Question #7f13f

1 Answer
Apr 23, 2017

The population mean may be 66.

Explanation:

Our initial hypothesis is
H_0: mu = 66H0:μ=66
Since we do not claim to have knowledge either way, our alternative hypothesis is
H_1: mu ne 66H1:μ66
n = 10n=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¯x=67.8 .
The sample standard deviation is s ~~ 3.011s3.011 .
Our test statistic is t = (barx - mu)/(s/sqrt(n))t=¯xμsn.
In this case t is approximately 1.89041.8904 .

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 = 0.1α=0.1, but that is a relatively weak significance level. For the more commonly-used tolerance specification of alpha = 0.05α=0.05, we would be unable to reject the null hypothesis and would have to conclude that muμ 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 ~~ 0.0913P0.0913.