1. MM150 Unit 9 Seminar

2. 4 Measures of Central Tendency Mean – To find the arithmetic mean, or mean, sum the data scores and then divide by the number of data scores. Example: Find the mean of the data scores 5, 6, 2, 9, 8 5 + 6 + 2 + 9 + 8 = 30 = 6 5 5 Median – To find the median, put the data scores in ascending or descending order and then find the middle data score. If there are an even number of data scores, after ranking the scores, find the mean of the middle two. Example: Find the median of the data scores 5, 6, 2, 9, 8 Put the scores in ascending order: 2, 5, 6, 8, 9 Example: Find the median of the data scores 4, 7, 2, 9 Put the scores in ascending order: 2, 4, 7, 9 Find the mean of 4 and 7: (4 + 7)/2 = 11/2 = 5.5

3. 4 Measures of Central Tendency Mode – The mode is the data score that occurs most frequently. Example: Find the mode of the data scores 6, 4, 9, 8, 6, 5 It may help to put the scores in ascending order: 4, 5, 6, 6, 8, 9 You can see that the data score 6 occurs most often. You can have data sets that don’t have a mode (each score occurs once) and you can have data sets that are bimodal – which means they have 2 modes. Midrange – The midrange is the value halfway between the greatest and least data score. To find it, take the mean of the greatest and least data score. Example: Find the midrange of the data scores 6, 4, 9, 8, 6, 5 It may help to put the scores is ascending order: 4, 5, 6, 6, 8, 9 The midrange is (4 + 9)/2 = 13/2 = 6.5 *Please read on page 362 of your text when each is the ‘better’ average.

4. Mean Example Todd is taking a math class where his end of term grade is based on 4 exams, each having the same number of points and weighted the same. He scored 98, 82, and 87 on the first three exams in his class. What does he need to score on the 4 th exam to get at least a 90% for the final grade? The instructor uses mean as the average. 98 + 82 + 87 + x = 90 4 267 + x = 90 4 267 + x = 360 X = 93 Todd must score a 93 or higher on the test.

5. EVERYONE: consider data 10, 11, 11, 12, 13, 15 Find Mean: Median: Mode: Midrange:

6. EVERYONE Solution Mean: 10 + 11 + 11 + 12 + 13 + 15 = 72 = 12 6 6 Median: 10, 11, 11, 12, 13, 15 There are an even number of data scores: 11 + 12 = 23 = 11.5 2 2 Mode: 10, 11, 11, 12, 13, 15 The data score 11 occurs most often. Midrange: 10 + 15 = 25 = 12.5 2 2

7. 2 Measures of Position Percentile – There are 99 percentiles that divide the data up into 100 equal parts. Quartile – Quartiles divide data into 4 equal parts, called quartiles. The first quartile is at 25%, the second at 50%, and the third at 75%. Determine Q1, Q2, and Q3 of the data below: 15, 10, 19, 18, 11, 15, 13, 18, 19, 17, 19, 15, 16, 13, 15, 16, 13, 12, 14 First put the data in ascending order 10, 11, 12, 13, 13, 13, 14, 15, 15, 15, 15, 16, 16, 17, 18, 18, 19, 19, 19 Find the median, 15 is Q2 Find the median of the lower half for Q1, 13 Find the median of the upper half for Q3, 18

8. 2 Measures of Dispersion Range – The range is the difference between the greatest and least data score. Example: Find the range of the data scores 55, 59, 51, 64, 60 Put the data scores in ascending order 51, 55, 59, 60, 64 The range is 64 – 51 = 13 Standard deviation – The standard deviation tells us how much the data differ from the mean. See the next powerpoint slide for an example of standard deviation.

9. Standard Deviation Example Find the standard deviation of 11, 15, 18, 9, 12 1.Find the mean of the data scores. 11+15+18+9+12 = 65 = 13 5 5 2. Make a chart with 3 columns DataData – Mean(Data – Mean) 2 1111- 13 = -24 1515 – 13 = 24 1818 – 13 = 525 99 – 13 = -416 1212 – 13 = -11 3.4.5. 6. 50 7. Divide 50 by n – 1, where n is the number of data scores. So divide 50 by 4, which is 12.5 8. Find the square root of the number found in step 7. √12.5 ≈ 3.5355

10. EVERYONE 1.Find the mean. 15, 16, 20, 13 2. Make a 3-column table. DataData – Mean(Data – Mean) 2 15 16 20 13 6. ? 3.4.5. 7. ?/? = 8.667 8. √8.667 ≈ 2.944

11. EVERYONE solution 1.Find the mean. 15 + 16 + 20 + 13 = 64 = 16 4 4 2. Make a 3-column table. DataData – Mean(Data – Mean) 2 151 1600 20416 13-39 6. 26 3.4.5. 7. 26/3 = 8.667 8. √8.667 ≈ 2.944

12. Histogram and Frequency Polygon

13. Rectangular Distribution Frequency Observed Values

14. 14 J-Shaped Distributions Constantly IncreasingConstantly Decreasing

15. 15 Skewed Distributions Mode Median Mean

16. 16 Normal Distributions

17. 17 Normal Distribution PROPERTIES OF A NORMAL DISTRIBUTION 1. The graph of a normal distribution is called a normal curve. 2. The normal curve is bell-shaped and symmetric about the mean. 3. The mean, median, and mode of a normal distribution all have the same value and all occur a the center of the distribution. EMPIRICAL RULE In any normal distribution 1. Approximately 68% of all the data lies within one standard deviation of the mean (in both directions). 2. Approximately 95% of all the data lies within two standard deviations of the mean (in both directions). 3. Approximately 99.7% of all the data lies within three standard deviations of the mean (in both directions).

18. 18 Z-Score z = value of a given data score - mean standard deviation Example: A normal distribution has a mean of 50 and a standard deviation of 10. Determine the z-score for the data value of 60. z 60 = value - mean standard dev. z 60 = 60 - 50 10 z 60 = 10 = 1 The score of 60 is one standard deviation 10 above the mean.

19. 19 EVERYONE A normal distribution has a mean of 50 and a standard deviation of 10. Determine the z-score for the data value of 45. z 45 = value - mean standard dev.

20. 20 EVERYONE Solution A normal distribution has a mean of 50 and a standard deviation of 10. Determine the z-score for the data value of 45. z 45 = value - mean standard dev. z 45 = 45 - 50 10 z 45 = -5 = -1 The score of 45 is 0.5 standard deviation 10 2 below the mean.

21. 21 Linear Correlation Coefficient r = n(∑xy) - (∑x)(∑y) √[n(∑x 2 )- (∑x) 2 ]√[n(∑y 2 ) - (∑y) 2 ] Let’s do Page 407, #24. The first thing to do is plot the points. Here we have (6,13), (8,11), (11,9), (14,10) and (17,7).

22. 22 xyx2x2 y2y2 xy 6133616978 8116412188 1191218199 1410196100140 17728949119 5650706520524 r = n(∑xy) - (∑x)(∑y) √[n(∑x 2 )- (∑x) 2 ]√[n(∑y 2 ) - (∑y) 2 ] r = 5(524) - (56)(50) √[(5)(706) - 56 2 ]√[(5)(520) - 50 2 ] r = 2620 - 2800 √[3530 - 3136]√[2600 - 2500] r = -180 = -180 = -180 = -0.907 √394 √100 19.8494 * 10 198.494 |-0.907| = 0.907 Remember n = 5 c) 0.907 > 0.878 so a correlation exists. d) 0.907 < 0.959 so a correlation does not exist.

23. 23 Linear Regression Slope-Intercept form of a line: y = mx + b where m is the slope and (0,b) is the y-intercept. Linear Regression Formula y = mx + b m = n(∑xy) - (∑x)(∑y) and b = ∑y - m(∑x) n(∑x 2 ) - (∑x) 2 n

24. 24 Linear Regression Example Let’s use the data from Page 407 #24 to save us some work! n∑xy - (∑x)(∑y) = -180 n(∑x 2 ) - (∑x) 2 = 394 ∑x = 56 ∑y = 50 m = -180/394 = -0.457 b = 50 - (-0.457)(56) = 50 + 25.584 = 15.117 5 5 y = -0.457x + 15.117 y = mx + b m = n(∑xy) - (∑x)(∑y) and b = ∑y - m(∑x) n(∑x 2 ) - (∑x) 2 n