How much does a sleeping bag cost? Let's say you want a sleeping bag that should keep you warm in temperatures from 20°F to 45°F. A random sample of prices ($) forsleeping bags in this temperature range is given below. Assume that the population of x values has an approximately normal distribution.115 60 35 100 120 60 30 23 100 110105 95 105 60 110 120 95 90 60 70A USE SALT(a) Use a calculator with mean and sample standard deviation keys to find the sample mean price x and sample standard deviation s. (Round your answers to four decimalplaces.)(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 w of all summer sieepingbags. (Round your answers to two decimal places.)lower limitupper limit

Given,sample size(n)=20degrees of freedom(df)=n-1=20-1=19α=1-0.90=0.10α2=0.05t0.05(19)=1.729

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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How much does a sleeping bag cost? Let's say you want a sleeping bag that should keep you warm in temperatures from 20°F to 45°F. A random sample of prices ($) for
sleeping bags in this temperature range is given below. Assume that the population of x values has an approximately normal distribution.
115 60 35 100 120 60 30 23 100 110
105 95 105 60 110 120 95 90 60 70
A USE SALT
(a) Use a calculator with mean and sample standard deviation keys to find the sample mean price x and sample standard deviation s. (Round your answers to four decimal
places.)
(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 w of all summer sieeping
bags. (Round your answers to two decimal places.)
lower limit
upper limit

Expert Solution

Step 1

$G i v e n, s a m p l e s i z e (n) = 20 d e g r e e s o f f r e e d o m (d f) = n - 1 = 20 - 1 = 19 α = 1 - 0.90 = 0.10 \frac{α}{2} = 0.05 t_{0.05} (19) = 1.729$

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