Question

In: Statistics and Probability

How much does a sleeping bag cost? Lets say you want a sleeping bag that should...

How much does a sleeping bag cost? Lets say you want a sleeping bag that should keep you warm in temperatures from 20 degrees farenheight to 45 degrees farenheight. 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.
35 90 105 45 120 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.
x=
s=

(b) Using the given data as representative of the popularion of prices of all summer sleeping bags, find a 90% confidence interval for the mean price of all summer sleeping bags

lower limit=
upper limit=

Solutions

Expert Solution

Solution:

x x2
35 1225
90 8100
105 11025
45 2025
120 14400
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=1633 x2=151329

The sample mean is

Mean   = (x / n) )
=35+90+105+45+120+65+30+23+100+110+105+95+105+60+110+120+95+90+60+7020

=163320

=81.65

Mean   = 81.65

The sample standard is S

  S =( x2 ) - (( x)2 / n ) n -

=√(151329-(1633)220) 19

=√(151329-133334.45) 19

=√17994.55 /19

=√947.0816

=30.7747

sample Standard deviation S= 30.77

n = 20

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* (30.77 / 20)

= 11.90

Margin of error = 11.90

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

- E < < + E

81.65 - 11.90 < < 81.65 + 11.90

69.75 < < 93.55

lower limit= 69.75

upper limit= 93.55


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