When \(\sigma^2\) is unknown, the pivot for \(\mu\),
\[ \frac{\overline{X} - \mu}{\diagfrac{S}{\sqrt{n}}} \;\;\sim\;\; \TDistn(n-1 \text{ df}) \]can be used to obtain a 95% confidence interval for \(\mu\). Writing the 2½th and 97½th percentiles of the \(\TDistn(n-1\text{ df})\) as \(-t_{n-1,\;0.975}\) and \(t_{n-1,\;0.975}\),
\[ P\left(-t_{n-1,\;0.975} \;\;\lt\;\; \frac{\overline{X} - \mu}{\diagfrac{S}{\sqrt{n}}} \;\;\lt\;\; t_{n-1,\;0.975} \right) \;\;=\;\;0.95 \]This leads to the following 95% confidence interval for \(\mu\)
\[ \overline{x} -t_{n-1,\;0.975} \frac{s}{\sqrt{n}} \;\;\lt\;\; \mu \;\;\lt\;\; \overline{x} +t_{n-1,\;0.975} \frac{s}{\sqrt{n}} \]In general, a \((1-\alpha)\) confidence interval is
\[ \overline{x} -t_{n-1,\;1-\diagfrac{\alpha}{2}} \frac{s}{\sqrt{n}} \;\;\lt\;\; \mu \;\;\lt\;\; \overline{x} +t_{n-1,\;1-\diagfrac{\alpha}{2}} \frac{s}{\sqrt{n}} \]Question
An experiment was conducted about the grazing behaviour of dairy cows. This table gives the grass intake rate (in grams dry mass per bite) in the 48 plots of grass used.
1.09 | 1.41 | 1.20 | 1.04 | 1.07 | 1.39 | 1.06 | 1.14 |
0.88 | 0.92 | 1.07 | 1.07 | 1.18 | 0.57 | 0.01 | 0.31 |
1.14 | 1.18 | 0.58 | 0.74 | 0.14 | 0.48 | 0.91 | 0.37 |
2.19 | 1.17 | 2.34 | 1.69 | 1.97 | 1.04 | 1.76 | 1.26 |
1.62 | 0.81 | 1.81 | 2.06 | 2.27 | 1.24 | 0.02 | 1.46 |
2.29 | 2.28 | 1.40 | 0.60 | 1.41 | 0.49 | 1.06 | 1.58 |
Assuming that the data come from a \(\NormalDistn(\mu,\;\sigma^2)\) distribution, find a 95% confidence interval for the mean grass intake per bite, \(\mu\).
(Solved in full version)
Robustness
The above confidence interval was based on the assumption that the data were a random sample from a \(\NormalDistn(\mu\;, \sigma^2)\) distribution, but we are rarely certain about the shape of the underlying distribution in practical problems. However
Provided the shape of the underlying distribution is not far from normal, a confidence interval based on the t-distribution has approximately the correct confidence level.