Week 9 Mon Oct 25 (1)

# Week 9 Mon Oct 25 (1) - Posted before class WEEK 9 – Mon...

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Unformatted text preview: Posted before class WEEK 9 – Mon, Oct 25 GRAPHICAL DISPLAYS (Background material is in Chapter 1) Numerical Data Distributions – Shapes of Distributions (Chap 1, Sec 5) Discuss Skewness only, not Kurtosis. Delay discussion until after we have covered graphical displays in Section 9. Numerical Data Distributions – Graphical Displays (Chap 1, Sec 9) Example 1. Numerical variable named C3 (n = 26) . C3 23 18 16 26 9 27 10 19 15 22 24 14 15 26 25 17 16 14 24 25 7 25 20 22 24 23 1 C3 Stem-and-Leaf Displays n = 26 Stem Leaf We see Median = 21 0| 0| 0|97 0|79 1|044 1|044 1|8695576 1|5566789 2|32440243 2|02233444 2|676555 2|555667 Minitab Software was used for the calculations and displays below. Variable n Mean StDev Min Q1 Median Q3 Max C3 26 19.46 5.73 7.00 15.00 21.00 24.25 27.00 Dotplot of C3 27 24 21 18 15 12 9 C3 Dotplot of C3 2 Histogram of C3 25 20 15 10 9 8 7 6 5 4 3 2 1 C3 Frequency Histogram of C3 Box Plot of C3 The box plot reveals that the data are skewed-to-the left (which can be stated skewed-toward-the small-values ). 30 25 20 15 10 C3 Boxplot of C3 3 Example 2. Consider the same data as shown above but with score 7 replaced by 1 . Variable n Mean StDev Min Q1 Median Q3 Max C5 26 19.23 6.34 1.00 15.00 21.00 24.25 27.00 30 25 20 15 10 5 C5 Boxplot of C5 Skewed to the left. 4 Example 3. x = No. Correct out of 30 on Exam 1 (n = 604) . Variable n Mean StDev Min Q1 Median Q3 Max x 604 21.199 4.678 5.000 18.000 22.000 25.000 30.000 30 25 20 15 10 5 x Boxplot of x Skewed to the left. 28 24 20 16 12 8 4 x Dotplot of x Each symbol represents up to 2 observations. 5 28 24 20 16 12 8 70 60 50 40 30 20 10 x Frequency Histogram of x Note. One should investigate outliers in an attempt to find reasons for the extremeness. Sometimes simple errors in recording data explain the extremeness. In the case of the score x = 5, the exam was scored by the wrong key. When scored with the correct key, the score for the student was found to be x = 18. 6 Exercise 1-18 . Discuss Exercise 1-35 . You may use the fact that 3197 16 1 = ∑ = i i x and 2291949 16 1 2 = ∑ = i i x EXER 1-48 (n+1)P x x sorted Position P Position Quartiles 7.0 5.5 1 0.25 8.75 6.525 Q1 6.9 5.6 2 0.50 17.5 7.2 Q2 8.2 5.8 3 0.75 26.25 7.6 Q3 7.8 5.8 4 7.7 6.0 5 7.3 6.1 6 6.8 6.2 7 6.7 6.3 8 8.2 6.6 9 8.4 6.7 10 7.0 6.7 11 6.7 6.7 12 7.5 6.8 13 7.2 6.9 14 7.9 7.0 15 7.6 7.0 16 6.7 7.2 17 6.6 7.2 18 6.3 7.2 19 5.6 7.3 20 7.8 7.3 21 5.5 7.3 22 6.2 7.4 23 5.8 7.5 24 5.8 7.5 25 6.1 7.6 26 6.0 7.6 27 7.3 7.7 28 7.3 7.8 29 7.5 7.8 30 7.2 7.9 31 7.2 8.2 32 7.4 8.2 33 7.6 8.4 34 Some additional comments in regard to Graphical Displays.Some additional comments in regard to Graphical Displays....
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## This note was uploaded on 12/11/2010 for the course STT 315 taught by Professor Anderson during the Spring '08 term at Michigan State University.

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Week 9 Mon Oct 25 (1) - Posted before class WEEK 9 – Mon...

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