Question

Question #7f13f

Answer 1

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 `barx = 67.8` .
The sample standard deviation is `s ~~ 3.011` .
Our test statistic is `t = (barx - mu)/(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 ~~ 0.0913`.