One sample z test for the mean - overview

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One sample $z$ test for the mean
$z$ test for the difference between two proportions
Chi-squared test for the relationship between two categorical variables
Independent variableIndependent/grouping variableIndependent /column variable
NoneOne categorical with 2 independent groupsOne categorical with $I$ independent groups ($I \geqslant 2$)
Dependent variableDependent variableDependent /row variable
One quantitative of interval or ratio levelOne categorical with 2 independent groupsOne categorical with $J$ independent groups ($J \geqslant 2$)
Null hypothesisNull hypothesisNull hypothesis
H0: $\mu = \mu_0$

Here $\mu$ is the population mean, and $\mu_0$ is the population mean according to the 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: there is no association between the row and column variable

More precisely, if there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable:
  • H0: the distribution of the dependent variable is the same in each of the $I$ populations
If there is one random sample of size $N$ from the total population:
  • H0: the row and column variables are independent
Alternative hypothesisAlternative hypothesisAlternative hypothesis
H1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
H1 two sided: $\pi_1 \neq \pi_2$
H1 right sided: $\pi_1 > \pi_2$
H1 left sided: $\pi_1 < \pi_2$
H1: there is an association between the row and column variable

More precisely, if there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable:
  • H1: the distribution of the dependent variable is not the same in all of the $I$ populations
If there is one random sample of size $N$ from the total population:
  • H1: the row and column variables are dependent
AssumptionsAssumptionsAssumptions
  • Scores are normally distributed in the population
  • Population standard deviation $\sigma$ is known
  • Sample is a simple random sample from the population. That is, observations are independent of one another
  • 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
  • Sample size is large enough for $X^2$ to be approximately chi-squared distributed under the null hypothesis. Rule of thumb:
    • 2 $\times$ 2 table: all four expected cell counts are 5 or more
    • Larger than 2 $\times$ 2 tables: average of the expected cell counts is 5 or more, smallest expected cell count is 1 or more
  • There are $I$ independent simple random samples from each of $I$ populations defined by the independent variable, or there is one simple random sample from the total population
Test statisticTest statisticTest statistic
$z = \dfrac{\bar{y} - \mu_0}{\sigma / \sqrt{N}}$
Here $\bar{y}$ is the sample mean, $\mu_0$ is the population mean according to the null hypothesis, $\sigma$ is the population standard deviation, and $N$ is the sample size.

The denominator $\sigma / \sqrt{N}$ is the standard deviation of the sampling distribution of $\bar{y}$. The $z$ value indicates how many of these standard deviations $\bar{y}$ is removed from $\mu_0$.
$z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
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 $\pi_2 > \pi_1.$
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here for each cell, the expected cell count = $\dfrac{\mbox{row total} \times \mbox{column total}}{\mbox{total sample size}}$, the observed cell count is the observed sample count in that same cell, and the sum is over all $I \times J$ cells.
Sampling distribution of $z$ if H0 were trueSampling distribution of $z$ if H0 were trueSampling distribution of $X^2$ if H0 were true
Standard normal distributionApproximately the standard normal distributionApproximately the chi-squared distribution with $(I - 1) \times (J - 1)$ degrees of freedom
Significant?Significant?Significant?
Two sided: Right sided: Left sided: Two sided: Right sided: Left sided:
  • Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
  • Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
$C\%$ confidence interval for $\mu$Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$n.a.
$\bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}}$
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 $\mu$ can also be used as significance test.
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 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)
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 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)
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Effect sizen.a.n.a.
Cohen's $d$:
Standardized difference between the sample mean and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{\sigma}$$ Cohen's $d$ indicates how many standard deviations $\sigma$ the sample mean $\bar{y}$ is removed from $\mu_0.$
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Visual representationn.a.n.a.
One sample z test
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n.a.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 contextExample context
Is the average mental health score of office workers different from $\mu_0 = 50$? Assume that the standard deviation of the mental health scores in the population is $\sigma = 3.$Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic.Is there an association between economic class and gender? Is the distribution of economic class different between men and women?
n.a.SPSSSPSS
-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
Analyze > Descriptive Statistics > Crosstabs...
  • Put one of your two categorical variables in the box below Row(s), and the other categorical 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
n.a.JamoviJamovi
-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
Frequencies > Independent Samples - $\chi^2$ test of association
  • Put one of your two categorical variables in the box below Rows, and the other categorical variable in the box below Columns
Practice questionsPractice questionsPractice questions