Goodness of fit test - overview

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Goodness of fit test
McNemar's test
$z$ test for a single proportion
Spearman's rho
Independent variableIndependent variableIndependent variableVariable 1
None2 paired groupsNoneOne of ordinal level
Dependent variableDependent variableDependent variableVariable 2
One categorical with $J$ independent groups ($J \geqslant 2$)One categorical with 2 independent groupsOne categorical with 2 independent groupsOne of ordinal level
Null hypothesisNull hypothesisNull hypothesisNull hypothesis
  • H0: the population proportions in each of the $J$ conditions are $\pi_1$, $\pi_2$, $\ldots$, $\pi_J$
or equivalently
  • H0: the probability of drawing an observation from condition 1 is $\pi_1$, the probability of drawing an observation from condition 2 is $\pi_2$, $\ldots$, the probability of drawing an observation from condition $J$ is $\pi_J$

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.
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.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
  • H1: the population proportions are not all as specified under the null hypothesis
or equivalently
  • H1: the probabilities of drawing an observation from each of the conditions are not all as specified under the null hypothesis

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: $\pi < \pi_0$
H1 two sided: $\rho_s \neq 0$
H1 right sided: $\rho_s > 0$
H1 left sided: $\rho_s < 0$
AssumptionsAssumptionsAssumptionsAssumptions
  • Sample size is large enough for $X^2$ to be approximately chi-squared distributed. Rule of thumb: all $J$ expected cell counts are 5 or more
  • Sample is a simple random sample from the population. That is, 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 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.
  • 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.
Test statisticTest statisticTest statisticTest statistic
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here the expected cell count for one cell = $N \times \pi_j$, the observed cell count is the observed sample count in that same cell, and the sum is over all $J$ cells.
$X^2 = \dfrac{(b - c)^2}{b + c}$
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.
$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.
Sampling distribution of $X^2$ if H0 were trueSampling distribution of $X^2$ if H0 were trueSampling distribution of $z$ if H0 were trueSampling distribution of $t$ if H0 were true
Approximately the chi-squared distribution with $J - 1$ 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 distributionApproximately the $t$ distribution with $N - 2$ degrees of freedom
Significant?Significant?Significant?Significant?
  • 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 $\alpha$
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: Two sided: Right sided: Left sided:
n.a.n.a.Approximate $C\%$ confidence interval for $\pi$n.a.
--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)
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n.a.Equivalent toEquivalent ton.a.
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  • 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.
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Example contextExample contextExample contextExample context
Is the proportion of people with a low, moderate, and high social economic status in the population different from $\pi_{low} = 0.2,$ $\pi_{moderate} = 0.6,$ and $\pi_{high} = 0.2$?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.Is there a monotonic relationship between physical health and mental health?
SPSSSPSSSPSSSPSS
Analyze > Nonparametric Tests > Legacy Dialogs > Chi-square...
  • Put your categorical variable in the box below Test Variable List
  • Fill in the population proportions / probabilities according to $H_0$ in the box below Expected Values. If $H_0$ states that they are all equal, just pick 'All categories equal' (default)
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
Analyze > Correlate > Bivariate...
  • Put your two variables in the box below Variables
  • Under Correlation Coefficients, select Spearman
JamoviJamoviJamoviJamovi
Frequencies > N Outcomes - $\chi^2$ Goodness of fit
  • Put your categorical variable in the box below Variable
  • Click on Expected Proportions and fill in the population proportions / probabilities according to $H_0$ in the boxes below Ratio. If $H_0$ states that they are all equal, you can leave the ratios equal to the default values (1)
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
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
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