Home page for accesible maths 4.2 Parametric confidence intervals 4.2.1 One-sample mean Justification

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

4.2.2 Confidence intervals and hypothesis tests

Using the example of the population mean, we can now establish the link between confidence intervals and hypothesis testing.

Generally for a parameter $\theta$ , suppose that interest is in testing the null hypothesis $H_{0}:\theta=\theta_{0}$ against a one- or two-tailed alternative at the $\alpha$ % level. Instead of calculating a test statistic and looking at a critical region or $p$ -value, we could instead calculate the $100(1-\alpha)$ % confidence interval. The null hypothesis can be rejected at the $100\alpha$ % level if

{mdframed}

1.

$H_{1}:\theta>\theta_{0}$ and $\theta_{0}$ lies below the $100(1-2\alpha)$ % confidence interval.
2.

$H_{1}:\theta<\theta_{0}$ and $\theta_{0}$ lies above the $100(1-2\alpha)$ % confidence interval.
3.

$H_{1}:\theta\neq\theta_{0}$ and $\theta_{0}$ lies outside the $100(1-\alpha)$ % confidence interval.

The additional benefits of this approach are

•

The confidence interval on its own is an informative measure of uncertainty in the estimate of $\theta$ .
•

Different alternative hypotheses can be tested from a single confidence interval, without needing to recalculate test statistics.

TheoremExample 4.2.2

For the sample of Arctic sea ice data, we wanted to test, at the 5% level,

\displaystyle H_{0}:\mu=6.5

vs.

\displaystyle H_{1}:\mu<6.5.

Carry out this test by calculating an appropriate confidence interval.

Since the alternative hypothesis is one-tailed, and we want to test at the 5% level, we need to calculate a 90% confidence interval.

Assume that $\sigma^{2}$ is unknown, we use the formula in equation (4.1). Since $t_{9}(0.95)=1.83$ , we have

\displaystyle\left(6.06-1.83\times 0.911/\sqrt{10},6.06+1.83\times 0.911/\sqrt% {10}\right)=(5.53,6.59)\times 10^{6}km^{2}.

As 6.5 $\times 10^{6}km^{2}$ lies within this confidence interval, there is no evidence to reject $H_{0}$ and we conclude, as before, that there is no evidence that the mean of the minimum Arctic sea ice extent is less than 6.5 million $km^{2}$ .

TheoremExample 4.2.3

In example 3.2.3, we tested the null hypothesis that the mean November rainfall in Durham is 70mm, against the two-tailed alternative.

Using the same data as in this example, calculate an appropriate confidence interval and use this to carry out the above test at the 5% level.

Because we want to test at the 5% level, and the alternative hypothesis is two-tailed, we will construct a 95% confidence interval. From example 3.2.3, $\bar{x}=51.3$ and $s^{2}=760$ . The degrees of freedom are $n-1=16-1=15$ .

Since $t_{15}(0.975)=2.13$ , the 95% confidence interval is given by

\displaystyle\left(51.3-2.13\times\sqrt{760}/\sqrt{16},51.3+2.13\times\sqrt{76% 0}/\sqrt{16}\right)=(36.6,66.0)\text{mm}.

Since 70mm lies outside of the confidence interval, we conclude that there is enough evidence to reject $H_{0}$ , i.e. there is evidence that the mean rainfall differs from 70mm. The confidence interval provides the additional information that the mean rainfall is less than 70mm.