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Chapter 4

Chapter 4. Frequency Distributions. Discrete Data Variables. Family size variable, Table 2.2 “X” ranges from 2 to 10, m = 9 (k = 1,…,9), the 100 observations can be arranged into nine groups, one for each family size. Discrete Data Variables.

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Chapter 4

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  1. Chapter 4 Frequency Distributions

  2. Discrete Data Variables • Family size variable, Table 2.2 • “X” ranges from 2 to 10, m = 9 (k = 1,…,9), the 100 observations can be arranged into nine groups, one for each family size

  3. Discrete Data Variables • Table 4.1 is an example of a Frequency Table for the variable: family size

  4. Discrete Data Variables

  5. Continuous Data Variables • m class intervals are created, so that each observation can be placed into only one of them • The 3 principles to follow when creating these intervals are: • The number of classes (m) should be between 5 & 15 • The range (width) of each interval should be the same • The mean point of each interval should be a convenient number

  6. Continuous Data Variables • X is family income from Table 2.2 • Lets set up m = 9 intervals starting from 0 with an interval range of 4.0 (4,000 dollars) • Observations that lie right on the boundary between two classes should be divided between lower and higher classes

  7. Continuous Data Variables • Table 4.2 is the constructed frequency table for family income • The most common shapes taken by histograms or bar charts: • Unimodal • Bimodal • Unimodal skewed to the right or to the left

  8. Continuous Data Variables • The mean and standard deviation of a continuous variable calculated form a frequency table (using the formulas given in the case of discrete variables) are only approximations. • There is a correspondence between relative frequencies and areas under histogram: • The ratio of the area of the kth bar to the total area of the histogram: wfk/ Σwfk= fk

  9. Proportions • What proportion of the observations have X values between Xaand Xb? Prop (Xa≤ X ≤Xb) = ? • Proportion of observations that lie in the one-standard-deviation interval 10.120 ± 5.755 (X =Income, Xa= 4.365 and Xb= 15.875 • Uniform distribution assumption – X values of the observations in any class interval are spread smoothly throughout it • Prop (4.365 ≤ X≤15.875) = {(8-4.365)/4.0}f2 + f3 + {(15.875-12)/4.0}f4 = 0.823

  10. Proportions • Determine the median: Prop(X≤Xmed) = 0.5 Xmed = 8 + (0.10/0.35) (4.0) = 9.14

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