$Z$-score

The z-score for an observation $x$ of random variable $X$ is

$$\begin{equation*} z_{x}=\frac{x-\mu}{\sigma}. \end{equation*}$$ For $z$-scores it holds that $\mu_{z}=0$ en $\sigma_{z}^2=1$. If $X$ is normally distributed, then $Z$ follows a standard normal distribution and for observations $z_x$ of $Z$ it holds that $$\begin{equation*} P(X\ge x)=P(Z\ge z_{x}). \end{equation*}$$ The $p$-values for the standard normal distribution can be found in Table 1.

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Proportions

Standard error for a proportion with known population value $p$

$$\begin{equation*} se_p=\sqrt{\frac{p(1-p)}{n}} \end{equation*}$$

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Standard error for a proportion with unknown population value $p$

$$\begin{equation*} se_{\hat{p}}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \end{equation*}$$

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$100(1-\alpha)\%$ - confidence interval for one proportion

$$\begin{equation*}\hat{p}-z_{\alpha/2}\cdot se_{\hat{p}}\le p\le \hat{p}+z_{\alpha/2}\cdot se_{\hat{p}}, \end{equation*}$$ where $P(Z\ge z_{\alpha/2})= \alpha/2$ ($z$-score corresponding to the selected confidence level (for example $z=1.96$ for a confidence level of 95%).

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Minimal sample size to estimate a population proportion

$$\begin{equation*} n=\frac{\hat{p}(1-\hat{p})z^2}{m^2}, \end{equation*}$$ where $\hat{p}$ is the estimated proportion, $m$ the margin of error and $z$ the $z$-score corresponding to the selected confidence level (for example $z=1.96$ for a confidence level of 95%).

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$z$-test for one proportion

$$\begin{equation*} z=\frac{p-p_0}{se_0}=\frac{p-p_0}{\sqrt{\displaystyle\frac{p_0(1-p_0)}{n}}}, \end{equation*}$$ where $p_0$ is the expected proportion under the null hypothesis.

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Standard error for the difference between two proportions

$$\begin{equation*} se_{\hat{p}_1-\hat{p}_2}=\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1}+ \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}, \end{equation*}$$ where $\hat{p}_1$ is the observed proportion based on $n_1$ observations in sample 1 and $\hat{p}_2$ the observed proportion based on $n_2$ observations in sample 2.

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$100(1-\alpha)\%$ confidence interval for the difference between two proportions

$$\begin{equation*} (\hat{p}_1-\hat{p}_2)-z_{\alpha/2}\cdot se_{\hat{p}_1-\hat{p}_2}\le (p_1-p_2) \le (\hat{p}_1-\hat{p}_2)+z_{\alpha/2}\cdot se_{\hat{p}_1-\hat{p}_2}, \end{equation*}$$ where $P(Z\ge z_{\alpha/2})= \alpha/2$ ($z$-score corresponding to the selected confidence level (for example $z=1.96$ for a confidence level of 95%).

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$z$-test for the difference between two independent proportions

$$\begin{equation*} z= \frac{\hat{p}_1-\hat{p}_2 - (p_1-p_2)}{se_0}, \end{equation*}$$ where $se_0$ is the standard error under the null hypothesis. If the null hypothesis assumes $p_1=p_2$, then $$\begin{equation*}se_0= \sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+ \frac{\hat{p}(1-\hat{p})}{n_2}}= \sqrt{\hat{p} (1- \hat{p}) \Bigg(\frac{1}{n_1}+\frac{1}{n_2}\Bigg)}, \end{equation*}$$ where $\hat{p} = \frac{n_1\hat{p}_1+ n_2\hat{p}_2}{n_1+n_2}$ (referred to as the pooled proportion).

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$z$-test for the difference between two dependent proportions - McNemar's test

With $n_{01}$ denoting the number of observations with a score of 0 on variable $A$ and a score of 1 on variable $B$, and $n_{10}$ denoting the number of observations with a score of 1 onvariable $A$ and a score of 0 on variable $B$:

$$\begin{equation*} z= \frac{n_{01}-n_{10}}{\sqrt{n_{01}+n_{10}}}, \end{equation*}$$ see:

	$B$
$A$	$n_{00}$	$n_{01}$
	$n_{10}$	$n_{11}$

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Means

Expected value for a mean

$$\begin{equation*} E(\overline{X})=\mu_{\overline{x}}=\mu \end{equation*}$$

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Standard error for a mean with known population variance $\sigma$

$$\begin{equation*} se_{\overline{X}}=\sigma/\sqrt{n} \end{equation*}$$

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Standard error for a mean with unknown population variance $\sigma$

$$\begin{equation*} se_{\overline{x}}= s/\sqrt{n} \end{equation*}$$

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$100(1-\alpha)\%$ confidence interval for one mean

$$\begin{equation*} \overline{x}-t_{\alpha/2}\cdot se_{\overline{x}}\le\mu\le\overline{x}+ t_{\alpha/2}\cdot se_{\overline{x}}, \end{equation*}$$ where $P(T\ge t_{\alpha/2})= \alpha/2$ for a $t$-distribution with $df=n-1$ degrees of freedom ($t$-score corresponding to the selected confidence level).

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Minimal sample size to estimate a population mean

$$\begin{equation*} n=\frac{\sigma^2z^2}{m^2}, \end{equation*}$$ where $\sigma$ is the (expected) standard deviation in the population, $m$ the margin of error and $z$ the $z$-score corresponding to the selected confidence level (for example $z=1.96$ for a confidence level of 95%).

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$t$-test for one independent mean

$$\begin{equation*} t_{\overline{x}}=\frac{\overline{x}-\mu_0}{se_{\overline{x}}}=\frac{\overline{x}-\mu_0}{s/\sqrt{n}} \end{equation*}$$ $\mu_0$ is the mean population value expected under the null hypothesis. If the null hypothesis holds and $X$ is normally distributed, then $t_{\overline{x}}$ follows a $t$-distribution with $df=n-1$ degrees of freedom.

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Standard error for the difference between two means

$$\begin{equation*} se_{\overline{x}_1 - \overline{x}_2}= \sqrt{\frac{s_1^2}{n_1} \frac{s_2^2}{n_2}} \end{equation*}$$

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$t$-test for the difference between two independent means with unequal population variances

$$\begin{equation*} t_{\overline{x}_1-\overline{x}_2} = \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)} {se_{\overline{x}_1-\overline{x}_2}}. \end{equation*}$$ If $X_1$ and $X_2$ are independent and normally distributed, $t_{\overline{x}_1-\overline{x}_2}$ approximately follows a $t$-distribution with degrees of freedom $$\begin{equation*} df=\frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2} {\frac{1}{n_1-1}\left(\frac{s_1^2}{n_1}\right)^2+ \frac{1}{n_2-1}\left(\frac{s_2^2}{n_2}\right)^2}. \end{equation*}$$

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$t$-test for the difference between two independent means with equal population variances

$$\begin{equation*} t_{\overline{x}_1-\overline{x}_2} = \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)} {se_{\overline{x}_1-\overline{x}_2}}, \end{equation*}$$ with $$\begin{equation*} se_{\overline{x}_1-\overline{x}_2} = s\cdot\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}, \end{equation*}$$ where $$\begin{equation*} s = \sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{(n_1-1)+(n_2-1)}} \end{equation*}$$ is referred to as the pooled standard deviation. If $X_1$ and $X_2$ are independent and normally distributed with equal variances, $t_{\overline{x}_1-\overline{x}_2}$ follows a $t$-distribution with $n_1+n_2-2$ degrees of freedom.

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$100(1-\alpha)\%$ confidence interval for the difference between two independent means

$$\begin{equation*} (\overline{x}_1-\overline{x}_2)-t_{\alpha/2}\cdot se_{\overline{x}_1-\overline{x}_2}\le (\mu_1-\mu_2) \le(\overline{x}_1-\overline{x}_2)+t_{\alpha/2}\cdot se_{\overline{x}_1-\overline{x}_2} \end{equation*}$$ The degrees of freedom and $se_{\overline{x}_1-\overline{x}_2}$ depend on whether the population variances are assumed to be equal or not, see the previous formulas for the appropriate method of calculation.

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Standardized effect sizes for the difference between two means

$$\begin{equation*} d = \frac{\overline{x}_1-\overline{x}_2}{s}, \end{equation*}$$ where $s$ can refer to the pooled standard deviation or the standard deviation of either one of the samples.

• $d=0.2$ small effect
• $d=0.5$ medium effect
• $d=0.8$ large effect

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$t$-test for paired samples (two dependent means)

$$\begin{equation*} t_{\overline{x}_d} = \frac{\overline{x}_d-\mu_d} {se_{\overline{x}_d}}, \end{equation*}$$ where $$\begin{equation*} se_{\overline{x}_d} = \sqrt{\frac{s_d^2}{n_d}} = \frac{s_d}{\sqrt{n_d}} \end{equation*}$$ and $n_d-1$ degrees of freedom.

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$100(1-\alpha)\%$ - confidence interval for paired samples

$$\begin{equation*} \overline{x}_d-t_{\alpha/2}\cdot se_{\overline{x}_d}\le \mu_d \le\overline{x}_d+t_{\alpha/2}\cdot se_{\overline{x}_d}, \end{equation*}$$ with $n_d-1$ degrees of freedom.