Maria and Eric meet z-scores

Maria and Eric

One of the problems many students have when first learning statistics is deciding when to reject the null hypothesis. Z is small and low probability means it is not likely to occur so you reject, right? (Wrong!). P > .86 and when you have a large z-score you reject the null hypothesis, so with p = .86 you reject, right? (Wrong!)

Enter Maria and Eric to help us explain z-scores. Eric is 6 foot 4, or 76 inches tall. That is a high number, both in mathematical terms and off of the ground. I want to determine if Eric’s height is significantly different from the mean. I use the heart data set included with the SAS Web Editor to compute mean and standard deviation for an adult male, as so:

proc sort data=sashelp.heart out=temp ;
by sex ;
proc univariate data=temp;
var height  ;
by sex ;

I find that the mean is 67.6 and the standard deviation is 2.7. I then compute my z-score which is the obtained value  of 76 inches, minus the mean value of 67.6 divided by the standard deviation of 2.7. This gives me a z-score of 3.1 which tells me that Eric is 3.1 standard deviations above the mean.

Listen carefully here — there is a SMALL probability of LARGE differences from the mean.

A z-score of 1.96 occurs less than 5% of the time, that is about two standard deviations from the mean. How often does a z-score of 3.1 occur? p < .002. So, even though he is a LARGE difference from the average height,  people who are that tall represent a small proportion of the population.

We would therefore REJECT that null hypothesis that there is NO difference between Eric’s height and the average and conclude that he is significantly taller than average.

Since Maria just sniffed disrespectfully,

I could have told you that!

(I can hear you over the Internet) … we will now examine Maria.

She is 5 foot 4, or 64 inches. The average height for a woman is 62.6 inches and the standard deviation is 2.5.  Her z-score is (64-62.6)/ 2.5 = .56 and the probability of a z-score that high or larger is almost 60%,  p> .59 . So, she differs a SMALL amount from the average and that will happen a LARGE proportion of the time.

SO … would you accept or reject the null hypothesis that Maria is no different than the average height for women? Discuss.

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  1. If I understand properly the terms, a z-score indicates how different a data point is from the standard deviation and P indicates how likely it is for a data point to be this different.

    z-score = 0 : perfectly average
    -1< z-score < 1 : different from average but within normal deviation: in engineering terms, I think we call that "pretty average"
    z-score1 : outside the norm

    For maria, with a z-score at 0.56, her height is well within the standard deviation and P indicates that it is pretty common, so the null hypothesis should be accepted:

    Maria’s height is not unusual at all (and I could have told you that without learning what a z-score is, thank you very much 😉

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