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4.2.3 Confidence intervals: two-sample tests

Recall from Chapter 3 that we can also use hypothesis testing to compare the means of two separate populations; for example the mean reaction times of individuals before and after a full night of sleep. We saw how to find critical regions and calculate $p$ -values for these tests, but the confidence interval method discussed above can also be used.

For unpaired data, the $100(1-\alpha)\%$ confidence interval is

	$\displaystyle\left((\bar{x}-\bar{y})-t_{n+m-2}(1-\alpha/2)\times s_{p}\sqrt{1/% m+1/n},\right.$
	$\displaystyle\left.(\bar{x}-\bar{y})+t_{n+m-2}(1-\alpha/2)\times s_{p}\sqrt{1/% m+1/n}\right)$

TheoremExample 4.2.4 Loaves of bread

Recall the baker who wanted to test whether Jack and Jill were baking loaves of a consistent weight.

Calculate a 90% confidence intervals for the difference in mean loaf weights between the two bakers. Use your confidence interval to test the hypothesis

H_{0}:\mu_{Jill}=\mu_{Jack}

vs.

H_{1}:\mu_{Jill}<\mu_{Jack}

State the significance level of the test.

Using the equation above, and the sample means and variances calculated in Chapter 3, the lower end-point of the confidence interval is

	$\displaystyle(\bar{x}-\bar{y})-t_{16}(0.95)\times s_{p}\sqrt{1/n+1/m}$	$\displaystyle=(498-501)-t_{16}(0.95)\times 5.82\times\sqrt{1/10+1/8}$
		$\displaystyle=-3-1.75\times 5.82\times 0.474$
		$\displaystyle=-7.82$

Similar calculations give the upper end point as

\displaystyle(\bar{x}-\bar{y})+t_{16}(0.95)\times s_{p}\times\sqrt{1/n+1/m}=1.% 82.

The 90% confidence interval for $\mu_{X}-\mu_{Y}$ is $(-7.82,1.82)$ grams. Since zero lies inside the confidence interval, there is no evidence to reject the null hypothesis at the 5% level. We conclude that there is no evidence that the mean weight of loaves baked by Jill is less than the mean weight of loaves baked by Jack.

For paired data, the $100(1-\alpha)\%$ confidence interval for the mean of the differences between the two groups is

\displaystyle(\bar{d}-t_{n-1}(1-\alpha/2)\times s_{d}/\sqrt{n},\bar{d}+t_{n-1}% (1-\alpha/2)\times s_{d}/\sqrt{n})

TheoremExample 4.2.5 Drug comparison

Recall the asthma drug trial in Chapter 3. The effects of two drugs, code-named $F$ and $S$ , on Peak Expiratory Flow (PEF) are compared. Calculate an appropriate confidence interval to test, at the 5% level, the hypothesis

\displaystyle H_{0}:\mu_{f}=\mu_{s}

vs.

\displaystyle H_{1}:\mu_{f}\neq\mu_{s}.

Because the test is two-tailed, we will use a 95% confidence interval. We calculate this using the above formula:

	$\displaystyle\mu_{d}\pm t_{n-1}(0.975)\times\frac{s_{d}}{\sqrt{n}}$	$\displaystyle=45.38\pm 2.18\times\frac{40.59}{3.61}$
		$\displaystyle=(20.84,69.93).$

Since 0 lies outside the confidence interval, we would conclude that there is evidence to reject $H_{0}$ at the 5% level.