Statistical Graphics

Transcription

1 Idea: Instant impression Statistical Graphics Bad graphics abound: From newspapers, magazines, Excel defaults, other software. 1 Color helpful: if used effectively. Avoid "chartjunk." Keep level/interests of audience in mind. Data types matter. Here are some common choices Nominal: Pie chart, sometimes bar chart Ordinal: Bar chart, sometimes pie (with extra cues) Discrete: Bar chart, sometimes histogram; For small sample only, maybe or dotplot (stripchart) Continuous: Histogram, boxplot Bivariate data: C = Categorical (Nominal or Ordinal), N = Numerical Data Types: Types of plots (Inferential methods later in course) C C: Table, multi-level barchart (Are categorical variables independent) C N: Multiple dotplots or boxplots (2-sample t-test, or one-way ANOVA) N N: Scatterplot or time-series plot (Correlation, regression, time series)

2 Simulated (fake) data to illustrate histogram and stripchart. Small normal sample. R software. 2 x = round(rnorm(25, 100, 15), 1) # normal data rounded to 1 place hist(x, ylim=c(-.5,10), main="test Scores") stripchart(x, at=-.5, add=t, pch=" ")

3 > sort(x) [1] [11] [21] > summary(x) Min. 1st Qu. Median Mean 3rd Qu. Max > boxplot(x) 3

4 Small normal sample. Minitab software. 4 MTB > rand 25 c1; SUBC> norm MTB > let c1 = round(c1 * 10)/10 MTB > name c1 'x' MTB > hist 'x' 6 Histogram of x 5 Frequency x

5 MTB > dotp 'x' 5 Dotplot of x x MTB > gstd # Starts 'standard' (character) graphics mode MTB > dotp 'x' Dotplot: x... ::..... :... :. : x MTB > gpro # Toggles back to 'professional' graphics mode

6 6 MTB > sort c1 c2 MTB > print c2 Data Display C MTB > desc c1 Descriptive Statistics: x Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 x Variable Maximum x MTB > gstd MTB > boxp c1 Boxplot I + I x

7 MTB > gpro MTB > boxp c1 7 Boxplot of x x

8 Large normal sample. R. 8 y = rnorm(1000, 100, 15) hist(y, prob=t) yy = seq(50, 150, by=.01) lines(yy, dnorm(yy, 100, 15), col="blue")

9 summary(y) Min. 1st Qu. Median Mean 3rd Qu. Max boxplot(y) 9

10 Large normal sample. Minitab. 10 MTB > rand 1000 c3; SUBC> norm MTB > name c3 'y' MTB > hist 'y' # Default algorithm--too many bins 80 Histogram of y Frequency y

11 Easy to restructure bins (boundaries called breakpoints or cutpoints) in Minitab. I have no idea how to overlay a theoretical normal curve on a Minitab histogram (but see below). 11 Histogram of y Frequency y

12 MTB > desc c3 12 Descriptive Statistics: y Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 y Variable Maximum y MTB > gstd MTB > boxplot 'y' Boxplot ** I + I * y

13 Caution: In Minitab character and professional graphics boxplots use different definitions of "quartile," so the "fences" that define outliers may also differ. (Here the two types of graph happen to agree on three outliers.) Also, the exact definition of quartile can differ from one statistical package to another. MTB > gpro Boxplot of y 125 y

14 About Outliers 14 "Detached" observations. In some sense, "far from other observations." Possible causes: Process error, data entry error Inherent feature of population 'Signal' that occasionally rises above 'noise' Difficulties: May indicate nonnormal distribution; thus call for alternative method of analysis. As above, even data known (simulated) to be exactly normal can show "outliers." Temptation: Throw them out. (Not always a good idea.) Only with good reason, and then report the deletion Caution: sometimes "outliers" may be main message of the data! More on outliers later.

15 MTB > rand 1000 c1; SUBC> norm MTB > GSummary c1. 15 Summary for C1 A nderson-darling N ormality Test A-S quared 0.20 P-V alue Mean StDev V ariance S kew ness Kurtosis N 1000 M inimum st Q uartile M edian rd Q uartile M aximum % Confidence Interval for Mean % Confidence Interval for Median % Confidence Intervals 95% Confidence Interval for StDev Mean Median The normal curve through the data is based on sample mean and standard deviation.

16 Standard Deviation 16 > x = c(1, 2, 6, 3) > mean(x) [1] 3 > d = x - 3 > d.sq = d^2 > x.sq = x^2 > sum(x) [1] 12 > sum(x.sq) [1] 50 > sum(d.sq) [1] 14 > cbind(x, d, d.sq, x.sq) x d d.sq x.sq [1,] [2,] [3,] [4,] > mean(x) [1] 3 > var(x) [1] > sd(x) [1] Illustrating computation of S: Two methods. i x i (xi - 3) (xi - 3) 2 x i Tot Definition: S 2 = 14/3 = 4.67 S = 2.16 "Computational Formula": S 2 = ( /4)/3 = (50-36)/3 = 14/3 = 4.67 Prove: For nonnegative data, the sample mean exceeds its standard error (SE = S/ n). Hints: Use the computational formula for S 2. What relationship between Σ i X i 2 and (Σ i X i ) 2?

17 Interpretations of Standard Deviation 17 Empirical Rule Within 1 standard deviation of mean: "About" 68% of observations Within 2 standard deviations of mean: "About" 95% of observations Within 3 standard deviations of mean: "All or almost all" observations Not a theorem. When does it work? For "normal" population: would be exactly true. (But we have a sample.) For large samples from a normal or "nearly normal" population (several hundred at least): often useful approximation. For small samples, even from normal: extremely rough guide. More on Empirical Rule Later Chebyshev's Inequality. Theorem (true for all samples and populations) Too general to be of much use except in proving limit theorems in probability. Within 2 standard deviations of mean: 1 1/2 2 = 75% Within 3 standard deviations of mean: 1 1/3 2 = 88.9% "ACB Rule." (Almost Completely Bogus Simply not true unless n 25 or n 100) "To estimate σ or approximate S divide sample range by 4 (O/L p91) or by 5 (other books)." DON'T USE THIS!

18 Simulation to disprove the "ACB Rule": m = n = c(5:10, 20, 25, 50, 100, 250, 500) k = numeric(length(n)) for (i in 1:length(n)) { DTA = matrix(rnorm(m*n[i]), nrow=m) x.sd = apply(dta, 1, sd) x.mx = apply(dta, 1, max) x.mn = apply(dta, 1, min) x.rg = x.mx - x.mn k[i] = mean(x.rg)/mean(x.sd) } plot(n, k) cbind(n, round(k,2)) > cbind(n, round(k,2)) n [1,] [2,] [3,] [4,] [5,] [6,] [7,] [8,] [9,] [10,] [11,] [12,] Notes on simulation: Each value k below (each dot on graph) is based on m = 10,000 simulated samples of the specified size. Specified sizes are n = 5, 6,..., 10, 20, 25, 50, 100, 250, 500. Go through a loop, using a different value of n on each pass through the loop. Within the loop, the DTA matrix has m = 10,000 rows and n columns. Each row of the matrix is one sample of the specified size. The apply function is used to find the standard deviation, maximum, and minimum for each sample (row). The correct divisor k is found for each n. After the loop the results are presented in a table and (next page) a graph. This demonstration is only for normal data. 18

19 Maybe it's reasonable to divide by 3 for n 8 or 9, by 4 for n 25, and by 5 for n 100. Unlike many results in stat/prob, approx. of S using range gets worse as n increases. 19 n = 5 (divide by 2.5). Range/2.5 gives pretty good approximation of S. (Find upper and lower linear boundaries for n = 5. What exact relationship for n = 2?)

20 20

21 21 # Boundary lines for n = 5 m = DTA = matrix(rnorm(m*5), nrow=m) x.sd = apply(dta, 1, sd) x.mx = apply(dta, 1, max) x.mn = apply(dta, 1, min) x.rg = x.mx - x.mn k = mean(x.rg)/mean(x.sd) plot(x.rg, x.sd, pch=".") abline(a=0, b=.3536, col="darkred") abline(a=0, b=.5477, col="darkred") # Exactly linear for n = 2 m = DTA = matrix(rnorm(m*2), nrow=m) x.sd = apply(dta, 1, sd) x.mx = apply(dta, 1, max) x.mn = apply(dta, 1, min) x.rg = x.mx - x.mn k = mean(x.rg)/mean(x.sd) plot(x.rg, x.sd, pch=".") > k [1]

22 n = 25 (divide by 4). Plot is not very nearly linear. Mediocre estimation of S Range/4 22

23 n = 500 (divide by 6) Plot is not at all linear: Very poor estimation of S Range/6. 23 Note: The required divisor goes to with increasing n.

24 Simulation in Minitab for n = 5. (Relatively rare case where simulation is easy in Minitab.) Try it for yourself. Look at first few rows and verify values of SD and range by hand. Note: When finished, delete HUGE worksheet before continuing to use Minitab for other purposes. MTB > random c1-c5; SUBC> norm 0 1. MTB > name c6 'sd' c7 'mx' c8 'mn' c9 'rng' c10 'k' MTB > rstd c1-c5 c6 MTB > rmax c1-c5 c7 MTB > rmin c1-c5 c8 MTB > let c9 = c7 - c8 MTB > let c10 = c9/c6 MTB > desc c10 Descriptive Statistics: k Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 k Variable Maximum k # Slightly different procedure: divided before averaging. # Result nearly the same as in R for n = 5: k approx. 2.5 MTB > plot c6 * c9 # click on large red dots, change to small red dots. # result is similar to plot done above in R. # Note y-axis mentioned first in Minitab. 24

25 In-Lab Problems: 25 Dataset: Gender Height Weight (inches) (pounds) College M Bsns F Sci F Bsns M Sci M Bsns F Eng M Eng M Sci 1. Input the data into Minitab system. 2. Summarizing Gender: find its empirical distribution and graph it. 3. Summarizing Height: find its mean and median. Graph its distribution; which measure best summarizes the center of the distribution? 4. Make a table to show the association between Gender and College. 5. Make a plot to illustrate the association between Height and Weight. 6. Make a plot to illustrate the association between Height and Gender.

26 Brief Instructions Basic descriptive statistics: Stat> Basic Statistics> Display Descriptive Statistics, Variables, Graphs 2. Drawing graphs: Graph> the wanted graph Use one of the two datasets presented on page 125 of the text (in Exercise 3.30), to do the following: (a) Make a histogram by hand. Explain your choice of intervals. Why might Minitab make a histogram that looks different? (b) Find the sample mean. On the horizontal axis (bottom scale) of your histogram, mark the position of the sample mean. Does the histogram seem to "balance" there? (c) Find the interval that extends two sample standard deviations on either side of the sample mean. Sketch this interval on the bottom scale of your histogram. What proportion of all observations actually falls in this interval? (You may find the sample mean and standard deviation either with software or with a calculator.)