Lecture 16

ConficenceConficenceIntervalsIntervalsforforTheTheMeanMean--SmallSmallSampleSample

Remark

When $n<30$, the sample standard deviation, $s$ is no longer a good enough estimate for the population parameter $\sigma$. But if the distribution in the population is normal, then we can compensate by replacing the $z$-score with a $t-$score.

The $t$-distribution is similar to the $z$-distribution, but is more spread out and has heavier tails. The spread increases with the degrees of freedom, and the tails are heavier for small degrees of freedom. As the degrees of freedom increases, the $t$-distribution approaches the $z$-distribution. In the drawing above, the red curve is the $t$-distribution with $1$ degree of freedom, the orrange curve is the $t$-distribution with $3$ degrees of freedom, and the purple curve is the $t$-distribution with $300$ degrees of freedom. This last (purple) curve is very close to the standard normal $z$-distribution.

Formula

A $100(1-\alpha)$% confidence interval for the population mean $\mu$ is given by$$\overline X - t_{\alpha/2}\frac{s}{\sqrt{n}} < \mu < \overline X + t_{\alpha/2}\frac{s}{\sqrt{n}}$$where $t_{\alpha/2}$ has $n-1$ degrees of freedom. This confidence interval holds for samples from normal population and is used for small samples of size $n \leq 30$.

Example 1

When the snow starts melting, field mice emerge from their nests in the ground. A researcher caught $10$ field mice in April and measured their weights (in g):$$ \begin{array}{10*c} 19.4 & 21.4 & 22.3 & 22.1 & 20.1 & 23.8 & 24.6 & 199 & 21.5 & 19.1 \end{array} $$Assuming that the population is normal construct a $95\%$ confidence interval for the population mean weight.

$\begin{aligned} \bar{x} &=19.4+\cdots+19.1=21.42 \\ &\\ s^2&=\frac{1}{9}\left((19.4 - 21 .42)^2+\cdots\right)=3.386 \\ &\\ s&=1.84 \\ &\\ t&=2.262 \quad (df=10-1=9) \end{aligned}$

$21.42 \pm 2.262 \frac{1.84}{\sqrt{10}} \quad \Rightarrow \quad 20.10 \leq \mu \leq 22.74 \mathrm{~g}$ with $95 \%$ confidence.

Example 2

10 years ago the average lifespan for females in Quebec was $83.4$ years. You gather at random $21$ death notices for females in Quebec within the last 6 months. Your sample shows $\bar{x}=86.45$ years, with sample standard deviation $s=2.8 $ years. Assuming the population lifespan is approxmately normally distributed construct a $90\%$ confidence interval for the average lifespan of females in Quebec now. Can you claim that the female lifespan has changed (increased) compared to 10 years ago with $90 \%$ confidence?

$ \begin{aligned} &86.45 \pm 1.725\frac{2.8}{\sqrt{21}} \\ &\\ \Rightarrow \quad &86.45 \pm 1.05 \end{aligned} $

$85.40 \leq \mu \leq 87.50 \text{ years}$ with $90 \%$confidence.

Since $83.4 \mathrm{~yrs}$ is not in the interval, we can claim that the average lifespan, has increased with $90 \%$ confidence.

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Lecture 15 Lecture 17