z Scores, Type I and II Errors, Hypothesis Testing.

 The Standard Normal Distribution and z Scores A student receives an intelligence quotient (IQ) score of 115 on a standardized intelligence test. What is his or her percentile rank? To calculate the percentile rank, you must understand the logic and application of the standard normal distribution. The standard normal distribution is introduced in Chapter 1 of the Warner text (Figure 1.4, p. 28). The advantage of the standard normal distribution is that the proportional area under the curve is constant. This constancy allows for the calculation of a percentile rank of an individual score X for a given distribution of scores. When a population mean (µ) and population standard deviation (σ) are known, such as a distribution of scores for a standardized intelligence test, the standard normal distribution determines the percentile rank of a given X score (for example, 50th percentile on IQ). Note in Figure 1.4 (Warner, 2013) that around two-thirds of standard scores (68.26%) fall within ±1 population standard deviation from the population mean. Approximately 95% of standard scores fall within ±2 population standard deviations. Approximately 99% of standard scores fall within ±3 population standard deviations. Knowing this, we can begin the process of determining a percentile rank from an individual score. Consider a standardized IQ test where µ = 100 and σ = 15. A standard score, or z score, is calculated with individual X scores rescaled to µ = 0 and σ = 1. The formula for a z score is [( X − µ) ÷ σ]. Refer to Figure 2.13 in the Warner text (p. 61). For example, an IQ score of X = 100 would be rescaled to z = 0.00 [(100 − 100) ÷ 15 = 0]. A z score of 0 means that the X score is 0 standard deviations above or below the mean. An IQ score of X = 85 would be rescaled to z = −1.00 [(85 − 100) ÷ 15 = −1.00]. A z score of −1 represents an IQ score 1 standard deviation below the mean. An IQ score of 130 would be rescaled to z = 2.00 [(130 − 100) ÷ 15 = 2.00]. A z score of 2 represents an IQ score 2 standard deviations above the mean. In short, a negative z score falls to the left of the population mean, whereas a positive z score falls to the right of the population mean. Once a given z score is calculated for a given X score, its percentile rank can be determined. Refer to Appendix A of the Warner text (pp. 1051–1055). For example, an IQ score of 115 is z = 1.00 (Column A). The area under the curve between z = 0.00 and z = 1.00 is .3413 (Column B), and the area under the curve to the right of z = 1.00 is .1587 (Column C). If we add the area under the curve of .3413 to the other half of the distribution (.50), we know that an IQ score of 115 is at about the 84th percentile (.50 + .3413). We can say that a student who receives an IQ score of 115 outperforms about 84% of students on the IQ test, whereas about 16% of students outperform him or her on the IQ test. Conversely, an IQ score of 85 is z = −1.00 (Column A at z = 1.00; only positive z scores are listed). In this case, we know that the area under the curve between z = 0.00 and z = −1.00 is also .3413 (Column B). The area under the curve to the left of z = −1.00 is .1587 (Column C). An IQ score of 85 is therefore at about the 16th percentile. A student who scores an 85 on an IQ test outperforms about 16 percent of other students on the IQ test, whereas about 84% of students outperform him or her on the IQ test. An important z score is ±1.96, where 95% of scores fall under the area of the curve, whereas 2.5% fall to the left and 2.5% fall to the right (2.5% + 2.5% = 5%). A z score beyond these cutoffs is typically considered to be “extreme” (Warner, 2013). In addition, we will see below that most inferential statistics set “statistical significance” to obtained probability values of less than 5%. Hypothesis Testing Probability is crucial for hypothesis testing. In hypothesis testing, you want to know the likelihood that your results occurred by chance. No matter how unlikely, there is always the possibility that your results have occurred by chance, even if that probability is less than 1 in 20 (5%). However, you are likely to feel more confident in your inferences if the probability that your results occurred by chance is less than 5% compared to, say, 50%. Most researchers in the social sciences find it reasonable to designate less than a 5% chance as a cutoff point for determining statistical significance. This cutoff point is referred to as the alpha level. An alpha level is set to determine when a researcher will reject or fail to reject a null hypothesis (discussed next). The alpha level is set before data are analyzed to avoid “fishing” for statistical significance. In high-stakes research (such as testing a new cancer drug), researchers may want to be even more conservative in designating an alpha level, such as less than 1 in 100 (1%) that the results are due to chance.