Question #7f13f

Apr 23, 2017

The population mean may be 66.

Explanation:

Our initial hypothesis is
${H}_{0} : \mu = 66$
Since we do not claim to have knowledge either way, our alternative hypothesis is
${H}_{1} : \mu \ne 66$
$n = 10$ is the sample size.

Since we do not know the population standard deviation, although the population is normal, we must use a t-test with 9 = 10-1 degrees of freedom.
The sample mean is $\overline{x} = 67.8$ .
The sample standard deviation is $s \approx 3.011$ .
Our test statistic is $t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$.
In this case t is approximately $1.8904$ .

If we consult a t-table, we see that the value of 67.8 deviates enough from 66 for us to reject the null hypothesis if our tolerance is $\alpha = 0.1$, but that is a relatively weak significance level. For the more commonly-used tolerance specification of $\alpha = 0.05$, we would be unable to reject the null hypothesis and would have to conclude that $\mu$ could be equal to 66.

If you are doing this work on a TI-83/84 calculator, the T-Test will reveal that the area of the rejection region is approximately $P \approx 0.0913$.