27.8 Hypothesis testing for one mean: A summary
Let’s recap the decision-making process seen earlier, in this context about body temperatures:
- Step 1: Assumption: Write the null hypothesis about the parameter (based on the RQ): \(H_0\): \(\mu=37.0^\circ\text{C}\). In addition, write the alternative hypothesis \(H_1\): \(\mu\ne 37.0^\circ\text{C}\). (This alternative hypothesis is two-tailed.)
- Step 2: Expectation: The sampling distribution describes what to expect from the sample statistic if the null hypothesis is true: under certain circumstances, the sample means will vary with an approximate normal distribution around a mean of \(\mu=37.0^\circ\text{C}\) with a standard deviation of \(\text{s.e.}(\bar{x}) = 0.03572\) (Fig. 27.3).
- Step 3: Observation: Compute the \(t\)-score: \(t=-5.45\). The \(t\)-score can be computed by software, or using the general equation (27.1).
- Step 4: Consistency?: Determine if the data are consistent with the assumption, by computing the \(P\)-value. Here, the \(P\)-value is much smaller than 0.001. The \(P\)-value can be computed by software, or approximated using the 68–95–99.7 rule.
The conclusion is that there is very strong evidence that \(\mu\) is not \(37.0^\circ\text{C}\):
Example 27.1 (Mean driving speeds) A study of driving speeds in Malaysia (Azwari and Hamsa 2021) recorded the speeds of vehicles on various roads.
One RQ of interest was whether the mean speed of cars on one road was the posted speed limit of 90 km.h-1, or whether it was higher. The parameter of interest is \(\mu\), the mean speed in the population.
The hypotheses are:
- \(H_0\): \(\mu = 90\); and
- \(H_1\): \(\mu > 90\) (since the researchers were interested in whether the mean speed was higher than the posted speed limit).
The researchers recorded the speed of \(n = 400\) vehicles on this road, and found \(\bar{x} = 96.56\), but this value is likely to vary from sample to sample. The sample standard deviation was \(s = 13.874\), so that
\[ \text{s.e.}(\bar{x}) = \frac{s}{\sqrt{n}} = \frac{13.874}{\sqrt{400}} = 0.6937. \] Hence, the test statistic is
\[ t = \frac{\bar{x} - \mu}{\text{s.e.}(\bar{x})} = \frac{96.56 - 90}{0.6937} = 9.46, \] where (as usual) the value of \(\mu\) is taken from the null hypothesis (which we always assume to be true).
This is a huge value, suggesting that the (one-tailed) \(P\)-value is very small.
We write (remembering the alternative hypothesis is one-tailed):
Of course, this statement refers to the mean speed; there may be individual vehicles travelling below the speed limit.There is very strong evidence (\(t = 9.46\); one-tailed \(P<0.001\)) that the mean speed of vehicles on this road (sample mean: 96.56; standard deviation: 13.874) is greater than 90 km.h-1.