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. 50 80 95 70 100 40 30 23 100 110 105 95 105 60 110 120 95 90 60 70a) 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

Answer:

a) x = 80.40

s = 29.54

b) lower limit = 69.53

upper limit = 91.27

Step-by-step explanation:

The mean price is:

$\bar X=\frac{1}{n}\sum x_i=\frac{1}{20}(50+80+...+70)=\frac{1608}{20} =80.40$

The standard deviation is

$s=\sqrt{\frac{1}{n-1}\sum(x_i-\bar X)^2}=\sqrt{\frac{1}{19}[(50-80.4)^2+...+(70-80.4)^2]} \\\\s=\sqrt{\frac{15,471}{19} }=\sqrt{814.25} =29.54$

We have to calculate a 90% CI for the mean.

The z-value for a 90% confidence interval is z=1.645.

The margin of error is:

$E=z\cdot \sigma / \sqrt{n}=1.645*29.54/\sqrt{20}=48.59/4.47=10.87$

The 90% CI is

$\bar X-z\sigma/\sqrt{n} \leq \mu \leq \bar X-z\sigma/\sqrt{n}\\\\80.40-10.87 \leq \mu \leq 80.40+10.87\\\\69.53\leq \mu \leq 91.27$