Statistical Analysis for Education and Psychology Researchers

deviate—

4.7

For a pulse rate of 84 then, Z=(84−72)/12=1.

For a pulse rate of 72 then, Z=(72−72)/12=0.

A linear transformation does not alter the shape of a distribution, that is the observations
remain in the same relative position to each other. Therefore, if the shape of the
distribution was not normal to begin with then the Z transformation will not normalize (in
the sense of producing a normal distribution) the data.
The total area under the standard normal curve is 1 and that areas under the curve
translate to probabilities. To find the area under the normal curve that is above two
standard deviations, we would look for the area above Z=2.
To answer the first question in Example 4.7, the probability of observing a male with a
pulse rate of 96 or more, first convert 96 to a Z score giving the value +2.0 then go to the
table of the normal distribution (Table 1, Appendix A4), and find the column headed Z.
Move down this column until the value 2.0 is reached. Go along the row until under the
column headed .00 (Z to two decimal places), the value is p=0.0228. This means that 2.28
per cent of the area under the curve is beyond the Z value of 2, that is above two standard
deviations. The larger portion to the left of Z comprises 97.72 per cent of the area under
the normal curve. The probability of observing a pulse rate of 96 or more is therefore
p=0.0228.
To answer the second question, that is to find the probability of observing a pulse rate
of 105 or more, we again transform this score into a Z score, =2.75, and look up this
value in the body of the normal distribution Table (Table 1, Appendix A4). The answer is
p=0.0030. The probability is very small, p<.05 (5 per cent). In fact, a Z score of only 1.65
would be needed to give a probability of p<.05. We would conclude that the male with a
pulse rate of 105 is unlikely to have come from the same population of males because it
is so different from the population mean of 72.
A Z score or standard score indicates where any particular score lies in relation to the
mean score of its distribution. We can convert any raw score, xi, into a Z score, all that is
required is the mean and standard deviation of the distribution of scores. A Z score will
show a score’s relative position above or below the mean of a distribution of scores. Z
scores may have negative values, namely all those below the mean. Since the normal
distribution is symmetrical we can use it to evaluate negative Z scores. For example, the
probability of a Z score <−2 is equivalent to the probability of a Z score of >2, namely
0.0228.

Example 4.8: Transforming Raw Scores into Z Scores

As part of a selection procedure a company requires graduates to take three tests
consisting of Test 1, Numeracy; Test 2, Computer Programming Aptitude; and Test 3,
Verbal Reasoning. Given the following test results for a candidate, and assuming

a normal distribution for each of the test scores with mean and standard deviation given

Statistical analysis for education and psychology researchers 106