Week 2 - Central Tendency PDF

Preview

Summary

In depth lecture notes...

Description

Week 2 Part 1: Explain what is meant by ‘central tendency’ Central Tendency: Specific measures of central tendency allow us to represent the midpoint of a frequency distribution. Differentiate between mean, median and mode Mode: the most common score in a data set Median: the score that corresponds to the point at or below 50% of the scores fall when the data are arranged in numerical order Mean: the average score in a data set

the mean is influenced dramatically by extreme scores and will be a poor representation of central tendency in such circumstances. One might therefore prefer to report the median when dealing with data which includes extreme values. Part 2: Understand the concept of variability After we receive our measure of central tendency we need some indication of how representative this value is.

Simply reporting the mean of a set of scores does not tell us how well this average value reflects the data. We want to know if the mean we have a calculated is a good indication of the set of scores Measures of Variability: tell us the degree to which scores vary from the mean are scores dispersed across a wide range of values or are they tightly clustered around the mean? Understand and calculate variance and standard deviation The variance is a fundamental concept which is calculated in a number of statistical techniques used in psychological research This statistic uses squared deviations to deal with the problem positive/negative deviations A negative value, once squared, becomes positive. So we calculate the sum of the squared deviations from the mean (or sums of squares, SS)

This value is then divided by N - 1 to provide the variance:

However, the variance cannot be directly interpreted because it is in squared form. The mean for this set of data is 6.13 with a variance of 4.15 squared units. As noted by Howell, this has little meaning. We need to convert the variance into a form that is interpretable. The Standard Deviation (s): is simply the square root of the variance it provides a measure of the average deviation from the mean and is the most

commonly reported statistic of variability Once the s^2 (s squared) has been calculated, it can be converted into the SD (standard deviation) by simply taking the square root of s^2; thus:

For the above example we could say that we found a mean of 6.13 and a standard deviation of 2.04. This indicates that, on average, scores deviated from the mean by 2 units

Howell presents definitional and computational formulae for the variance and standard deviation. The way we have calculated the variance and s in Table 2.2 follows the definitional formulae. In accordance with Howell, we would like you to focus on the definitional formulae as these give you a better understanding of what it is you are actually doing with the data when you are calculating various statistics. (Computational formulae can help when there are very large data sets.)

Part 3: Describe the concepts of normal distribution and standard normal distribution Normal distribution: Assumption is that dependent variables are normally distributed in the population If we could sample the entire population, scores would fall into a normal distribution This assumption allows us to answer research questions Probability!

The Standard Normal Distribution has a mean of 0 and a standard deviation of 1 We can convert any distribution into a standard normal distribution by applying a simple formula (This is the formula for z-scores) By applying this formula we are converting each score into standard deviation

units Thus, a z-score of +2 indicates that the score is 2 standard deviations above the mean of the data set In contrast a score of -1.5 indicates that the score is 1.5 standard deviations below the mean The formula for z-score is

Use the tables of the standard normal distribution

Using the z‑tables [1] Consider an intelligence test which has a mean of 100 and a standard deviation of 15 in the general population. From this information we can work out how much of the area under the normal curve is above 1 standard deviations from the mean. That is, we can work out the probability of obtaining a score of 115 or more (+1 standard deviation) on this intelligence test. Given that z‑scores represent standard deviation units we want to find the area above z = 1. To answer this we need to consult the z‑table (or the tables of the standard normal distribution; Appendix z). From Appendix z you need to find the row corresponding to z = 1 and locate the ‘smaller portion’ column, as we are interested in the area above that score. You should find that the smaller portion equals 0.1587. So, the probability that a person drawn randomly from the population will score 115 or above on this intelligence test is 0.1587 or approximately 16%. (multiply 0.1587 by 100 to get percentage) Understand and calculate confidence limits around z.

Setting probable limits on z [1]

Setting limits on an observation provides researchers with a way of stating the confidence of their results. That is, having set limits on a score we can specify with a certain degree of confidence that the score will lie between two values. Usually researchers employ 95% confidence intervals so that they can say with 95% confidence that a score will lie between two values. The z‑tables help us here also. We need to remove 5% of the area under the curve, which means we need to find the value that cuts off 2.5% at each end. Inspection of the z‑tables (Appendix z) indicates that z = ±1.96 corresponds to this value. This means that we can be 95% confident that a score on the intelligence test will fall between 1.96 standard deviations below and 1.96 standard deviations above the mean. So what does this mean for our intelligence test? This is simply a matter of calculating scores that correspond to 1.96 standard deviations from the mean. A way to calculate this is provided by Howell on page 74: X = μ ±1.96σ where μ = mean, and σ = standard deviation. Limits = 100 ± 1.96(15) = 100 ± 29.4 = 70.6 and 129.4 This indicates that we can be 95% confident that a person selected at random from the population will score between 70.6 and 129.4 on this intelligence test. Thus, any score higher than 130 (i.e. 2 standard deviations above the mean) is a rare occurrence as only 2% of the population are expected to score this highly....