Given that examination of the data does not find these problems, the test statistic
z = (x-bar - mu) / (sigma / sqrt(n))
has a standard normal distribution when the null hypothesis is true. Unfortunately, we do not know the value of sigma. We can substitute in the sample standard deviation s. This adds variability to the denominator as well as the numerator, and the resulting sampling distribution is no longer standard normal. It has a t distribution with n-1 degrees of freedom. These distributions are symmetric, bell-shaped, and centered at 0 as the standard normal distribution is, but are more spread out. the extra spread decreases as the sample size increases.
The test statistic we use is
t = (x-bar - mu) / (s / sqrt(n))
For our data, t = (132 - 120) / (20/sqrt(16)) = 2.40.
In the example, the result is not significant at the 1% level. A user of fixed significance levels would not reject the null hypothesis with this data. This does not mean, however, that there is strong evidence that the null hypothesis is actually true. It only means that the data was insufficient to rule out chance alone as the reason for a difference between the observed sample mean and the assumed population mean. You cannot prove the null hypothesis is true. You can say with high confidence that the true population mean is in a small interval containing an assumed value if you have sufficient data.
Bret Larget, email@example.com