Two sample z test - overview
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Two sample $z$ test | Sign test |
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Independent/grouping variable | Independent variable | |
One categorical with 2 independent groups | 2 paired groups | |
Dependent variable | Dependent variable | |
One quantitative of interval or ratio level | One of ordinal level | |
Null hypothesis | Null hypothesis | |
H0: $\mu_1 = \mu_2$
Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2. |
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Alternative hypothesis | Alternative hypothesis | |
H1 two sided: $\mu_1 \neq \mu_2$ H1 right sided: $\mu_1 > \mu_2$ H1 left sided: $\mu_1 < \mu_2$ |
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Assumptions | Assumptions | |
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Test statistic | Test statistic | |
$z = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}}$
Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $\sigma^2_1$ is the population variance in population 1, $\sigma^2_2$ is the population variance in population 2, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to the null hypothesis. The denominator $\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}}$ is the standard deviation of the sampling distribution of $\bar{y}_1 - \bar{y}_2$. The $z$ value indicates how many of these standard deviations $\bar{y}_1 - \bar{y}_2$ is removed from 0. Note: we could just as well compute $\bar{y}_2 - \bar{y}_1$ in the numerator, but then the left sided alternative becomes $\mu_2 < \mu_1$, and the right sided alternative becomes $\mu_2 > \mu_1$. | $W = $ number of difference scores that is larger than 0 | |
Sampling distribution of $z$ if H0 were true | Sampling distribution of $W$ if H0 were true | |
Standard normal distribution | The exact distribution of $W$ under the null hypothesis is the Binomial($n$, $P$) distribution, with $n =$ number of positive differences $+$ number of negative differences, and $P = 0.5$.
If $n$ is large, $W$ is approximately normally distributed under the null hypothesis, with mean $nP = n \times 0.5$ and standard deviation $\sqrt{nP(1-P)} = \sqrt{n \times 0.5(1 - 0.5)}$. Hence, if $n$ is large, the standardized test statistic $$z = \frac{W - n \times 0.5}{\sqrt{n \times 0.5(1 - 0.5)}}$$ follows approximately the standard normal distribution if the null hypothesis were true. | |
Significant? | Significant? | |
Two sided:
| If $n$ is small, the table for the binomial distribution should be used: Two sided:
If $n$ is large, the table for standard normal probabilities can be used: Two sided:
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$C\%$ confidence interval for $\mu_1 - \mu_2$ | n.a. | |
$(\bar{y}_1 - \bar{y}_2) \pm z^* \times \sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval). The confidence interval for $\mu_1 - \mu_2$ can also be used as significance test. | - | |
Visual representation | n.a. | |
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n.a. | Equivalent to | |
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Two sided sign test is equivalent to
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Example context | Example context | |
Is the average mental health score different between men and women? Assume that in the population, the standard devation of the mental health scores is $\sigma_1 = 2$ amongst men and $\sigma_2 = 2.5$ amongst women. | Do people tend to score higher on mental health after a mindfulness course? | |
n.a. | SPSS | |
- | Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
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n.a. | Jamovi | |
- | Jamovi does not have a specific option for the sign test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the two sided $p$ value that would have resulted from the sign test. Go to:
ANOVA > Repeated Measures ANOVA - Friedman
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Practice questions | Practice questions | |