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I am wondering if somebody could explain to me why the Z-statistic formula (for hypothesis testing) is derived as it is. Correct me if I'm wrong about my deductions: the numerator portion of the Z-statistic is the measure of the deviation of the estimated mean value of X from the actual mean value of X. However, I do not get the denominator portion. Is it for scaling purposes so that the deviation can be expressed in "relative" error terms instead of "absolute" error terms? Why do we have to divide by the standard error/standard deviation of sampling distribution of X? Thanks a lot!

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Let $z$ be a random variable with standard normal distribution. That is

$$ z \sim \mathcal{N}(0,1)$$

Multiply $z$ by a positive number $\sigma$. The distribution of $\sigma z$ is

$$ \sigma z \sim \mathcal{N}(0,\sigma^2)$$

Now, add a constant $\mu$ to $z\sigma$. The distribution of $\mu + z\sigma$ is

$$ \mu + \sigma z \sim \mathcal{N}(\mu,\sigma^2)$$

Now, to your question. Suppose I draw $x$ from a $\mathcal{N}(\mu,\sigma^2)$. What happens if I subtract $\mu$ and divide by $\sigma$? I get

$$ \dfrac{x-\mu}{\sigma} \sim \mathcal{N}(0,1) $$

The reason why we scale the difference between our draw and our hypothesized population mean by the standard deviation is to put the test statistic on a scale which facilitates calculating p-values. Back when critical regions for hypothesis tests were put in tables, it was much easier to choose a particular mean and standard deviation for the normal and just transform the test statistic appropriately rather than have multiple tables for different means and standard deviations.

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I am an undergrad, but from my interpretation (and I may be entirely wrong), with any statistics, mathematicians use standard deviation to measure the error.

You know how ratio works where you have part/whole? And the bigger the numerator, the bigger the part is when compared to the whole?

Well standard deviation works kinda like that, but a little differently. With statistic/deviation, we find our data with statistic, but then we "normalize" it by dividing it by standard deviation. This way we can tell how much deviation there is in the test statistic, and also compare it to other things measured in standard deviation. Again, I may be wrong so someone correct me if I am.

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    $\begingroup$ You seem to be conflating the standard error and the standard deviation $\endgroup$ – Frans Rodenburg Jan 26 at 4:16

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