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 degrees F to 45 degrees . A random sample of prices for sleeping bags in this temperature range was taken from Backpacker Magazine: Gear Guide (Vol.25 Issue 157, No 2) Brand names include American Camper, Cabela's Camp 7, Caribou, Cascade, and Coleman 80,90,100,120,75,37,30,23,100,110 105,95,105,60,110,120,95,90,60,70 A) Use a calculator with mean and sample standard deviation keys to verify that sample mean is around $83.75 and S is around $28.97 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. C) What does the confidence interval mean in the context of this problem?

Answer 1

Answer:

a)$\bar X =\frac{\sum_{i=1}^n x_i}{n}=\frac{1675}{20}=83.75$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=28.97$

b) The 90% confidence interval would be given by (72.551;94.949)    

c)We are 90% confident that the true mean temperature for the sleeping bags it's between 72.551 and 94.949

Step-by-step explanation:

Data set given

80,90,100,120,75,37,30,23,100,110 105,95,105,60,110,120,95,90,60,70

Part a

We can calculate the sample mean and the sample deviation with the following formulas:

$\bar X =\frac{\sum_{i=1}^n x_i}{n}=\frac{1675}{20}=83.75$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=28.97$

Part b

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=20-1=19$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,19)".And we see that $t_{\alpha/2}=1.73$

Now we have everything in order to replace into formula (1):

$83.75-1.73\frac{28.97}{\sqrt{20}}=72.551$    

$83.75+1.73\frac{28.97}{\sqrt{20}}=94.949$

So on this case the 90% confidence interval would be given by (72.551;94.949)    

Part c

We are 90% confident that the true mean temperature for the sleeping bags it's between 72.551 and 94.949