One sample z test for the mean - overview
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One sample $z$ test for the mean | One sample Wilcoxon signed-rank test |
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Independent variable | Independent variable | |
None | None | |
Dependent variable | Dependent variable | |
One quantitative of interval or ratio level | One of ordinal level | |
Null hypothesis | Null hypothesis | |
$\mu = \mu_0$
$\mu$ is the unknown population mean; $\mu_0$ is the population mean according to the null hypothesis | $m = m_0$
$m$ is the unknown population median; $m_0$ is the population median according to the null hypothesis | |
Alternative hypothesis | Alternative hypothesis | |
Two sided: $\mu \neq \mu_0$ Right sided: $\mu > \mu_0$ Left sided: $\mu < \mu_0$ |
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Assumptions | Assumptions | |
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Test statistic | Test statistic | |
$z = \dfrac{\bar{y} - \mu_0}{\sigma / \sqrt{N}}$
$\bar{y}$ is the sample mean, $\mu_0$ is the population mean according to H0, $\sigma$ is the population standard deviation, $N$ 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; 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:
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Sampling distribution of $z$ if H0 were true | Sampling distribution of $W_1$ and of $W_2$ if H0 were true | |
Standard normal | 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 a 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 a 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: the formula for the standard deviations $\sigma_{W_1}$ and $\sigma_{W_2}$ is more complicated if ties are present in the data. | |
Significant? | Significant? | |
Two sided:
| For large samples, the table for standard normal probabilities can be used: Two sided:
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$C\%$ confidence interval for $\mu$ | n.a. | |
$\bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}}$
where $z^*$ 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. | - | |
Effect size | n.a. | |
Cohen's $d$: Standardized difference between the sample mean and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{\sigma}$$ Indicates how many standard deviations $\sigma$ the sample mean $\bar{y}$ is removed from $\mu_0$ | - | |
Visual representation | n.a. | |
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Example context | Example 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 mental health score different from 50? | |
n.a. | SPSS | |
- | Specify the measurement level of your variable on the Variable View tab, in the column named Measure. Then go to:
Analyze > Nonparametric Tests > One Sample...
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n.a. | Jamovi | |
- | T-Tests > One Sample T-Test
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Practice questions | Practice questions | |