Question

Please help solve problem #14. Thank you!

383 Section 7.2 Estimating u When a Is Unknown In Problems 13-19, assume that the population of x values has an approximately normal distribution 13. | Archaeology: Tree Rings At Burnt Mesa Pueblo, the method of tree-ring dating gave the following years A.D. for an archaeological excavation site (Bandelier Archaeological Excavation Project: Summer 1990 Excavations at Burnt Mesa Pueblo, edited by Kohler, Washington State University): 1275 1317 1275 1316 1268 1271 1267 1272 1189 (a) Use a calculator with mean and standard deviation keys to verify that the sample mean year is x = 1272, with sample standard deviation S37 years. (b) Find a 90% confidence interval for the mean of all tree-ring dates from this archaeological site (c) Interpretation What does the confidence interval mean in the context of this problem? Camping: Cost of a Sleeping Bag How much does a sleeping bag cost? Let's say you want a sleeping bag that should keep you warm in tempera tures from 20°F to 45°F. A random sample of prices ($) for sleeping bags in this temperature range was taken from Backpacker Magazine: Gear Guide (Vo. 25, Issue 157, No. 2). Brand names include American Camper, Cabela's, Camp 7, Caribou, Cascade, and Coleman 14 110 100 30 23 100 105 120 75 37 90 80 60 70 120 95 90 110 60 105 95 (a) Use a calculator with mean and sample standard deviation keys to verify that x$83.75 and s $28.97 (b) Using the given data as representative of the population of prices of all summer sleeping bags, find a 90% confidence interval for the mean price u of all summer sleeping bags (c) Interpretation What does the confidence interval mean in the context of this problem?

Answer 1

@ Mean ( 20 Ση solution - 14 meam (X) = n (ni's are 3 observation) [80 +105+90 +95+ ... + 110+70) = 1675 = 83.75 standard deviation - - sminar S = AX 80-83.75)'#(105-8375) 7 - - 4(70-85.7554 7 28.97 the formular for confidence intercial form is ñ I tse wherer; ñ = sample mean tat statistic determined by confidence Sea standand enror a tend. Here, ñ=83.75 t=1.73 [fron't table I se = $28.97/20 = 6.48 :.M= 83.75 + 1.73x648 = 83.75 + 11.2011 :. confidence interval = [ 72.5489, 94.95115 © Interpretation! - you can be qoy. confident that the population mean em falls between 72.5489 and 94.9511