Question

Question Help 0 Arandom sample of 20 purchases showed the amounts in the table in $). The mean is $47.86 and the standard deviation is $19.25. a) Construct a 80% confidence interval for the mean purchases of all customers, assuming that the assumptions and conditions for the confidence interval have been met. b) How large is the margin of error? c) How would the confidence interval change if you had assumed that the standard deviation was known to be $20? 80.22 81.81 34.62 41.25 23,89 35,02 41.47 64.35 55.88 61.67 45.04 55.84 41.82 46.32 38.90 42.05 9.91 79.70 23.44 53.92 a) What is the confidence interval? (Round to two decimal places as needed.) b) What is the margin of error? The margin of error is (Round to two decimal places as needed.)

Answer 1

n = 20 = 47.86 5 - 19.25 la) 80 % C. I for mean The critical value ofor a = 0.2 is zo = 21-212 = 1.282 C. I = (7-2ct / Y + 2(x 0 ) 47.86 +1.282x 19.25 V20 19.25 20 - (47.86 -1.282x = 142.344 53.376) . (42.34, 53.38) (b) margin of error = Critical value x 8.0.10) population 8.0 is known otherwise 8.0 l I samples Sm margin of error = Critical value x = 1.282x 19.25 120 =S.5182 An (C) 20 is known to be standard deviation 94 othen C.I = (7-2c. In x+26 m) 20 (x - 1.282 x 20 - 1.282 São J2O 20 20 = (47.86-1.282 x 47.86 - 1.282 x $20 520) = 1 42.129, 53.591) 1 42.13, 53.59) Hence A option is correct. l.e. que new confidence interval (42.13,53.59) is wider than the interval from part a (42.34, 53.381