Home page for accesible maths 4.2 Parametric confidence intervals 4.2.2 Confidence intervals and hypothesis tests 4.2.3 Confidence intervals: two-sample tests

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Justification

We now justify the confidence interval approach in the case of the mean of a Normal distribution.

Suppose that we have observed realisations of IID random variables $X_{1},\ldots,X_{n}$ which have a $\operatorname{Normal}(\mu,\sigma^{2})$ distribution. To test, at the $\alpha\%$ level,

\displaystyle H_{0}:\mu=\mu_{0}

vs.

\displaystyle H_{1}:\mu\neq\mu_{0}

we calculate the appropriate test statistic

\displaystyle T=\frac{\bar{X}-\mu_{0}}{S/\sqrt{n}}

Then $H_{0}$ is rejected if $T<t_{n-1}(\alpha/2)$ or $T>t_{n-1}(1-\alpha/2)$ .

By the symmetry of the $t$ -distribution, our rejection criterion could be written instead as,

\displaystyle|T|>t_{n-1}(1-\alpha/2).

as shown in Figure 4.1.

Fig. 4.1: Probability density function of the $t_{10}$ distribution with critical region for a two-tailed test at the 5% level shaded (blue).

The probability that we accept $H_{0}$ , given that $H_{0}$ is true, is

	$\displaystyle 1-\alpha$	$\displaystyle=1-\Pr(\|T\|>t_{n-1}(1-\alpha/2))$
		$\displaystyle=\Pr(\|T\|<t_{n-1}(1-\alpha/2))$
		$\displaystyle=\Pr(-t_{n-1}(1-\alpha/2)<T<t_{n-1}(1-\alpha/2))$
		$\displaystyle=\Pr(-t_{n-1}(1-\alpha/2)S/\sqrt{n}<\bar{X}-\mu_{0}<z(1-\alpha/2)% \sigma/\sqrt{n})$
		$\displaystyle=\Pr(\bar{X}-t_{n-1}(1-\alpha/2)S/\sqrt{n}<\mu_{0}<\bar{X}+t_{n-1% }(1-\alpha/2)S/\sqrt{n})$

But this is exactly the definition of the $100(1-\alpha)$ % confidence interval given in equation (4.1).

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To test at an $\alpha\%$ level

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For a two-tailed alternative hypothesis, we reject $H_{0}$ if the hypothesised value $\mu_{0}$ lies outside the $100(1-\alpha)\%$ confidence interval.
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For a one-tailed alternative hypothesis $\mu<\mu_{0}$ , we reject $H_{0}$ if and only if the hypothesised value $\mu_{0}$ lies above the $100(1-2\alpha)\%$ confidence interval.
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For a one-tailed alternative hypothesis $\mu>\mu_{0}$ , we reject $H_{0}$ if and only if the hypothesised value $\mu_{0}$ lies below the $100(1-2\alpha)\%$ confidence interval.

Some things to note:

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The size of a confidence interval depends on both the sample size and $\alpha$ .
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  If all else is equal, what should happen to the confidence interval as the sample size increases?
  
  A larger sample size $n$ will lead to narrower confidence intervals as there is more information and the uncertainty in the point estimate is less.
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  If all else is equal, what should happen to the confidence interval if $\alpha$ is decreased?
  
  A smaller value of $\alpha$ will lead to wider confidence intervals.
  
  Since $\alpha$ is the probability of rejecting the null hypothesis, given that it is in fact true i.e. the probability of a Type I error, the smaller this probability, the fewer times we should incorrectly reject $H_{0}$ and therefore the larger the confidence interval needs to be.
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In the case of the population mean, the size of the confidence interval also depends on the size of the population variance. All else being equal, a larger variance will lead to a wider confidence interval. The intuition for this is that, for a larger variance, there is less chance that a sample of size $n$ will capture the full population behaviour.