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. 105 80 45 35 105 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 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 answers to two decimal places.) lower limit $ upper limit $

Answer 1

Solution:

x	x2
105	11025
80	6400
45	2025
35	1225
105	11025
65	4225
30	900
23	529
100	10000
110	12100
105	11025
95	9025
105	11025
60	3600
110	12100
120	14400
95	9025
90	8100
60	3600
70	4900
∑x=1608	∑x2=146254

a ) Mean ˉx=∑xn

=105+80+45+35+105+65+30+23+100+110+105+95+105+60+110+120+95+90+60+70/20

=1608/20

=80.4


Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√146254-(1608)220/19

=√146254-129283.2/19

=√16970.8/19

=√893.2

=29.8865

b ) Degrees of freedom = df = n - 1 = 20 - 1 = 19

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t /2,df = t0.05,19 =1.729

Margin of error = E = t/2,df * (s /n)

= 1.729* (29.89 / 20)

= 11.56

Margin of error = 11.56

The 90% confidence interval estimate of the population mean is,

- E < < + E

80.4 - 11.56 < < 80.4 +11.56

68.84 < < 91.96

lower limit =68.84

Upper limit =91.96