1. Statics – Part II Chapter 9

2. Mean The mean, is the sum of the data divided by the number of pieces of data. The formula for calculating the mean is where represents the sum of all the data and n represents the number of pieces of data. ( indicates “Summation”) is called “X Bar”.

3. Example-find the mean Find the mean amount of money parents spent on new school supplies and clothes if 5 parents randomly surveyed replied as follows: $327 $465 $672 $150 $230

4. Median The median is the value in the middle of a set of ranked data. Example: Determine the median of $327 $465 $672 $150 $230. Rank the data from smallest to largest. $150 $230 $327 $465 $672 middle value (median)

5. Example: Median (even data) Determine the median of the following set of data: 8, 15, 9, 3, 4, 7, 11, 12, 6, 4. Rank the data: 3 4 4 6 7 8 9 11 12 15 There are 10 pieces of data so the median will lie halfway between the two middle pieces the 7 and 8. The median is (7 + 8)/2 = 7.5 3 4 4 6 9 11 12 15 7 8 (median) middle value

6. Mode The mode is the piece of data that occurs most frequently. Example: Determine the mode of the data set: 3, 4, 4, 6, 7, 8, 9, 11, 12, 15. The mode is 4 since it occurs twice and the other values only occur once.

7. Midrange The midrange is the value halfway between the lowest (L) and highest (H) values in a set of data. Example: Find the midrange of the data set $327, $465, $672, $150, $230.

8. Example The weights of eight Labrador retrievers rounded to the nearest pound are 85, 92, 88, 75, 94, 88, 84, and 101. Determine the a) mean b) median c) mode d) midrange e) rank the measures of central tendency from lowest to highest.

9. Example--dog weights 85, 92, 88, 75, 94, 88, 84, 101 a. Mean b. Median-rank the data 75, 84, 85, 88, 88, 92, 94, 101 The median is 88.

10. Example--dog weights 85, 92, 88, 75, 94, 88, 84, 101 c. Mode-the number that occurs most frequently. The mode is 88. d. Midrange = (L + H)/2 = (75 + 101)/2 = 88 e. Rank the measures, lowest to highest 88, 88, 88, 88.375

11. To Find the Quartiles of a Set of Data 1.Order the data from smallest to largest. 2.Find the median, or 2 nd quartile, of the set of data. If there are an odd number of pieces of data, the median is the middle value. If there are an even number of pieces of data, the median will be halfway between the two middle pieces of data.

12. To Find the Quartiles of a Set of Data continued 3.The first quartile, Q 1, is the median of the lower half of the data; that is, Q 1, is the median of the data less than Q 2. 4.The third quartile, Q 3, is the median of the upper half of the data; that is, Q 3 is the median of the data greater than Q 2.

13. Example: Quartiles The weekly grocery bills for 23 families are as follows. Determine Q 1, Q 2, and Q 3. 170210270270280 33080170240270 22522521531050 75 1601307481 95172190

14. Example: Quartiles continued Order the data: 50 75 74 80 81 95130 160170170172190210215 225225240270270270280 310330 Q 2 is the median of the entire data set which is 190. Q 1 is the median of the numbers from 50 to 172 which is 95. Q 3 is the median of the numbers from 210 to 330 which is 270.

15. Example: Range Nine different employees were selected and the amount of their salary was recorded. Find the range of the salaries. $24,000 $32,000 $26,500 $56,000 $48,000 $27,000 $28,500 $34,500 $56,750 Range = $56,750  $24,000 = $32,750

16. To Find the Standard Deviation of a Set of Data 1. Find the mean of the set of data. 2. Make a chart having three columns: Data Data  Mean (Data  Mean) 2 3. List the data vertically under the column marked Data. 4. Subtract the mean from each piece of data and place the difference in the Data  Mean column.

17. To Find the Standard Deviation of a Set of Data continued 5.Square the values obtained in the Data  Mean column and record these values in the (Data  Mean) 2 column. 6.Determine the sum of the values in the (Data  Mean) 2 column. 7.Divide the sum obtained in step 6 by n  1, where n is the number of pieces of data. 8.Determine the square root of the number obtained in step 7. This number is the standard deviation of the set of data.

18. Example Find the standard deviation of the following prices of selected washing machines: $280, $217, $665, $684, $939, $299 Find the mean.

19. Example continued, mean = 514 421,5160 180,625425939 28,900170684 22,801151665 46,225  215299 54,756  234280 (  297) 2 = 88,209  297217 (Data  Mean) 2 Data  MeanData

20. Example continued, mean = 514 The standard deviation is $290.35.

21. Find the range & standard deviation of the set of data. 43, 51, 44, 49, 46, 48, 48 43,44, 46, 48, 48, 49, 51 Range = 51-43=8 Х = 329/7 = 47

22. Standard Deviation continued: Data Data-Mean (D-M)² 4343-47 =-4(-4)²=16 4444-47 =-3(-3)²=9 4646-47 =-1(-1)²=1 4848-47 =1(1)²=1 4949-47 =2(2)²=4 5151-47 =4(4)²=16 48

23. Standard Deviation continued: S = 48 s= 8 7-1 48s= 2.83 6

24. Correlation Absolute Value 3 = 3 -5 = 5

25. Correlation continued: If r is › the value given in the Table Value under the specified α & appropriate sample size “n”, assume that a correlation “DOES” exist between the variable. If r is ‹ the Table Value, assume that “NO” correlation exist.

26. Examples Correlation: Correlation Coefficient, r Use table to right for specified sample n α =.01 size & level of significance to determine 47.372 whether a linear correlation exists. 52.354 62.325 72.302 r = -0.28 when n=62 at α =.01 82.283 92.267 102.254 -.28 =.28.28 ‹.325 A linear correlation DOES NOT Exist

27. Use table below for the specified sample size & level of significance to determine whether a linear correlation exists. n α =.05 α =.01 4.950.990 5.878.959 6.811.917 7.754.875 8.707.834 9.666.798 10.632.765 r =.77 when n=7 at α =.05.77 =.77.77 ›.754 A linear correlation DOES exists