Two sample t test - equal variances not assumed - overview

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Two sample $t$ test - equal variances not assumed
$z$ test for the difference between two proportions
Marginal Homogeneity test / Stuart-Maxwell test
Sign test
One sample $z$ test for the mean
Independent/grouping variableIndependent/grouping variableIndependent variableIndependent variableIndependent variable
One categorical with 2 independent groupsOne categorical with 2 independent groups2 paired groups2 paired groupsNone
Dependent variableDependent variableDependent variableDependent variableDependent variable
One quantitative of interval or ratio levelOne categorical with 2 independent groupsOne categorical with $J$ independent groups ($J \geqslant 2$)One of ordinal levelOne quantitative of interval or ratio level
Null hypothesisNull hypothesisNull hypothesisNull hypothesisNull 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.
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: for each category $j$ of the dependent variable, $\pi_j$ for the first paired group = $\pi_j$ for the second paired group.

Here $\pi_j$ is the population proportion in category $j.$
  • H0: P(first score of a pair exceeds second score of a pair) = P(second score of a pair exceeds first score of a pair)
If the dependent variable is measured on a continuous scale, this can also be formulated as:
  • H0: the population median of the difference scores is equal to zero
A difference score is the difference between the first score of a pair and the second score of a pair.
H0: $\mu = \mu_0$

Here $\mu$ is the population mean, and $\mu_0$ is the population mean according to the null hypothesis.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
H1 two sided: $\mu_1 \neq \mu_2$
H1 right sided: $\mu_1 > \mu_2$
H1 left sided: $\mu_1 < \mu_2$
H1 two sided: $\pi_1 \neq \pi_2$
H1 right sided: $\pi_1 > \pi_2$
H1 left sided: $\pi_1 < \pi_2$
H1: for some categories of the dependent variable, $\pi_j$ for the first paired group $\neq$ $\pi_j$ for the second paired group.
  • H1 two sided: P(first score of a pair exceeds second score of a pair) $\neq$ P(second score of a pair exceeds first score of a pair)
  • H1 right sided: P(first score of a pair exceeds second score of a pair) > P(second score of a pair exceeds first score of a pair)
  • H1 left sided: P(first score of a pair exceeds second score of a pair) < P(second score of a pair exceeds first score of a pair)
If the dependent variable is measured on a continuous scale, this can also be formulated as:
  • H1 two sided: the population median of the difference scores is different from zero
  • H1 right sided: the population median of the difference scores is larger than zero
  • H1 left sided: the population median of the difference scores is smaller than zero
H1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
AssumptionsAssumptionsAssumptionsAssumptionsAssumptions
  • Within each population, the scores on the dependent variable are normally distributed
  • 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 $z$ 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 of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
  • Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
  • 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
Test statisticTest statisticTest statisticTest statisticTest statistic
$t = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{\sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}}$
Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $s^2_1$ is the sample variance in group 1, $s^2_2$ is the sample variance in group 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 $\sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}$ 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 $\mu_2 > \mu_1$.
$z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
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 $\pi_2 > \pi_1.$
Computing the test statistic is a bit complicated and involves matrix algebra. Unless you are following a technical course, you probably won't need to calculate it by hand.$W = $ number of difference scores that is larger than 0$z = \dfrac{\bar{y} - \mu_0}{\sigma / \sqrt{N}}$
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 $\mu_0$.
Sampling distribution of $t$ if H0 were trueSampling distribution of $z$ if H0 were trueSampling distribution of the test statistic if H0 were trueSampling distribution of $W$ if H0 were trueSampling distribution of $z$ if H0 were true
Approximately the $t$ distribution with $k$ degrees of freedom, with $k$ equal to
$k = \dfrac{\Bigg(\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}\Bigg)^2}{\dfrac{1}{n_1 - 1} \Bigg(\dfrac{s^2_1}{n_1}\Bigg)^2 + \dfrac{1}{n_2 - 1} \Bigg(\dfrac{s^2_2}{n_2}\Bigg)^2}$
or
$k$ = the smaller of $n_1$ - 1 and $n_2$ - 1

First definition of $k$ is used by computer programs, second definition is often used for hand calculations.
Approximately the standard normal distributionApproximately the chi-squared distribution with $J - 1$ degrees of freedomThe 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.
Standard normal distribution
Significant?Significant?Significant?Significant?Significant?
Two sided: Right sided: Left sided: Two sided: Right sided: Left sided: If we denote the test statistic as $X^2$:
  • 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 $\alpha$
If $n$ is small, the table for the binomial distribution should be used:
Two sided:
  • Check if $W$ observed in sample is in the rejection region or
  • Find two sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$
Right sided:
  • Check if $W$ observed in sample is in the rejection region or
  • Find right sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$
Left sided:
  • Check if $W$ observed in sample is in the rejection region or
  • Find left sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$

If $n$ is large, the table for standard normal probabilities can be used:
Two sided: Right sided: Left sided:
Two sided: Right sided: Left sided:
Approximate $C\%$ confidence interval for $\mu_1 - \mu_2$Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$n.a.n.a.$C\%$ confidence interval for $\mu$
$(\bar{y}_1 - \bar{y}_2) \pm t^* \times \sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}$
where the critical value $t^*$ is the value under the $t_{k}$ 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.
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)
--$\bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}}$
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.
n.a.n.a.n.a.n.a.Effect size
----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 $\mu_0.$
Visual representationn.a.n.a.n.a.Visual representation
Two sample t test - equal variances not assumed
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One sample z test
n.a.Equivalent ton.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.- Two sided sign test is equivalent to -
Example contextExample contextExample contextExample contextExample context
Is the average mental health score different between men and women?Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic.Subjects are asked to taste three different types of mayonnaise, and to indicate which of the three types of mayonnaise they like best. They then have to drink a glass of beer, and taste and rate the three types of mayonnaise again. Does drinking a beer change which type of mayonnaise people like best?Do people tend to score higher on mental health after a mindfulness course?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 $\sigma = 3.$
SPSSSPSSSPSSSPSSn.a.
Analyze > Compare Means > Independent-Samples T Test...
  • Put your dependent (quantitative) variable in the box below Test Variable(s) and your independent (grouping) variable in the box below Grouping Variable
  • Click on the Define Groups... button. If you can't click on it, first click on the grouping variable so its background turns yellow
  • Fill in the value you have used to indicate your first group in the box next to Group 1, and the value you have used to indicate your second group in the box next to Group 2
  • Continue and click OK
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 > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
  • Put the two paired variables in the boxes below Variable 1 and Variable 2
  • Under Test Type, select the Marginal Homogeneity test
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
  • Put the two paired variables in the boxes below Variable 1 and Variable 2
  • Under Test Type, select the Sign test
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JamoviJamovin.a.Jamovin.a.
T-Tests > Independent Samples T-Test
  • Put your dependent (quantitative) variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable
  • Under Tests, select Welch's
  • Under Hypothesis, select your alternative hypothesis
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
-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
  • Put the two paired variables in the box below Measures
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