z test for the difference between two proportions - overview

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$z$ test for the difference between two proportions
Two sample $z$ test
Independent variableIndependent variable
One categorical with 2 independent groupsOne categorical with 2 independent groups
Dependent variableDependent variable
One categorical with 2 independent groupsOne quantitative of interval or ratio level
Null hypothesisNull hypothesis
$\pi_1 = \pi_2$
$\pi_1$ is the unknown proportion of "successes" in population 1; $\pi_2$ is the unknown proportion of "successes" in population 2
$\mu_1 = \mu_2$
$\mu_1$ is the unknown mean in population 1, $\mu_2$ is the unknown mean in population 2
Alternative hypothesisAlternative hypothesis
Two sided: $\pi_1 \neq \pi_2$
Right sided: $\pi_1 > \pi_2$
Left sided: $\pi_1 < \pi_2$
Two sided: $\mu_1 \neq \mu_2$
Right sided: $\mu_1 > \mu_2$
Left sided: $\mu_1 < \mu_2$
AssumptionsAssumptions
  • Sample size is large enough for $z$ to be approximately normally distributed. Rule of thumb:
    • Significance test: number of successes and number of failures are each 5 or more in both sample groups
    • Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures are each 10 or more in both sample groups
    • Plus four 90%, 95%, or 99% confidence interval: sample sizes of both groups are 5 or more
  • Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
  • Within each population, the scores on the dependent variable are normally distributed
  • Population standard deviations $\sigma_1$ and $\sigma_2$ are known
  • Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
Test statisticTest statistic
$z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
$p_1$ 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, $n_2$ 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 $\pi_2 > \pi_1$
$z = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}}$
$\bar{y}_1$ 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, $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to H0.

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 $\mu_2 > \mu_1$
Sampling distribution of $z$ if H0 were trueSampling distribution of $z$ if H0 were true
Approximately standard normalStandard normal
Significant?Significant?
Two sided: Right sided: Left sided: Two sided: Right sided: Left sided:
Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$$C\%$ confidence interval for $\mu_1 - \mu_2$
Regular (large sample):
  • $(p_1 - p_2) \pm z^* \times \sqrt{\dfrac{p_1(1 - p_1)}{n_1} + \dfrac{p_2(1 - p_2)}{n_2}}$
    where $z^*$ 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)
With plus four method:
  • $(p_{1.plus} - p_{2.plus}) \pm z^* \times \sqrt{\dfrac{p_{1.plus}(1 - p_{1.plus})}{n_1 + 2} + \dfrac{p_{2.plus}(1 - p_{2.plus})}{n_2 + 2}}$
    where $p_{1.plus} = \dfrac{X_1 + 1}{n_1 + 2}$, $p_{2.plus} = \dfrac{X_2 + 1}{n_2 + 2}$, and $z^*$ 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)
$(\bar{y}_1 - \bar{y}_2) \pm z^* \times \sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}$
where $z^*$ 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.
n.a.Visual representation
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Two sample z test
Equivalent ton.a.
When testing two sided: chi-squared test for the relationship between two categorical variables, where both categorical variables have 2 levels-
Example contextExample context
Is the proportion smokers different between men and women? 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 devation of the mental health scores is $\sigma_1$ = 2 amongst men and $\sigma_2$ = 2.5 amongst women.
SPSSn.a.
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...
  • Put your independent (grouping) variable in the box below Row(s), and your dependent variable in the box below Column(s)
  • Click the Statistics... button, and click on the square in front of Chi-square
  • Continue and click OK
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Jamovin.a.
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
  • Put your independent (grouping) variable in the box below Rows, and your dependent variable in the box below Columns
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Practice questionsPractice questions