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.
40 | 105 | 70 | 70 | 70 | 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 | $ |
a)
x = 79.15
s = 28.46
b)
sample mean, xbar = 79.15
sample standard deviation, s = 28.46
sample size, n = 20
degrees of freedom, df = n - 1 = 19
Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.729
ME = tc * s/sqrt(n)
ME = 1.729 * 28.46/sqrt(20)
ME = 11.003
CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (79.15 - 1.729 * 28.46/sqrt(20) , 79.15 + 1.729 *
28.46/sqrt(20))
CI = (68.15 , 90.15)
lower limit = 68.15
upper limit = 90.15
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