3: Key Odd

 Review Questions 

1. The three measures of central location covered in the mean, median, and mode. The mean is the arithmetic average (gravitational center). The median is the value in the middle of the ordered array. The mode is the most frequently occurring value.
2. The standard deviation is the square root of the variance. The variance is the mean "sum of square squared deviations" .The inter-quartile range captures the middle 50% of the distribution. 
3. This suggests that the distribution has a long right-tail or high outlier. 
4. The sum of the deviations will always equal 0. The sum of the squared deviations will be a positive value.
5. median
6. upper inside value.
7. Height of the box (IQR), whisker-to-whisker length, and range. (Range is whisker to whisker if there are no outside values.)
8. (1) Expected value of observation selected at random from the sample. (2) Expected value of an observation selected at random from the population. (3) Expected value of the population mean.
9. Variances carry units squared, so it is difficult to interpret. 
10. IQR.
11. min, Q1, median, Q3, maximum 
12. bottom
13. No; the fences do not appear on the plot. They are only used to determine if there are any outside values.
14. n . . . n-1.
15. µ represents  the population mean; represents the sample mean.
16.   = population standard deviation; s = sample standard deviation.
17. Group A has lower values on the average. It also has less spread. 
18. Symmetrical distributions

3.1 Serum polyphenols and red wine consumption

(A) Plot data as a stemplot. 

0.|0
0t|3
0f|445
0s|77
0*|8
×10

[Data are symmetrical with a center at about 5 and range of about 0 to 8.]

(B) Calculate the distribution's mean and standard deviation. = 5.50 and s = 2.517.  [The mean is the balancing point of the distribution and the standard deviation measures distributional spread in terms of the "root mean square deviation."]

3.3 Gravitational center - This exercise is intended to show the mean as the balancing point (but it tells you nothing about the shape or spread of the distribution). 

(A) mean = 3 (marked with a ^)

X                   X
1----2----3----4----5
          ^                     

(B) mean = 3.7 (marked with a ^)

                    X
X                   X 
1----2----3----4----5
              ^

(C) mean = 3

         XXX
1----2----3----4----5
          ^

(D) mean = 3.7

          X 
          X         X
1----2----3----4----5
              ^

3.5 Spread #1.Which has the greatest variability? (Arithmetic not required.) Distribution A has the greatest spread. It may help to plot out these points if you can't see this fact from just looking at the numbers:

            . . . . .             Batch C
         .   .  .   .   .         Batch B 
.       .       .       .       . Batch A
|-------|-------|-------|-------|
0        50      100       150     200

3.7 Heights of 11-year old boys (68-95-99.7 rule)

(A) 146 ± 8 = 138 to 154 centimeters.
(B) 146 ± (2)(8) = 146 ± 16 = 130 to 162 centimeters
(C) 146 ± (3)(8) = 146 ± 24 = 122 to 170 centimeters

3.9 Forced expiratory volume comparisons. Data are FEV measured in liters per second. Group 1 has a higher mean (3.0 vs. 2.4) and slightly more  variability (standard deviations: 1.0 vs. 0.9). Calculations are shown below:

Group 1

• n = 7 
• = 20.78 / 7 = 2.9686 = 2.969
• SS = (3.94 - 2.9686)² + (1.47 - 2.9686)² + (2.06 - 2.9686)² + (2.36 - 2.9686)² + (3.74 - 2.9686)² + (3.43 - 2.9686)² + (3.78 - 2.9686)²  
       =    0.9428           +         2.2470     +        0.8263      +         0.3709      +      0.5944         +   0.2125            +     0.6578
       =  5.8516
• s² = (5.8516) / (7 - 1) = 0.9753 0.975
• s = s² = 0.9753 = 0.9876 0.988 

Group 2

• n = 6
• = 14.26 / 6 = 2.3767 2.377 
• SS = 3.6400 
• s² = (3.6400) / (6 - 1) = 0.7280
• s = s² = 0.7280 = 0.8532 0.853 

3.11 The median is more robust than the mean

(A) Calculate the mean and median of these observations. = 120.9 and median = 117.5
(B) Draw a stemplot and identify the two outliers.  Stemplot is below. The outliers are 152 and 155.

09|9
10|17
11|469
12|15
13|
14|
15|25
×10

(C) Exclude the two outliers and recalculate the mean and median. = 112.8;  median = 115.
What impact did removing the outliers have... Removal of the outliers brings the mean down about 8 points while bringing the median down only 2 points. Outliers have much less impact on medians than means. 

3.13 Body weight expressed as a percentage of ideal

(A) Provide a 5-point summary for the data. 88, 101, 114, 120, 152

(B) Draw a boxplot of these data. Show all work. Use graph paper when drawing your plot. The boxplot is shown below. Notice that  IQR = 120 - 101 = 19, F_L = 101 - (1.5)(19) = 72.5 and F_U = 120 + (1.5)(19) = 148.5. Therefore, there are no outside values on the bottom but there is one outside value on the top.

3.15 Treatment for tachycardia (HEARTRATE). Data are "beats per minute." Before treatment  = 70 and s =  14.1. After treatment  = 70 and s = 1.4. The decrease in the variability of the heart rate shows the medication has  been effective in stabilizing the heart rate of the patient.