Question

manufacturer samples 11 bags of pretzels off the assembly line and weighs their contents. If the sample mean is 15.2 oz. and the sample 0.50 oz., find the 95% confidence interval of the true mea 95% confidence interval of the true mean. X=15.2,0-2,5o, noll 2 A) 15.0<u< 15.4 B) 12.9<x< 17.5 1 74.4 Luz 14.9<u< 15.5 D) 14.1 Su< 16.3 C 15.2 -1. 90 <us 15.a +auG ) > 15.2 -0.30-14.94 14) The winning team's score in 13 high school basketball games was recorded. If the sample mean is 54.3 points and the sample standard deviation is 13.0 points, find the 98% confidence interval of the true mean. A) 44.6<< 64.0 C) 45.9<x<62.7 B) 35.0<x< 73.6 D) 50.7<u<57.9 15) In a study of 100 new cars, 29 are white. Find p and, where p is the proportion of new cars that are white. n = 100 X=29 A) p = 0.29, q=0.29 § = B ) P = 0.71, 9-0.71 a = C) P = 0.71, q=0.29 29 0 = 0.29, ĝ=0.71 16) Find the point estimate for the given values of x and n. x= 122, n=201 A) 0.607 B) 0.393 C) 122 D) 0.034 17) Use the given data to construct a confidence interval of the requested level. -0.01 x=98, n = 223, confidence level 99% 2:.095 A) 0.219 <p<0.660 X 98 Ada B) 0.406 <p <0.473 0.439-2. Pon 223011700.354<p < 0.525 05-0.450) 0.362 <p<0.517 - 2.57 9 = -0.439 = 0.5l 0.434 -0,0 18) A recent study of 750 internet users in Europe found that 35% of internet users were women. What is the 95% confidence interval of the true proportion of women in E who use the internet? A) 0.309<p <0.391 B) 0.316<p <0.384 C) 0.321 <p <0.379 D) 0.349 <p<0.351 CO 19) A random sample of 50 voters found that 46% were going to vote for a certain car Find the 95% limit for the population pronortion of yotare who will

please help with number 14

Answer 1

given n=11 Somple mean = X Sample SD = Find 95%. C. The 95 -. = 15.2 = 0.50 For tove C.I is, mean. (x - tconan & L41+ tcm-2412 5 ) =) ( 7 I trnoodles here, t 10,0.05 = to.os, lo = 2.228 = ( 15:2 + 2.228. (0:50) ) ( 15.2 to 3359) (15.2- 0.3359 Cu 15.2 + 0.3359) C 14.864 < s < 15.586) - (14.9 L M < 15.5 ) Ans - @ 14.9 CM (15.5 given that - n = 13 7 = 54.3 points s = 13 points. a 981. confidence foo t oue mean.. find interval

The 98 %. C'I is, - ( 4 & tcm-es Here to.o2 (13-1) t0.02,12 = - - = -2.681 J13 - ( 54.3 + 2.681 (18) ) (5431 965648) =) ( 54:3-9.6665 CM (54.3 + 9.6665) ( 44.6335 cu < 63.9665) - ( 44.6 cuc 69.96)... =) ( 44.6 24 < 64.0) Ang 44.6 <i < 64.0 | Hest, given n = 100 x = 29 and P & = 29 100

Fr.: e = 0.29 q=t- = 1-0.29 q=0.41 Ans - p = 0.29, § = 0.71 Hose; x = 122 n:20) the point Find estimate. → P = x = 122 201 P = 0.6069 point estimate = 0.607 3 Ans - ③ 0.607 Hese, given x=98, 99%. that n. 223 C. I. find = 98 223

(4) = 0.4395 Ć 1-P - 0.4395 = 0.5605 The 991. (.I is, 0.4395 + 2.58 0.4395 (0:5605) 223 ZO.005 - 2.58 1 = ( 0.4395 + 0.0858) a = (0.4395 -0-0858 <p (0:4395 +0.0858) =) ( 0.3537 cp ( 0.5253) (0:354 <p < 0:525) Ans 0.354 (p <0.525

↑ Hose, given that = 750 P = 35 . =0.35 1- P = 0.65 find: 95 % CI + 2x12 ] P (168) - 750 = ( 0.35 1.96 0:35 (065) - (0.35 + 0.03414) -) ( 0.85 - 0.0341 < p < 0:35 + 0.0341) 3) ( 0-3159 <p < 0.3841) 5) CO. BIG <p c 0:384) Ans - ® [ 0.316 (p <0.384] Heon, n=50 P = 461 = 0.46 1-P = -0.46 = 0.54 The 951. C.I is . 0.46 + 1.96 0.46x0.54 50

=) ( 0.46 0.188148) → (0.46 + 0.138 ) =) ( 0:46 - 0:138 cp ( 0:46+0:139) =) (0-322 (p <0.698) Ans - [ 0.322 (p <0.598]