# T-Intervals and % Confidence Interval

The question is the following:

" A random sample of six 2009 sports cars is taken and their "in the city" miles per gallon is recorded. The results are as follows: 23 19 24 17 16 22. Assuming the population distribution is normal, calculate the 99% confidence interval for μ, the population mean "in city" mpg for 2009 sports cars.

Any help?!

I am so stuck :(

1. Calculate the sample mean $\bar x$, which is the sum of the data divided by the sample size $n = 6$.
2. Calculate the sample standard deviation $$s_x = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar x)^2}.$$
3. Calculate the standard error $s_x/\sqrt{n}$.
4. Find the critical value of the Student's $t$ distribution with $n-1 = 5$ degrees of freedom corresponding to a 99% confidence interval--this is the 99.5th percentile.
5. The confidence interval margin of error is the critical value $t_{n-1,\alpha/2}$ that you found in step 4 multiplied by the standard error you found in step 3.
6. The desired confidence interval for the population mean $\mu$ is $[\bar x - ME, \bar x + ME]$, where $ME$ is the margin of error you found in step 5.