Two sample t test - equal variances not assumed - overview

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Two sample $t$ test - equal variances not assumed
One way ANOVA
Friedman test
Pearson correlation
Wilcoxon signed-rank test
Independent/grouping variableIndependent/grouping variableIndependent/grouping variableVariable 1Independent variable
One categorical with 2 independent groupsOne categorical with $I$ independent groups ($I \geqslant 2$)One within subject factor ($\geq 2$ related groups)One quantitative of interval or ratio level2 paired groups
Dependent variableDependent variableDependent variableVariable 2Dependent variable
One quantitative of interval or ratio levelOne quantitative of interval or ratio levelOne of ordinal levelOne quantitative of interval or ratio 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.
ANOVA $F$ test:
  • H0: $\mu_1 = \mu_2 = \ldots = \mu_I$
    $\mu_1$ is the population mean for group 1; $\mu_2$ is the population mean for group 2; $\mu_I$ is the population mean for group $I$
$t$ Test for contrast:
  • H0: $\Psi = 0$
    $\Psi$ is the population contrast, defined as $\Psi = \sum a_i\mu_i$. Here $\mu_i$ is the population mean for group $i$ and $a_i$ is the coefficient for $\mu_i$. The coefficients $a_i$ sum to 0.
$t$ Test multiple comparisons:
  • H0: $\mu_g = \mu_h$
    $\mu_g$ is the population mean for group $g$; $\mu_h$ is the population mean for group $h$
H0: the population scores in any of the related groups are not systematically higher or lower than the population scores in any of the other related groups

Usually the related groups are the different measurement points. Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
H0: $\rho = \rho_0$

Here $\rho$ is the Pearson correlation in the population, and $\rho_0$ is the Pearson correlation in the population according to the null hypothesis (usually 0). The Pearson correlation is a measure for the strength and direction of the linear relationship between two variables of at least interval measurement level.
H0: $m = 0$

Here $m$ is the population median of the difference scores. A difference score is the difference between the first score of a pair and the second score of a pair.

Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
H1 two sided: $\mu_1 \neq \mu_2$
H1 right sided: $\mu_1 > \mu_2$
H1 left sided: $\mu_1 < \mu_2$
ANOVA $F$ test:
  • H1: not all population means are equal
$t$ Test for contrast:
  • H1 two sided: $\Psi \neq 0$
  • H1 right sided: $\Psi > 0$
  • H1 left sided: $\Psi < 0$
$t$ Test multiple comparisons:
  • H1 - usually two sided: $\mu_g \neq \mu_h$
H1: the population scores in some of the related groups are systematically higher or lower than the population scores in other related groups H1 two sided: $\rho \neq \rho_0$
H1 right sided: $\rho > \rho_0$
H1 left sided: $\rho < \rho_0$
H1 two sided: $m \neq 0$
H1 right sided: $m > 0$
H1 left sided: $m < 0$
AssumptionsAssumptionsAssumptionsAssumptions of test for correlationAssumptions
  • 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
  • Within each population, the scores on the dependent variable are normally distributed
  • The standard deviation of the scores on the dependent variable is the same in each of the populations: $\sigma_1 = \sigma_2 = \ldots = \sigma_I$
  • Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2, $\ldots$, group $I$ sample is an independent SRS from population $I$. That is, within and between groups, observations are independent of one another
  • Sample of 'blocks' (usually the subjects) is a simple random sample from the population. That is, blocks are independent of one another
  • In the population, the two variables are jointly normally distributed (this covers the normality, homoscedasticity, and linearity assumptions)
  • Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
Note: these assumptions are only important for the significance test and confidence interval, not for the correlation coefficient itself. The correlation coefficient just measures the strength of the linear relationship between two variables.
  • The population distribution of the difference scores is symmetric
  • Sample of difference scores is a simple random sample from the population of difference scores. That is, difference scores are independent of one another
Note: sometimes it considered sufficient for the data to be measured on an ordinal scale, rather than an interval or ratio scale. However, since the test statistic is based on ranked difference scores, we need to know whether a change in scores from, say, 6 to 7 is larger than/smaller than/equal to a change from 5 to 6. This is impossible to know for ordinal scales, since for these scales the size of the difference between values is meaningless.
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 $\mu_2 > \mu_1$.
ANOVA $F$ test:
  • $\begin{aligned}[t] F &= \dfrac{\sum\nolimits_{subjects} (\mbox{subject's group mean} - \mbox{overall mean})^2 / (I - 1)}{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2 / (N - I)}\\ &= \dfrac{\mbox{sum of squares between} / \mbox{degrees of freedom between}}{\mbox{sum of squares error} / \mbox{degrees of freedom error}}\\ &= \dfrac{\mbox{mean square between}}{\mbox{mean square error}} \end{aligned} $
    where $N$ is the total sample size, and $I$ is the number of groups.
    Note: mean square between is also known as mean square model, and mean square error is also known as mean square residual or mean square within.
$t$ Test for contrast:
  • $t = \dfrac{c}{s_p\sqrt{\sum \dfrac{a^2_i}{n_i}}}$
    Here $c$ is the sample estimate of the population contrast $\Psi$: $c = \sum a_i\bar{y}_i$, with $\bar{y}_i$ the sample mean in group $i$. $s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA, $a_i$ is the contrast coefficient for group $i$, and $n_i$ is the sample size of group $i$.
    Note that if the contrast compares only two group means with each other, this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). In that case the only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
$t$ Test multiple comparisons:
  • $t = \dfrac{\bar{y}_g - \bar{y}_h}{s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}}}$
    $\bar{y}_g$ is the sample mean in group $g$, $\bar{y}_h$ is the sample mean in group $h$, $s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA, $n_g$ is the sample size of group $g$, and $n_h$ is the sample size of group $h$.
    Note that this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). The only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
$Q = \dfrac{12}{N \times k(k + 1)} \sum R^2_i - 3 \times N(k + 1)$

Here $N$ is the number of 'blocks' (usually the subjects - so if you have 4 repeated measurements for 60 subjects, $N$ equals 60), $k$ is the number of related groups (usually the number of repeated measurements), and $R_i$ is the sum of ranks in group $i$.

Remember that multiplication precedes addition, so first compute $\frac{12}{N \times k(k + 1)} \times \sum R^2_i$ and then subtract $3 \times N(k + 1)$.

Note: if ties are present in the data, the formula for $Q$ is more complicated.
Test statistic for testing H0: $\rho = 0$:
  • $t = \dfrac{r \times \sqrt{N - 2}}{\sqrt{1 - r^2}} $
    where $r$ is the sample correlation $r = \frac{1}{N - 1} \sum_{j}\Big(\frac{x_{j} - \bar{x}}{s_x} \Big) \Big(\frac{y_{j} - \bar{y}}{s_y} \Big)$ and $N$ is the sample size
Test statistic for testing values for $\rho$ other than $\rho = 0$:
  • $z = \dfrac{r_{Fisher} - \rho_{0_{Fisher}}}{\sqrt{\dfrac{1}{N - 3}}}$
    • $r_{Fisher} = \dfrac{1}{2} \times \log\Bigg(\dfrac{1 + r}{1 - r} \Bigg )$, where $r$ is the sample correlation
    • $\rho_{0_{Fisher}} = \dfrac{1}{2} \times \log\Bigg( \dfrac{1 + \rho_0}{1 - \rho_0} \Bigg )$, where $\rho_0$ is the population correlation according to H0
Two different types of test statistics can be used, but both will result in the same test outcome. We will denote the first option the $W_1$ statistic (also known as the $T$ statistic), and the second option the $W_2$ statistic. In order to compute each of the test statistics, follow the steps below:
  1. For each subject, compute the sign of the difference score $\mbox{sign}_d = \mbox{sgn}(\mbox{score}_2 - \mbox{score}_1)$. The sign is 1 if the difference is larger than zero, -1 if the diffence is smaller than zero, and 0 if the difference is equal to zero.
  2. For each subject, compute the absolute value of the difference score $|\mbox{score}_2 - \mbox{score}_1|$.
  3. Exclude subjects with a difference score of zero. This leaves us with a remaining number of difference scores equal to $N_r$.
  4. Assign ranks $R_d$ to the $N_r$ remaining absolute difference scores. The smallest absolute difference score corresponds to a rank score of 1, and the largest absolute difference score corresponds to a rank score of $N_r$. If there are ties, assign them the average of the ranks they occupy.
Then compute the test statistic:

  • $W_1 = \sum\, R_d^{+}$
    or
    $W_1 = \sum\, R_d^{-}$
    That is, sum all ranks corresponding to a positive difference or sum all ranks corresponding to a negative difference. Theoratically, both definitions will result in the same test outcome. However:
    • tables with critical values for $W_1$ are usually based on the smaller of $\sum\, R_d^{+}$ and $\sum\, R_d^{-}$. So if you are using such a table, pick the smaller one.
    • If you are using the normal approximation to find the $p$ value, it makes things most straightforward if you use $W_1 = \sum\, R_d^{+}$ (if you use $W_1 = \sum\, R_d^{-}$, the right and left sided alternative hypotheses 'flip').
  • $W_2 = \sum\, \mbox{sign}_d \times R_d$
    That is, for each remaining difference score, multiply the rank of the absolute difference score by the sign of the difference score, and then sum all of the products.
n.a.Pooled standard deviationn.a.n.a.n.a.
-$ \begin{aligned} s_p &= \sqrt{\dfrac{(n_1 - 1) \times s^2_1 + (n_2 - 1) \times s^2_2 + \ldots + (n_I - 1) \times s^2_I}{N - I}}\\ &= \sqrt{\dfrac{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2}{N - I}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned} $

Here $s^2_i$ is the variance in group $i.$
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Sampling distribution of $t$ if H0 were trueSampling distribution of $F$ and of $t$ if H0 were trueSampling distribution of $Q$ if H0 were trueSampling distribution of $t$ and of $z$ if H0 were trueSampling distribution of $W_1$ and of $W_2$ 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.
Sampling distribution of $F$:
  • $F$ distribution with $I - 1$ (df between, numerator) and $N - I$ (df error, denominator) degrees of freedom
Sampling distribution of $t$:
  • $t$ distribution with $N - I$ degrees of freedom
If the number of blocks $N$ is large, approximately the chi-squared distribution with $k - 1$ degrees of freedom.

For small samples, the exact distribution of $Q$ should be used.
Sampling distribution of $t$:
  • $t$ distribution with $N - 2$ degrees of freedom
Sampling distribution of $z$:
  • Approximately the standard normal distribution
Sampling distribution of $W_1$:
If $N_r$ is large, $W_1$ is approximately normally distributed with mean $\mu_{W_1}$ and standard deviation $\sigma_{W_1}$ if the null hypothesis were true. Here $$\mu_{W_1} = \frac{N_r(N_r + 1)}{4}$$ $$\sigma_{W_1} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{24}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_1 - \mu_{W_1}}{\sigma_{W_1}}$$ follows approximately the standard normal distribution if the null hypothesis were true.

Sampling distribution of $W_2$:
If $N_r$ is large, $W_2$ is approximately normally distributed with mean $0$ and standard deviation $\sigma_{W_2}$ if the null hypothesis were true. Here $$\sigma_{W_2} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{6}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_2}{\sigma_{W_2}}$$ follows approximately the standard normal distribution if the null hypothesis were true.

If $N_r$ is small, the exact distribution of $W_1$ or $W_2$ should be used.

Note: if ties are present in the data, the formula for the standard deviations $\sigma_{W_1}$ and $\sigma_{W_2}$ is more complicated.
Significant?Significant?Significant?Significant?Significant?
Two sided: Right sided: Left sided: $F$ test:
  • Check if $F$ observed in sample is equal to or larger than critical value $F^*$ or
  • Find $p$ value corresponding to observed $F$ and check if it is equal to or smaller than $\alpha$ (e.g. .01 < $p$ < .025 when $F$ = 3.91, df between = 4, and df error = 20)

$t$ Test for contrast two sided: $t$ Test for contrast right sided: $t$ Test for contrast left sided:
$t$ Test multiple comparisons two sided:
  • Check if $t$ observed in sample is at least as extreme as critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
  • Find two sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons right sided
  • Check if $t$ observed in sample is equal to or larger than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
  • Find right sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons left sided
  • Check if $t$ observed in sample is equal to or smaller than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
  • Find left sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
If the number of blocks $N$ is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$:
  • 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 $\alpha$
$t$ Test two sided: $t$ Test right sided: $t$ Test left sided: $z$ Test two sided: $z$ Test right sided: $z$ Test left sided: For large samples, the table for standard normal probabilities can be used:
Two sided: Right sided: Left sided:
Approximate $C\%$ confidence interval for $\mu_1 - \mu_2$$C\%$ confidence interval for $\Psi$, for $\mu_g - \mu_h$, and for $\mu_i$n.a.Approximate $C$% confidence interval for $\rho$n.a.
$(\bar{y}_1 - \bar{y}_2) \pm t^* \times \sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}$
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.
Confidence interval for $\Psi$ (contrast):
  • $c \pm t^* \times s_p\sqrt{\sum \dfrac{a^2_i}{n_i}}$
    where the critical value $t^*$ is the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). Note that $n_i$ is the sample size of group $i$, and $N$ is the total sample size, based on all the $I$ groups.
Confidence interval for $\mu_g - \mu_h$ (multiple comparisons):
  • $(\bar{y}_g - \bar{y}_h) \pm t^{**} \times s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}}$
    where $t^{**}$ depends upon $C$, degrees of freedom ($N - I$), and the multiple comparison procedure. If you do not want to apply a multiple comparison procedure, $t^{**} = t^* = $ the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$. Note that $n_g$ is the sample size of group $g$, $n_h$ is the sample size of group $h$, and $N$ is the total sample size, based on all the $I$ groups.
Confidence interval for single population mean $\mu_i$:
  • $\bar{y}_i \pm t^* \times \dfrac{s_p}{\sqrt{n_i}}$
    where $\bar{y}_i$ is the sample mean in group $i$, $n_i$ is the sample size of group $i$, and the critical value $t^*$ is the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). Note that $n_i$ is the sample size of group $i$, and $N$ is the total sample size, based on all the $I$ groups.
-First compute the approximate $C$% confidence interval for $\rho_{Fisher}$:
  • $lower_{Fisher} = r_{Fisher} - z^* \times \sqrt{\dfrac{1}{N - 3}}$
  • $upper_{Fisher} = r_{Fisher} + z^* \times \sqrt{\dfrac{1}{N - 3}}$
where $r_{Fisher} = \frac{1}{2} \times \log\Bigg(\dfrac{1 + r}{1 - r} \Bigg )$ 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).
Then transform back to get the approximate $C$% confidence interval for $\rho$:
  • lower bound = $\dfrac{e^{2 \times lower_{Fisher}} - 1}{e^{2 \times lower_{Fisher}} + 1}$
  • upper bound = $\dfrac{e^{2 \times upper_{Fisher}} - 1}{e^{2 \times upper_{Fisher}} + 1}$
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n.a.Effect sizen.a.Properties of the Pearson correlation coefficientn.a.
-
  • Proportion variance explained $\eta^2$ and $R^2$:
    Proportion variance of the dependent variable $y$ explained by the independent variable: $$ \begin{align} \eta^2 = R^2 &= \dfrac{\mbox{sum of squares between}}{\mbox{sum of squares total}} \end{align} $$ Only in one way ANOVA $\eta^2 = R^2.$ $\eta^2$ (and $R^2$) is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population.

  • Proportion variance explained $\omega^2$:
    Corrects for the positive bias in $\eta^2$ and is equal to: $$\omega^2 = \frac{\mbox{sum of squares between} - \mbox{df between} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}$$ $\omega^2$ is a better estimate of the explained variance in the population than $\eta^2.$

  • Cohen's $d$:
    Standardized difference between the mean in group $g$ and in group $h$: $$d_{g,h} = \frac{\bar{y}_g - \bar{y}_h}{s_p}$$ Cohen's $d$ indicates how many standard deviations $s_p$ two sample means are removed from each other.
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  • The Pearson correlation coefficient is a measure for the linear relationship between two quantitative variables.
  • The Pearson correlation coefficient squared reflects the proportion of variance explained in one variable by the other variable.
  • The Pearson correlation coefficient can take on values between -1 (perfect negative relationship) and 1 (perfect positive relationship). A value of 0 means no linear relationship.
  • The absolute size of the Pearson correlation coefficient is not affected by any linear transformation of the variables. However, the sign of the Pearson correlation will flip when the scores on one of the two variables are multiplied by a negative number (reversing the direction of measurement of that variable).
    For example:
    • the correlation between $x$ and $y$ is equivalent to the correlation between $3x + 5$ and $2y - 6$.
    • the absolute value of the correlation between $x$ and $y$ is equivalent to the absolute value of the correlation between $-3x + 5$ and $2y - 6$. However, the signs of the two correlation coefficients will be in opposite directions, due to the multiplication of $x$ by $-3$.
  • The Pearson correlation coefficient does not say anything about causality.
  • The Pearson correlation coefficient is sensitive to outliers.
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Visual representationn.a.n.a.n.a.n.a.
Two sample t test - equal variances not assumed
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n.a.ANOVA tablen.a.n.a.n.a.
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ANOVA table

Click the link for a step by step explanation of how to compute the sum of squares.
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n.a.Equivalent ton.a.Equivalent ton.a.
-OLS regression with one categorical independent variable transformed into $I - 1$ code variables:
  • $F$ test ANOVA is equivalent to $F$ test regression model
  • $t$ test for contrast $i$ is equivalent to $t$ test for regression coefficient $\beta_i$ (specific contrast tested depends on how the code variables are defined)
-OLS regression with one independent variable:
  • $b_1 = r \times \frac{s_y}{s_x}$
  • Results significance test ($t$ and $p$ value) testing $H_0$: $\beta_1 = 0$ are equivalent to results significance test testing $H_0$: $\rho = 0$
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Example contextExample contextExample contextExample contextExample context
Is the average mental health score different between men and women?Is the average mental health score different between people from a low, moderate, and high economic class?Is there a difference in depression level between measurement point 1 (pre-intervention), measurement point 2 (1 week post-intervention), and measurement point 3 (6 weeks post-intervention)?Is there a linear relationship between physical health and mental health?Is the median of the differences between the mental health scores before and after an intervention different from 0?
SPSSSPSSSPSSSPSSSPSS
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 > Compare Means > One-Way ANOVA...
  • Put your dependent (quantitative) variable in the box below Dependent List and your independent (grouping) variable in the box below Factor
or
Analyze > General Linear Model > Univariate...
  • Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factor(s)
Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples...
  • Put the $k$ variables containing the scores for the $k$ related groups in the white box below Test Variables
  • Under Test Type, select the Friedman test
Analyze > Correlate > Bivariate...
  • Put your two variables in the box below Variables
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 Wilcoxon test
JamoviJamoviJamoviJamoviJamovi
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
ANOVA > ANOVA
  • Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factors
ANOVA > Repeated Measures ANOVA - Friedman
  • Put the $k$ variables containing the scores for the $k$ related groups in the box below Measures
Regression > Correlation Matrix
  • Put your two variables in the white box at the right
  • Under Correlation Coefficients, select Pearson (selected by default)
  • Under Hypothesis, select your alternative hypothesis
T-Tests > Paired Samples T-Test
  • Put the two paired variables in the box below Paired Variables, one on the left side of the vertical line and one on the right side of the vertical line
  • Under Tests, select Wilcoxon rank
  • Under Hypothesis, select your alternative hypothesis
Practice questionsPractice questionsPractice questionsPractice questionsPractice questions