z test for the difference between two proportions - overview
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$z$ test for the difference between two proportions | $z$ test for a single proportion | Two sample $t$ test - equal variances assumed |
You cannot compare more than 3 methods |
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Independent/grouping variable | Independent variable | Independent/grouping variable | |
One categorical with 2 independent groups | None | One categorical with 2 independent groups | |
Dependent variable | Dependent variable | Dependent variable | |
One categorical with 2 independent groups | One categorical with 2 independent groups | One quantitative of interval or ratio level | |
Null hypothesis | Null hypothesis | Null hypothesis | |
H0: $\pi_1 = \pi_2$
Here $\pi_1$ is the population proportion of 'successes' for group 1, and $\pi_2$ is the population proportion of 'successes' for group 2. | H0: $\pi = \pi_0$
Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the 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. | |
Alternative hypothesis | Alternative hypothesis | Alternative hypothesis | |
H1 two sided: $\pi_1 \neq \pi_2$ H1 right sided: $\pi_1 > \pi_2$ H1 left sided: $\pi_1 < \pi_2$ | H1 two sided: $\pi \neq \pi_0$ H1 right sided: $\pi > \pi_0$ H1 left sided: $\pi < \pi_0$ | H1 two sided: $\mu_1 \neq \mu_2$ H1 right sided: $\mu_1 > \mu_2$ H1 left sided: $\mu_1 < \mu_2$ | |
Assumptions | Assumptions | Assumptions | |
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Test statistic | Test statistic | Test statistic | |
$z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
Here $p_1$ is the sample proportion of successes in group 1: $\dfrac{X_1}{n_1}$, $p_2$ is the sample proportion of successes in group 2: $\dfrac{X_2}{n_2}$, $p$ is the total proportion of successes in the sample: $\dfrac{X_1 + X_2}{n_1 + n_2}$, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. Note: we could just as well compute $p_2 - p_1$ in the numerator, but then the left sided alternative becomes $\pi_2 < \pi_1$, and the right sided alternative becomes $\pi_2 > \pi_1.$ | $z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$
Here $p$ is the sample proportion of successes: $\dfrac{X}{N}$, $N$ is the sample size, and $\pi_0$ is the population proportion of successes according to the null hypothesis. | $t = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}}$
Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $s_p$ is the pooled standard deviation, $n_1$ 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 $s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}$ is the standard error of the sampling distribution of $\bar{y}_1 - \bar{y}_2$. The $t$ value indicates how many standard errors $\bar{y}_1 - \bar{y}_2$ 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 $\mu_2 > \mu_1$. | |
n.a. | n.a. | Pooled standard deviation | |
- | - | $s_p = \sqrt{\dfrac{(n_1 - 1) \times s^2_1 + (n_2 - 1) \times s^2_2}{n_1 + n_2 - 2}}$ | |
Sampling distribution of $z$ if H0 were true | Sampling distribution of $z$ if H0 were true | Sampling distribution of $t$ if H0 were true | |
Approximately the standard normal distribution | Approximately the standard normal distribution | $t$ distribution with $n_1 + n_2 - 2$ degrees of freedom | |
Significant? | Significant? | Significant? | |
Two sided:
| Two sided:
| Two sided:
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Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$ | Approximate $C\%$ confidence interval for $\pi$ | $C\%$ confidence interval for $\mu_1 - \mu_2$ | |
Regular (large sample):
| Regular (large sample):
| $(\bar{y}_1 - \bar{y}_2) \pm t^* \times s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}$
where the critical value $t^*$ is the value under the $t_{n_1 + n_2 - 2}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). The confidence interval for $\mu_1 - \mu_2$ can also be used as significance test. | |
n.a. | n.a. | Effect size | |
- | - | Cohen's $d$: Standardized difference between the mean in group $1$ and in group $2$: $$d = \frac{\bar{y}_1 - \bar{y}_2}{s_p}$$ Cohen's $d$ indicates how many standard deviations $s_p$ the two sample means are removed from each other. | |
n.a. | n.a. | Visual representation | |
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Equivalent to | Equivalent to | Equivalent to | |
When testing two sided: chi-squared test for the relationship between two categorical variables, where both categorical variables have 2 levels. |
| One way ANOVA with an independent variable with 2 levels ($I$ = 2):
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Example context | Example context | Example context | |
Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic. | Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$? Use the normal approximation for the sampling distribution of the test statistic. | Is the average mental health score different between men and women? Assume that in the population, the standard deviation of mental health scores is equal amongst men and women. | |
SPSS | SPSS | SPSS | |
SPSS does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Analyze > Descriptive Statistics > Crosstabs...
| Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
| Analyze > Compare Means > Independent-Samples T Test...
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Jamovi | Jamovi | Jamovi | |
Jamovi does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Frequencies > Independent Samples - $\chi^2$ test of association
| Frequencies > 2 Outcomes - Binomial test
| T-Tests > Independent Samples T-Test
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Practice questions | Practice questions | Practice questions | |