Question #7f13f

1 Answer
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 #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#.