Question

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.

45	65	105	110	120	55	30	23	100	110
105	95	105	60	110	120	95	90	60	70

(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 two decimal places.)

x =	$
s =	$

(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 μ of all summer sleeping bags. (Round your answers to two decimal places.)

lower limit	$
upper limit	$

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	45	1493.8225
	65	347.8225
	105	455.8225
	110	694.3225
	120	1321.3225
	55	820.8225
	30	2878.3225
	23	3678.4225
	100	267.3225
	110	694.3225
	105	455.8225
	95	128.8225
	105.0	455.8225
	60	559.3225
	110	694.3225
	120	1321.3225
	95	128.8225
	90	40.3225
	60	559.3225
	70	186.3225
Total	1673	17182.55

Part a)

Mean X̅ = Σ Xi / n
X̅ = 1673 / 20 = 83.65
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 17182.55 / 20 -1 ) = 30.07

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 20- 1 ) = 1.729
83.65 ± t(0.1/2, 20 -1) * 30.0723/√(20)
Lower Limit = 83.65 - t(0.1/2, 20 -1) 30.0723/√(20)
Lower Limit = 72.02
Upper Limit = 83.65 + t(0.1/2, 20 -1) 30.0723/√(20)
Upper Limit = 95.28
90% Confidence interval is ( 72.02 , 95.28 )