11
$\begingroup$

I read about the 68-95-99.7 rule that shows

probability levels

I also read that a Z value of 1.96 gives a confidence level of 95%

Is it correct to think of Z as the "number of standard deviations" ? Only I have never seen it mentioned that way.

$\endgroup$
1
  • 1
    $\begingroup$ For future readers of this page. Don't confuse the statistical ratio z (which this question refers to) with the Z-factor used to assess the signal-to-noise ratio of a screening assay. They are not at all related. graphpad.com/support/faq/… $\endgroup$ Jul 29 at 20:02
11
$\begingroup$

No. The z score is not 'the number of standard deviations'. Instead the z-score of a value is the number of standard deviations that value is above the mean. A z-score of 1.7 is 1.7 standard deviations above the mean. A z score of -1 is one standard deviation below the mean, and so on.

This is not mere nitpicking, it's essential to correctly conveying your meaning. I have seen exactly this imprecision in relation to z-scores lead to error on numerous occasion. Stats is not the place for woolly thinking and muddled words $-$ it is tricky enough when you say exactly what you mean.

$\endgroup$
6
  • 1
    $\begingroup$ Yes, thanks, fixed; my phone's not a great way to write answers and I do sometimes miss the typos (or in many cases, stray autocorrects). You're also free to edit answers when there's an annoying error as long as it doesn't alter the intent. $\endgroup$
    – Glen_b
    Jul 30 at 6:59
  • 1
    $\begingroup$ An edit needs 6 characters, so I didn't know what other 5 to put in ;-) $\endgroup$
    – Kirsten
    Jul 30 at 7:08
  • 14
    $\begingroup$ I know the answer claims it is not mere nitpicking, but it really seems like nitpicking to me. A z score is "not the number of standard deviations", but it is "the number of standard deviations ..." $\endgroup$
    – Curt F.
    Jul 30 at 15:41
  • 5
    $\begingroup$ @CurtF. Suppose the mean is 10, the SD is 2, and our score is 14. Interpreting z-score as "the number of standard deviations" leads to calculating the z-score as 14/2 = 7, whereas it should be (14-10)/2 = 2. For somebody familiar with stats that over-literal interpretation is so obviously wrong that we might not even consider it (I didn't until I read Glen's answer) but it's plausible that a novice might take it too literally. $\endgroup$ Jul 31 at 0:24
  • 2
    $\begingroup$ @GeoffreyBrent Thanks for the explanation, it actually didn't occur to me even after reading Glen_b's answer. Spelling it out could improve the answer. $\endgroup$
    – Gala
    Aug 1 at 8:39
22
$\begingroup$

Yes. A Z value of a particular data point tells you how many standard deviations it is from its mean. Z=0 means it has the same value as the population mean, Z=-1 means it is 1std lower than its mean etc. The probability that an observation will lie within the interval of its population mean plus/minus two times the standard deviation is 95%. This is the connection between z scores and confidence intervals.

$\endgroup$
4
  • 9
    $\begingroup$ "... given that it is sampled from a normal distribution" is not right. You can standardize any data with a defined mean and standard deviation, turning it into Z-scores. By definition, the Z-score represents number of standard deviations away form the mean, regardless of the distribution. This generally isn't terribly useful for non-normal distributions, but a Z-score of 1 always means that the value is 1 standard deviation above the mean, no matter what the distribution is. $\endgroup$ Jul 29 at 19:35
  • $\begingroup$ I edited my answer conform your suggestion, but now I am uncertain again. I agree that you can standardize any data with defined mean and standard deviation regardless of distribution. But this transformation does not necessarily result in a bell shaped curve. I included the sampling condition, because the questionner already mentioned the 68-95-99 rule. For this rule to have meaning the original dataset should be sampled from something approximately normal, right? $\endgroup$ Jul 30 at 20:37
  • $\begingroup$ @confusedstudent The 68-95-99 rule requires normalcy, but Z-scores can be useful even when we can't apply that rule. For instance, Chebychev's inequality](en.wikipedia.org/wiki/Chebyshev%27s_inequality) gives us a 0-75-89-94-... bound even for non-normal distributions. (That is, at least 3/4 of observations must have a Z-score of magnitude <= 2, 8/9 <= 3, 15/16 <= 4, and so on.) This is considerably looser than the 68-95-99 rule, but it can still be useful as a bound on rare events. $\endgroup$ Jul 31 at 0:35
  • 2
    $\begingroup$ @confusedstudent The 68-95-99 rule only applies to normally distributed variables. You've removed the normality requirement correctly in the Z-score standard deviation definition, but need to put it back in for the probability statement - "The probability that an observation will lie within the interval of its population mean plus/minus two times the standard deviation is 95%, for normally distributed variables." For non-normal distributions, Z=1 is still one standard deviation from the mean, but 68% of values won't be between Z=1 and Z=-1. $\endgroup$ Aug 2 at 17:10
3
$\begingroup$

no

Sometimes z-score refers to a quantile randomized z-score, where the quantiles of distribution are mapped to z of a standard normal, so by construction, z-score of one is bigger than 68.27% percent of values in the distribution, regardless of how many standard deviations from a mean a value is.

$\endgroup$
2
  • $\begingroup$ wouldn't it be exactly? not bigger? $\endgroup$
    – Kirsten
    Jul 30 at 20:20
  • 1
    $\begingroup$ @KirstenGreed I suppose I was a bit imprecise in my language. What I meant is that if you score for example on an IQ test in a 68th percintile, that would mean your score is bigger than 68% of people who took the test, and your IQ z-score would be 1 $\endgroup$
    – rep_ho
    Jul 30 at 20:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.