# 6b.5 - Statistical Inference - Hypothesis Testing

6b.5 - Statistical Inference - Hypothesis Testing

Statisticians construct the null hypothesis and the alternative hypothesis for statistical hypothesis testing such that the research hypothesis is the alternative hypothesis:

$$H_0: \left\{ \text{non-equivalence}\right\} \text{ vs. } H_1: \left\{ \text{equivalence}\right\}$$

or

$$H_0: \left\{ \text{inferiority}\right\} \text{ vs. } H_1: \left\{ \text{non-inferiority}\right\}$$

In terms of the population means, the hypotheses for testing equivalence are expressed as:

$$H_0: \left\{ \mu_{E} - \mu_{A} \le -\Psi \text{ or } \mu_{E} - \mu_{A} \ge \Psi \right\}$$

vs.

$$H_1: \left\{-\Psi < \mu_{E} - \mu_{A}< \Psi \right\}$$

also expressed as

$$H_0: \left\{|\mu_{E} - \mu_{A}| \ge \Psi \right\} \text{ vs. } H_1: \left\{|\mu_{E} - \mu_{A}| < \Psi \right\}$$

In terms of the population means, the hypotheses for testing non-inferiority are expressed as

$$H_0: \left\{\mu_{E} - \mu_{A} \le -\Psi \right\} \text{ vs. } H_1: \left\{\mu_{E} - \mu_{A} > -\Psi \right\}$$

The null and alternative hypotheses for an equivalence trial can be decomposed into two distinct hypothesis testing problems, one for non-inferiority:

$$H_{01}: \left\{\mu_{E} - \mu_{A} \le -\Psi \right\} \text{ vs. } H_{11}: \left\{\mu_{E} - \mu_{A} > -\Psi \right\}$$

and one for non-superiority

$$H_{02}: \left\{\mu_{E} - \mu_{A} \ge \Psi \right\} \text{ vs. } H_{12}: \left\{\mu_{E} - \mu_{A} <\Psi \right\}$$

The null hypothesis of non-equivalence is rejected if and only if the null hypothesis of non-inferiority $$\left(H_{01}\right)$$ is rejected AND the null hypothesis of non-superiority $$\left(H_{02}\right)$$ is rejected.

This rationale leads to what is called two one-sided testing (TOST). If the data are approximately normally distributed, then two-sample t tests can be applied. If normality is suspect, then Wilcoxon rank-sum tests can be applied.

With respect to two-sample t tests, reject the null hypothesis of inferiority if:

$$t_{inf}= \left(\bar{Y}_E - \bar{Y}_A + \Psi \right) / s \sqrt{\frac {1}{n_E}+\frac {1}{n_Z}}>t_{n_{E}+n_{A}-2, 1-\alpha}$$

and reject the null hypothesis of superiority if:

$$t_{sup}= \left(\bar{Y}_E - \bar{Y}_A + \Psi \right) / s \sqrt{\frac {1}{n_E}+\frac {1}{n_A}}< -t_{n_{E}+n_{A}-2, 1-\alpha}$$

where s is the pooled sample estimate of the standard deviation, calculated as the square-root of the pooled sample estimate of the variance:

$$s^2 = \left(\sum_{i=1}^{n_E}\left(Y_{Ei}-\bar{Y}_E\right)^2+\sum_{j=1}^{n_A}\left(Y_{Aj}-\bar{Y}_A\right)^2\right) / \left(n_E + n_A -2\right)$$

NOTE! Each one-sided t test is conducted at the $$\alpha$$ significance level.

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