Mathematics 375 – Probability and Statistics 2 Clarification of Class Notes from February 8

In our first example of using a “pivotal quantity” to derive confidence intervals this day, I wasn’t very clear about exactly how you derive the formulas for the endpoints. This sheet should clarify the algebra involved.

We considered the case of sampling from a N(μ, σ^{2}) population to estimate the target parameter μ (the population mean). From our work in section 7.2 of the text, we know that with n independent samples, the quantity

Q = \frac{\overline{Y} − μ}{s/\sqrt{n}},

(where s = \frac{1}{n−1} ∑_{i=1}^{n}(Y_{i} − \overline{Y})^{2}, the sample standard deviation) satisfies

• Q contains μ as the only unknown, and

• Q has a t(n − 1) distribution.

Hence we can use Q as a pivotal quantity to derive confidence intervals. Here are complete derivations of each of the three cases (being more careful about signs than I was in class):

Two-sided, (1 − α) × 100% confidence interval

Let t_{α/2} be the α/2-percentile point for the t-distribution: that is P (T > t_{α/2}) = α/2. Then by symmetry:

1 − α = P (−t_{α/2}(n − 1) < \frac{\overline{Y} − μ}{s/\sqrt{n}} < t_{α/2}(n − 1))

= P (−t_{α/2}(n − 1)\frac{s}{\sqrt{n}} < \overline{Y} − μ < t_{α/2}(n − 1)\frac{s}{\sqrt{n}})

= P (−\overline{Y} − t_{α/2}(n − 1)\frac{s}{\sqrt{n}} < −μ < −\overline{Y} + t_{α/2}(n − 1)\frac{s}{\sqrt{n}})

= P (\overline{Y} + t_{α/2}(n − 1)\frac{s}{\sqrt{n}} > μ > \overline{Y} − t_{α/2}(n − 1)\frac{s}{\sqrt{n}})

= P (\overline{Y} − t_{α/2}(n − 1)\frac{s}{\sqrt{n}} < μ < \overline{Y} + t_{α/2}(n − 1)\frac{s}{\sqrt{n}})

(Note: at the next-to-last step, we multiply through by −1, which reverses all inequalities, then we switch back to write the terms increasing left to right as usual).

(turn over)

One-sided, (1 − α) × 100% lower confidence bound

Note: The formula I gave for this in class was correct, but the way I suggested to derive it wasn’t (d’oh!). We start from the statement (with t_{α} instead of t_{α/2}):

1 − α = P (\frac{\overline{Y} − μ}{s/\sqrt{n}} < t_{α}(n − 1))

= P (\overline{Y} − μ < t_{α}(n − 1)\frac{s}{\sqrt{n}})

= P (−μ < −\overline{Y} + t_{α}(n − 1)\frac{s}{\sqrt{n}})

= P (μ > \overline{Y} − t_{α}(n − 1)\frac{s}{\sqrt{n}})

One-sided, (1 − α) × 100% upper confidence bound

Starting from

1 − α = P (−t_{α}(n − 1) < \frac{\overline{Y} − μ}{s/\sqrt{n}})

and proceeding as for the lower confidence bound,

1 − α = P (μ < \overline{Y} + t_{α}(n − 1)\frac{s}{\sqrt{n}})