Home page for accesible maths 2.11 Inference for other estimators 2.11.1 Confidence intervals for nearly normal point estimates 2.11.3 Non-normal point estimates

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2.11.2 Hypothesis testing for nearly normal point estimates

Just as the confidence interval method works with many other point estimates, we can generalize our hypothesis testing methods to new point estimates. Here we only consider the p-value approach, introduced in Section 2.9.4, since it is the most commonly used technique and also extends to non-normal cases.

Hypothesis testing using the normal model 1. First write the hypotheses in plain language, then set them up in mathematical notation. 2. Identify an appropriate point estimate of the parameter of interest. 3. Verify conditions to ensure the standard error estimate is reasonable and the point estimate is nearly normal and unbiased. 4. Compute the standard error. Draw a picture depicting the distribution of the estimate under the idea that $H_{0}$ is true. Shade areas representing the p-value. 5. Using the picture and normal model, compute the test statistic (Z score) and identify the p-value to evaluate the hypotheses. Write a conclusion in plain language.

Example 2.11.4

A drug called sulphinpyrazone was under consideration for use in reducing the death rate in heart attack patients. To determine whether the drug was effective, a set of 1,475 patients were recruited into an experiment and randomly split into two groups: a control group that received a placebo and a treatment group that received the new drug. What would be an appropriate null hypothesis? And the alternative?

Answer. The sceptic’s perspective is that the drug does not work at reducing deaths in heart attack patients ( $H_{0}$ ), while the alternative is that the drug does work ( $H_{A}$ ). We can formalize the hypotheses from Exercise 2.11.4 by letting $p_{control}$ and $p_{treatment}$ represent the proportion of patients who died in the control and treatment groups, respectively. Then the hypotheses can be written as

	$\displaystyle H_{0}:p_{control}=p_{treatment}\quad\text{(the drug doesn't work)}$
	$\displaystyle H_{A}:p_{control}>p_{treatment}\quad\text{(the drug works)}$

or equivalently,

	$\displaystyle H_{0}:p_{control}-p_{treatment}=0\quad\text{(the drug doesn't % work)}$
	$\displaystyle H_{A}:p_{control}-p_{treatment}>0\quad\text{(the drug works)}$

Strong evidence against the null hypothesis and in favour of the alternative would correspond to an observed difference in death rates,

\displaystyle\text{point estimate}=\hat{p}_{control}-\hat{p}_{treatment}

being larger than we would expect from chance alone. This difference in sample proportions represents a point estimate that is useful in evaluating the hypotheses.

Example 2.11.5

We want to evaluate the hypothesis setup from Exercise 2.11.4 using data from the actual study.²⁹²⁹Anturane Reinfarction Trial Research Group. 1980. Sulfinpyrazone in the prevention of sudden death after myocardial infarction. New England Journal of Medicine 302(5):250-256. In the control group, 60 of 742 patients died. In the treatment group, 41 of 733 patients died. The sample difference in death rates can be summarized as

\displaystyle\text{point estimate}=\hat{p}_{control}-\hat{p}_{treatment}=\frac% {60}{742}-\frac{41}{733}=0.025

This point estimate is nearly normal and is an unbiased estimate of the actual difference in death rates. The standard error of this sample difference is $SE=0.013$ . Evaluate the hypothesis test at a 5% significance level: $\alpha=0.05$ .

Answer. We would like to identify the p-value to evaluate the hypotheses. If the null hypothesis is true, then the point estimate would have come from a nearly normal distribution, like the one shown in Figure LABEL:sulphStudyFindPValueUsingNormalApprox. The distribution is centred at zero since $p_{control}-p_{treatment}=0$ under the null hypothesis. Because a large positive difference provides evidence against the null hypothesis and in favour of the alternative, the upper tail has been shaded to represent the p-value. We need not shade the lower tail since this is a one-sided test: an observation in the lower tail does not support the alternative hypothesis.

The p-value can be computed by using the Z score of the point estimate and pnorm.

\displaystyle Z=\frac{\text{point estimate}-\text{null value}}{SE_{\text{point% estimate}}}=\frac{0.025-0}{0.013}=1.92

(2.14)

Answer. Typing pnorm(1.92) in R, we find that the lower unshaded tail is about $\mathbb{P}(Z<1.92)=0.973$ . Thus, the upper shaded tail representing the p-value is

\displaystyle\text{p-value}=1-0.973=0.027.

Because the p-value is less than the significance level ( $\alpha=0.05$ ), we say the null hypothesis is implausible. That is, we reject the null hypothesis in favour of the alternative and conclude that the drug is effective at reducing deaths in heart attack patients. The Z score is called a test statistic. In most hypothesis tests, a test statistic is a particular data summary that is especially useful for computing the p-value and evaluating the hypothesis test. In the case of point estimates that are nearly normal, the test statistic is the Z score.

Test statistic A test statistic is a special summary statistic that is particularly useful for evaluating a hypothesis test or identifying the p-value. When a point estimate is nearly normal, we use the Z score of the point estimate as the test statistic. In later chapters we encounter situations where other test statistics are helpful.