One sample z test for the mean - overview

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One sample $z$ test for the mean
Wilcoxon signed-rank test
McNemar's test
You cannot compare more than 3 methods
Independent variableIndependent variableIndependent variable
None2 paired groups2 paired groups
Dependent variableDependent variableDependent variable
One quantitative of interval or ratio levelOne quantitative of interval or ratio levelOne categorical with 2 independent groups
Null hypothesisNull hypothesisNull hypothesis
H0: $\mu = \mu_0$

Here $\mu$ is the population mean, and $\mu_0$ is the population mean according to the null hypothesis.
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.

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)

Alternative hypothesisAlternative hypothesisAlternative hypothesis
H1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
H1 two sided: $m \neq 0$
H1 right sided: $m > 0$
H1 left sided: $m < 0$

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)

AssumptionsAssumptionsAssumptions
  • 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
  • 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.
  • Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
Test statisticTest statisticTest statistic
$z = \dfrac{\bar{y} - \mu_0}{\sigma / \sqrt{N}}$
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 $\mu_0$.
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.
$X^2 = \dfrac{(b - c)^2}{b + c}$
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.
Sampling distribution of $z$ if H0 were trueSampling distribution of $W_1$ and of $W_2$ if H0 were trueSampling distribution of $X^2$ if H0 were true
Standard normal distributionSampling 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.

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 $b$.

Significant?Significant?Significant?
Two sided: Right sided: Left sided: For large samples, the table for standard normal probabilities can be used:
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 $\alpha$
$C\%$ confidence interval for $\mu$n.a.n.a.
$\bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}}$
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.
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Effect sizen.a.n.a.
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 $\mu_0.$
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Visual representationn.a.n.a.
One sample z test
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n.a.n.a.Equivalent to
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Example contextExample contextExample context
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 $\sigma = 3.$Is the median of the differences between the mental health scores before and after an intervention different from 0?Does a tv documentary about spiders change whether people are afraid (yes/no) of spiders?
n.a.SPSSSPSS
-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
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
n.a.JamoviJamovi
-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
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
Practice questionsPractice questionsPractice questions