Paired sample t test - overview
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Paired sample $t$ test | Sign test |
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Independent variable | Independent variable | |
2 paired groups | 2 paired groups | |
Dependent variable | Dependent variable | |
One quantitative of interval or ratio level | One of ordinal level | |
Null hypothesis | Null hypothesis | |
H0: $\mu = \mu_0$
Here $\mu$ is the population mean of the difference scores, and $\mu_0$ is the population mean of the difference scores according to the null hypothesis, which is usually 0. A difference score is the difference between the first score of a pair and the second score of a pair. |
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Alternative hypothesis | Alternative hypothesis | |
H1 two sided: $\mu \neq \mu_0$ H1 right sided: $\mu > \mu_0$ H1 left sided: $\mu < \mu_0$ |
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Assumptions | Assumptions | |
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Test statistic | Test statistic | |
$t = \dfrac{\bar{y} - \mu_0}{s / \sqrt{N}}$
Here $\bar{y}$ is the sample mean of the difference scores, $\mu_0$ is the population mean of the difference scores according to the null hypothesis, $s$ is the sample standard deviation of the difference scores, and $N$ is the sample size (number of difference scores). The denominator $s / \sqrt{N}$ is the standard error of the sampling distribution of $\bar{y}$. The $t$ value indicates how many standard errors $\bar{y}$ is removed from $\mu_0$. | $W = $ number of difference scores that is larger than 0 | |
Sampling distribution of $t$ if H0 were true | Sampling distribution of $W$ if H0 were true | |
$t$ distribution with $N - 1$ degrees of freedom | The 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. | |
Significant? | Significant? | |
Two sided:
| If $n$ is small, the table for the binomial distribution should be used: Two sided:
If $n$ is large, the table for standard normal probabilities can be used: Two sided:
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$C\%$ confidence interval for $\mu$ | n.a. | |
$\bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}}$
where the critical value $t^*$ is the value under the $t_{N-1}$ 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 $\mu$ can also be used as significance test. | - | |
Effect size | n.a. | |
Cohen's $d$: Standardized difference between the sample mean of the difference scores and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{s}$$ Cohen's $d$ indicates how many standard deviations $s$ the sample mean of the difference scores $\bar{y}$ is removed from $\mu_0.$ | - | |
Visual representation | n.a. | |
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Equivalent to | Equivalent to | |
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Two sided sign test is equivalent to
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Example context | Example context | |
Is the average difference between the mental health scores before and after an intervention different from $\mu_0 = 0$? | Do people tend to score higher on mental health after a mindfulness course? | |
SPSS | SPSS | |
Analyze > Compare Means > Paired-Samples T Test...
| Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
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Jamovi | Jamovi | |
T-Tests > Paired Samples T-Test
| 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
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Practice questions | Practice questions | |