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.

90	65	95	70	55	65	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

a) For the given data set the mean is calculated as:

Mean = (90 + 65 + 95 + 70 + 55 + 65 + 30 + 23 + 100 + 110 + 105 + 95 + 105 + 60 + 110 + 120 + 95 + 90 + 60 + 70)/20
= 1613/20
Mean = 80.65

And the sample standard deviation as:

Standard Deviation s = √(1/20 - 1) x ((90 - 80.65)² + ( 65 - 80.65)² + ( 95 - 80.65)² + ( 70 - 80.65)² + ( 55 - 80.65)² + ( 65 - 80.65)² + ( 30 - 80.65)² + ( 23 - 80.65)² + ( 100 - 80.65)² + ( 110 - 80.65)² + ( 105 - 80.65)² + ( 95 - 80.65)² + ( 105 - 80.65)² + ( 60 - 80.65)² + ( 110 - 80.65)² + ( 120 - 80.65)² + ( 95 - 80.65)² + ( 90 - 80.65)² + ( 60 - 80.65)² + ( 70 - 80.65)²)
= √(1/19) x ((9.35)² + (-15.65)² + (14.35)² + (-10.65)² + (-25.65)² + (-15.65)² + (-50.65)² + (-57.65)² + (19.35)² + (29.35)² + (24.35)² + (14.35)² + (24.35)² + (-20.65)² + (29.35)² + (39.35)² + (14.35)² + (9.35)² + (-20.65)² + (-10.65)²)
= √(0.0526) x ((87.4225) + (244.9225) + (205.9225) + (113.4225) + (657.9225) + (244.9225) + (2565.4225) + (3323.5225) + (374.4225) + (861.4225) + (592.9225) + (205.9225) + (592.9225) + (426.4225) + (861.4225) + (1548.4225) + (205.9225) + (87.4225) + (426.4225) + (113.4225))
= √(0.0526) x (13740.55)
= √(722.75293)
= 26.89

b) The confidence interval is calcualted as:

where tc is computed using the excel formula for t-distribution which takes significance level and degree of freedom n-1= 20-1, thus the formula used is =T.INV.2T(0.10, 19), thus the tc computed as 1.729

Now the confidence interval is computed as:

So, the Lower Limit is 70.25

Upper Limit is 91.05.