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[KNR445: CLASS EXERCISES WEEK 1 (MEASUREMENT January 21, 2009 LEVELS, CENTRAL TENDENCY, SPREAD, NORMAL CURVES)] Class exercise set 1: Measurement Levels (or scales) A. Name the measurement level for each of these variables (and say why you chose your answer): 1. Average score at bowling for each player in a bowling league ratio 2. Placement in a bowling league ordinal 3. All the team names in the bowling league nominal 4. Average speed at which each player in the bowling league bowls the ball ratio 5. Bowling skill knowledge, as measured by a survey given to all bowling league members (survey was multiple choice, had 30 questions, and generated a score between 0 and 30). interval 6. The score on the above test (just as a score, not as a measure of knowledge) ratio B. Let’s say you have knowledge of ranks, and want to convert it to a ratio scale. Can you legitimately do this, without any extra information? no Class exercise set 2: Central tendency, spread, normal curves 1. What are the mean, median, and mode of the following set of numbers? 3, 4, 1, 2, 4, 4 Mean = 18/6 = 3 Mode = 4 [KNR445: CLASS EXERCISES WEEK 1 (MEASUREMENT January 21, 2009 LEVELS, CENTRAL TENDENCY, SPREAD, NORMAL CURVES)] Median = 3.5 (3+4/2) 2. Which set of numbers has the largest mean? The highest median? The highest standard deviation? 3, 4, 1, 2, 4, 4 or 1,1,1,7,7,7 Largest mean = 111777 = 4 Highest median = 111777 = 4 (1+7/2) Highest SD = 111777 (look at sum of absolute deviations from mean: for 123444, mean is 3, so sum of abs deviations is: 2+1+0+1+1+1 = 6. For the other one, sum of abs devn is: 3+3+3+3+3+3 = 18) 3. Which of the above sets is bimodal? 111777 4. If you have a distribution with a mean of 50 and a SD of 10, what’s the probability of getting these values: a. inside the range 40 to 60 40 to 60 = +/- 1SD, so inside is 68%, or 0.68 b. outside the range 40 to 60 40 to 60 = +/- 1SD, so inside is 68%, or 0.68. So outside is 1-0.68 = 0.32 or 32% c. outside the range 50 to 70 30 to 70 = +/- 2SD, so inside is 95%, or 0.95. 50 to 70 = ½ of 30 to 70 = 95%/2 = 47.5% So outside of 50 to 70 = 100% - 47.5% = 52.5% d. less than 50 50 is the mean or central point, so less than that is 50% or 0.5 e. more than 60 60 is 1SD above the mean. So from the mean to 60 is 0.34, or 34% (68%/2). So the area above that is 0.50 – 0.34 = 0.16, or 16% 5. You have 2 distributions. You analyze their structure, and find that the first (a) is normally distributed, and the second (b) is positively skewed. [KNR445: CLASS EXERCISES WEEK 1 (MEASUREMENT January 21, 2009 LEVELS, CENTRAL TENDENCY, SPREAD, NORMAL CURVES)] What would be the appropriate measures of central tendency and spread to choose for each of them? a. Mean, SD b. Median, Semi-interquartile range 6. In a normal distribution, what measure of central tendency occurs most frequently? What measure of central tendency has the greatest magnitude? All of them (same answer for both – they are all equal) 7. In a positively skewed distribution, what measure of central tendency occurs most frequently? What measure of central tendency has the greatest magnitude? Mode, Mean.