = -2.72;.Table 3 gives .0033. Y 155 - 145 Thus, .9967 - .0033 = .9934 or 99.34% of the groups. z = 5.5 = 1.82; Table 3 gives .9656. 135 - 145 z = 5.5 = -1.82; Table 3 gives .0344. Thus, .9656 - .0344 = .9312 or 93.12% of the groups. e6.20 = 1.20; s = .14; n = 50. The degrees of freedom are 50 - 1 = 49. From Table 4 with df = 50 (the df'value closest to 49) we find that 1.05 = 1.676. The 90% confidence interval for p is a34 (a) No. Ihe obrerved fiaquency djsttibution t highly skewed, which s u m that lb popularjan dist+bmman islocwad a (b) Even if the population distri ion ie not normal, Student's t method j6 appmxhately vaIid if tk sample size i a large. In thi8 cese, the sample size ia n = 242, which ir qdte I-. *&40 (a) pl n (69+2)/(339+4) = .207. The standard e r m ie Thc959bamMenxinterval is g * 1.96SEp 207 (1.96)(.022) 207 * -043 (.164,.250) or .I64 < p < -250. (b) We arc 95% d e n t that the probability of adverse reaction in infanta who receive thcir first injdm of vaccine is between ,164 and ,250. 6.46 4 ~ ( 1 - ~ ) is largest when = .5. Thus. the required n must satisfy the inequality 4 9 5 .Ol. 4 c 3 It follows that -jj!/- S M (.5)(.5) .012 < n+4 or WOO I n+4, so n 2 2496. . 95% confidence interval: -073 t (1 .%~.0069) m (.059,.087) a .0.59 p < ,087, 6.58 (a) 5.1679 2 (2.052)(.1237) (4.91,5.42) or 4.91 < p < 5.42 kg. (b) 5.1679 & (2.771)(. 1237) (4.83,551) or 4.83 < JA < 5.51 kg. (c) W e art 95% confidtnt that the avmgc birthweight uf all lambs that wtn bonn in A@, w- all the sarne tneed, and weae all single births, is between 4.91 kg and 5.43 kg.
