Finding Confidence Intervals for Mean, Variance, and Standard Deviation

I really need some help with this problem ASAP. Here is the problem word-for-word:

Find the 90% confidence interval for the mean, variance, and standard deviation for the price in dollars of an adult single-day lift ticket. The data represent a selected sample of nationwide ski resorts. Assume the variable is normally distributed.

$59 $54 $53 $52 $51

$39 $49 $46 $49 $48

Thanks in advance for your help!

Re: Finding Confidence Intervals for Mean, Variance, and Standard Deviation

Hi there, let's do these one at a time so you fully understand. (Happy)

Your data is normal, but you don't know the population standard deviation

Therefore the confidence interval for the mean is $\displaystyle \displaystyle \bar{x}\pm t_{1-\frac{\alpha}{2}, n-1}\times\frac{s}{\sqrt{n}}$

Where

$\displaystyle n$ is the number of samples

$\displaystyle \bar{x}$ is the sample mean

$\displaystyle s$ is the sample standard deviation

$\displaystyle t$ is the test statistic, and

$\displaystyle \alpha$ is the level of significance

What do you get?