# elementary t-test question

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?

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$H_1 : \mu > 4.11.$ Have you check to see that the criteria to use a t-test is met? –  Chris K. Caldwell May 7 '12 at 21:21
yes @ChrisK.Caldwell the criteria to use t-test is met. i.e. a population standard of deviation is unknown but can be approximated using the sample s.d. –  franklin May 7 '12 at 21:31
@ChrisK.Caldwell what criteria do you mean? –  Ronald May 7 '12 at 23:56
@franklin your working seems to be correct, apart from the alternative hypothesis already pointed out. –  Ronald May 7 '12 at 23:56
The t-test requires that the sampling distribution of the scaled value be (approximately) normal. One way to satisfy this is to have a sample of at least 30. That condition was not met, so unless you know something about the underlying distributions normality, the t-test is inappropriate. –  Chris K. Caldwell May 8 '12 at 22:22

In this case, a sample s.d. is used rather than the population deviation $\sigma$, also the sample size is small $n$ < 30. In this case a t-distribution would seem like a sound substitute for a standard normal curve.
Indeed I was, as that page states: The assumptions underlying a t-test are that Z [$T=Z/s$] follows a standard normal distribution under the null hypothesis... However, it is common for text books to ignore this in the exercises (even after warning about it in the text of the section). So your answer may be exactly what is required. –  Chris K. Caldwell May 8 '12 at 22:26