Two sample t test - equal variances not assumed - overview

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Two sample $t$ test - equal variances not assumed
Spearman's rho
McNemar's test
$z$ test for a single proportion
Independent/grouping variableVariable 1Independent variableIndependent variable
One categorical with 2 independent groupsOne of ordinal level2 paired groupsNone
Dependent variableVariable 2Dependent variableDependent variable
One quantitative of interval or ratio levelOne of ordinal levelOne categorical with 2 independent groupsOne categorical with 2 independent groups
Null 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: $\rho_s = 0$

Here $\rho_s$ is the Spearman correlation in the population. The Spearman correlation is a measure for the strength and direction of the monotonic relationship between two variables of at least ordinal measurement level.

In words, the null hypothesis would be:

H0: there is no monotonic relationship between the two variables in the population.

Let's say that the scores on the dependent variable are scored 0 and 1. Then for each pair of scores, the data allow four options:

  1. First score of pair is 0, second score of pair is 0
  2. First score of pair is 0, second score of pair is 1 (switched)
  3. First score of pair is 1, second score of pair is 0 (switched)
  4. First score of pair is 1, second score of pair is 1
The null hypothesis H0 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) = P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is the same as the probability that a pair of scores switches from 1 to 0.

Other formulations of the null hypothesis are:

  • H0: $\pi_1 = \pi_2$, where $\pi_1$ is the population proportion of ones for the first paired group and $\pi_2$ is the population proportion of ones for the second paired group
  • H0: for each pair of scores, P(first score of pair is 1) = P(second score of pair is 1)

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.
Alternative 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: $\rho_s \neq 0$
H1 right sided: $\rho_s > 0$
H1 left sided: $\rho_s < 0$

The alternative hypothesis H1 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) $\neq$ P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is not the same as the probability that a pair of scores switches from 1 to 0.

Other formulations of the alternative hypothesis are:

  • H1: $\pi_1 \neq \pi_2$
  • H1: for each pair of scores, P(first score of pair is 1) $\neq$ P(second score of pair is 1)

H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
AssumptionsAssumptionsAssumptionsAssumptions
  • 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 of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
Note: this assumption is only important for the significance test, not for the correlation coefficient itself. The correlation coefficient itself just measures the strength of the monotonic relationship between two variables.
  • Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
  • Sample size is large enough for $z$ to be approximately normally distributed. Rule of thumb:
    • Significance test: $N \times \pi_0$ and $N \times (1 - \pi_0)$ are each larger than 10
    • Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures in sample are each 15 or more
    • Plus four 90%, 95%, or 99% confidence interval: total sample size is 10 or more
  • Sample is a simple random sample from the population. That is, observations are independent of one another
If the sample size is too small for $z$ to be approximately normally distributed, the binomial test for a single proportion should be used.
Test 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 $\mu_2 > \mu_1$.
$t = \dfrac{r_s \times \sqrt{N - 2}}{\sqrt{1 - r_s^2}} $
Here $r_s$ is the sample Spearman correlation and $N$ is the sample size. The sample Spearman correlation $r_s$ is equal to the Pearson correlation applied to the rank scores.
$X^2 = \dfrac{(b - c)^2}{b + c}$
Here $b$ is the number of pairs in the sample for which the first score is 0 while the second score is 1, and $c$ is the number of pairs in the sample for which the first score is 1 while the second score is 0.
$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$
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.
Sampling distribution of $t$ if H0 were trueSampling distribution of $t$ if H0 were trueSampling distribution of $X^2$ 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 $t$ distribution with $N - 2$ degrees of freedom

If $b + c$ is large enough (say, > 20), approximately the chi-squared distribution with 1 degree of freedom.

If $b + c$ is small, the Binomial($n$, $P$) distribution should be used, with $n = b + c$ and $P = 0.5$. In that case the test statistic becomes equal to $b$.

Approximately the standard normal distribution
Significant?Significant?Significant?Significant?
Two sided: Right sided: Left sided: Two sided: Right sided: Left sided: For test statistic $X^2$:
  • 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 $b + c$ is small, the table for the binomial distribution should be used, with as test statistic $b$:
  • Check if $b$ observed in sample is in the rejection region or
  • Find two sided $p$ value corresponding to observed $b$ and check if it is equal to or smaller than $\alpha$
Two sided: Right sided: Left sided:
Approximate $C\%$ confidence interval for $\mu_1 - \mu_2$n.a.n.a.Approximate $C\%$ confidence interval for $\pi$
$(\bar{y}_1 - \bar{y}_2) \pm t^* \times \sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}$
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):
  • $p \pm z^* \times \sqrt{\dfrac{p(1 - p)}{N}}$
    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)
With plus four method:
  • $p_{plus} \pm z^* \times \sqrt{\dfrac{p_{plus}(1 - p_{plus})}{N + 4}}$
    where $p_{plus} = \dfrac{X + 2}{N + 4}$ 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)
Visual representationn.a.n.a.n.a.
Two sample t test - equal variances not assumed
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n.a.n.a.Equivalent toEquivalent to
--
  • When testing two sided: goodness of fit test, with a categorical variable with 2 levels.
  • When $N$ is large, the $p$ value from the $z$ test for a single proportion approaches the $p$ value from the binomial test for a single proportion. The $z$ test for a single proportion is just a large sample approximation of the binomial test for a single proportion.
Example contextExample contextExample contextExample context
Is the average mental health score different between men and women?Is there a monotonic relationship between physical health and mental health?Does a tv documentary about spiders change whether people are afraid (yes/no) of spiders?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.
SPSSSPSSSPSSSPSS
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
Analyze > Correlate > Bivariate...
  • Put your two variables in the box below Variables
  • Under Correlation Coefficients, select Spearman
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 McNemar test
Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
  • Put your dichotomous variable in the box below Test Variable List
  • Fill in the value for $\pi_0$ in the box next to Test Proportion
If computation time allows, SPSS will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution
JamoviJamoviJamoviJamovi
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
Regression > Correlation Matrix
  • Put your two variables in the white box at the right
  • Under Correlation Coefficients, select Spearman
  • Under Hypothesis, select your alternative hypothesis
Frequencies > Paired Samples - McNemar test
  • Put one of the two paired variables in the box below Rows and the other paired variable in the box below Columns
Frequencies > 2 Outcomes - Binomial test
  • Put your dichotomous variable in the white box at the right
  • Fill in the value for $\pi_0$ in the box next to Test value
  • Under Hypothesis, select your alternative hypothesis
Jamovi will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution
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