Q: A tire manufacturer wishes to compare the tread wear of tires made of a new material with that of conventional material. 10 Cars are driven 40,000 miles as the sample set. the following data is obtained:
(To avoid some confusion and messy tables I've done some of these computations myself)
$\mu_{conventional}$ =4.11
$\mu_{new}$ = 4.814
$s$ = .6699 (To clarify this is the sample standard deviation of the "new" material dataset)
The question then is to test at $\alpha$=0.05 that the true mean of the new material exceeds that of the old material.
This is a one-sided t-test. (Right?)
So I've set it up the following way:
$H_0 : \mu = 4.11$
$H_1 : \mu$ > 4.814
$ t = \frac{\bar{x}-\mu_0}{s/\sqrt{n}} = \frac{4.814-4.11}{.6699/\sqrt{10}}$ = 3.3233
Now to look for a .05 confidence using the t-test I obtained the value 1.833 from the t-distribution table. Since the value from our test statistic is 3.3233 > 1.833, I'd reject the Null hypothesis. Can anyone check my work to verify this?