If X is a random variable from a normal distribution with mean () and standard deviation (), its Z -score may be calculated by subtracting mean from X and dividing the whole by standard deviation. For the average of a sample from a population ‘n’, the mean is and the standard deviation is .
A z -score equal to 1 represents an element, which is 1 standard deviation greater than the mean; a z -score equal to 2 signifies 2 standard deviations greater than the mean; etc. A z -score equal to -1 represents an element, which is 1 standard deviation less than the mean; a z -score equal to -2 signifies 2 standard deviations less than the mean; etc.
The z score table helps to know the percentage of values below (to the left) a z -score in a standard normal distribution. A z score is simply defined as the number of standard deviation from the mean.
Another misuse may be to use the relatively sophisticated Z -score in a famine emergency, when mid upper arm circumference may be the more appropriate diagnostic tool. One clear misuse, which is fortunately rare, is the belief that standardizing measurements will also normalize them.
De Ones et al. (1999) and Ma et al. (1997) look at the use of mid upper-arm circumference for age and height as nutritional status screening indicators. Bible et al. (1987a) (1987b) describe the development of “normalized” curves for the international growth reference, whilst WHO Working Group (1986) looks at the various indices of weight and height and at the biological significance of wasting and stunting.
A z -score describes the position of a raw score in terms of its distance from the mean, when measured in standard deviation units. A standard normal distribution (SND) is a normally shaped distribution with a mean of 0 and a standard deviation (SD) of 1 (see Fig.
(a) it allows researchers to calculate the probability of a score occurring within a standard normal distribution; (b) and enables us to compare two scores that are from different samples (which may have different means and standard deviations). As the formula shows, the z -score is simply the raw score minus the population mean, divided by the population standard deviation.
The value of the z -score tells you how many standard deviations you are away from the mean. A positive z -score indicates the raw score is higher than the mean average.
A negative z -score reveals the raw score is below the mean average. Fig 3 illustrates the important features of any standard normal distribution (SND).
The SND (i.e. z -distribution) is always the same shape as the raw score distribution. The SND allows researchers to calculate the probability of randomly obtaining a score from the distribution (i.e. sample).
For example, there is a 68% probability of randomly selecting a score between -1 and +1 standard deviations from the mean (see Fig. Proportion of a standard normal distribution (SND) in percentages.
The probability of randomly selecting a score between -1.96 and +1.96 standard deviations from the mean is 95% (see Fig. If there is less than a 5% chance of a raw score being selected randomly, then this is a statistically significant result.
As the formula shows, the z -score and standard deviation are multiplied together, and this figure is added to the mean. Check your answer makes sense: If we have a negative z -score the corresponding raw score should be less than the mean, and a positive z -score must correspond to a raw score higher than the mean.
At this point, there's no way of telling because we don't know what people typically score on this test. A quick peek at some of our 100 scores on our first IQ test shows a minimum of 1 and a maximum of 6.
However, we'll gain much more insight into these scores by inspecting their histogram as shown below. This pattern is known as a uniform distribution, and we typically see this when we roll a die a lot of times: numbers 1 through 6 are equally likely to come up.
In a similar vein, if we had plotted scores versus squared scores, our line would have been curved; in contrast to standardizing, taking squares is a non-linear transformation. Due to the central limit theorem, this holds especially for test statistics.
However, if a variable also follows a standard normal distribution, then we also know that 1.5 roughly corresponds to the 95th percentile. A tutor sets a piece of English Literature coursework for the 50 students in his class.
We make the assumption that when the scores are presented on a histogram, the data is found to be normally distributed. However, this does not take into consideration the variation in scores amongst the 50 students (in other words, the standard deviation).
In the next academic year, he must choose which of his students have performed well enough to be entered into an advanced English Literature class. Whilst it is possible to calculate the answer to both of these questions using the existing mean score and standard deviation, this is very complex.
Judging by your R code, it seems like you think your data is Weibull distributed. In that case, I'd just use the Weibull statistic and not scale anything unless you absolutely have to.
Though theoretically we say we are using Student's-t but for higher values of n (sample size or degree of freedom), t distribution & Z distribution are nearly equal. TO CHECK THAT CARRY OUT AN F-TEST ON YOUR TWO DATASETS, AND IF YOUR VARIANCES ARE APPROXIMATELY EQUAL, THE Z TEST RESULT IS USEFUL.