Study Sheet. December 10, The course PDF has been updated (6/11). Read the new one.

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1 Study Sheet December 10, 2017 The course PDF has been updated (6/11). Read the new one. 1 Definitions to know The mode:= the class or center of the class with the highest frequency. The median : Q 2 is defined by Φ(Q 2 ).5 and for all x < Q 2, Φ(x) <.5. The arithmetic mean or average: x def = 1 n n x i. The range: R = x max x min. The interquartile range: IQR def = Q 3 Q 1. The variance: VarX = 1 n (x i x) 2. Standard deviation: σ X = Var X = 1 n (x i x) 2. Note that ax + b = a x + b and Var(aX + b) = a 2 Var X. For a discrete variable with K categories, {x i } n, we define: The class x k where x k < x k+1 for k 1, K 1. The number of elements in the class k: n k with n = K k=1 n k. The frequency: f k = n k /n with K k=1 f k = 1. A discrete variable can be represented in a table: (k, x k, n k ) K k=1. (1) 1

2 More completely: (k, x k, n k, f k ) K k=1 (2) The cumulative number of elements is s k = k l=1 n k with s K = n. The cumulative frequency Φ k = k l=1 f k. The empirical distribution function Φ(x) : R [0, 1] is defined by Φ(x) := the proportion of the range of X x. 2 Two variables For data (i, x(i), y(i)) n : The covariance: Cov (X, Y ) = 1 n n (x i x)(y i ȳ) Property: Var (X + Y ) = Var X+Var Y + 2Cov(X, Y ). The Bravais Pearson correlation coefficient. r(x, Y ) = Cov(X, Y ) σ X σ Y. 0 R 2 = Var(Ŷ )/Var(Y ) = Var(Ŷ )/(Var(Ŷ ) + 1 n (ŷi y i ) 2 ) 1, Linear regression: We suppose y i = f(x i ) = ax i + b + ɛ i and compute â and ˆb such that inf a,b R ɛ 2 i = inf a,b R = (y i (ax i + b)) 2 (y i (âx + ˆb)) 2. One obtains Cov (X,Y ) Var X â = ˆb = ȳ â x. 2.1 Time Series A simple time series is a data set of the form: (t, x t ) T t=1. Let T t := the trend (3) s t := seasonal variations (4) c t := cyclical variations (5) ɛ t := random variations. (6) 2

3 We consider the model: x t = T t + s t + c t + ɛ t. We suppose s t to be periodic with 0 average, that is, there is a k such that s t = s t+k t k s t+i = 0 t. The moving averages k : i=0 y t := x t p + x t p x t + + x t+p 2p + 1 (7) for k = 2p + 1, y t := x t p/2 + x t p x t+p 1 + x t+p /2 2p (8) for k = 2p. Forecasting: Step 1: linear regression of y t as a function of t, Step 2: estimate seasonal values, ŷ t = ât + ˆb. ŝ t = x t ˆT t = x t ŷ t Step 3: because we have estimated s t several times, we take the average ŝ t := 1 I I ŝ t+ik. Thus ˆx t = ŷ t + ŝ t. 2.2 Two nominal variables From the contingency table (cf 1.4) we may test whether two variables X and Y are independent or not with the χ 2 test of Independence s: χ 2 = L C (n lc E lc ) 2 E l=1 c=1 lc 3

4 where n lc := the observed number E lc := n l.n. c n. For large n, we approximate χ 2 with the limit distribution to test the hypothesis: H 0 : X and Y are independent. H 0 is rejected if the χ 2 statistic is sufficiently large. More precisely, we fix our tolerance of error, usually at either 5% or 1%. A bound is computed such that the probability of χ 2 exceeding the bound under H 0 is our tolerated error. If this bound is exceeded by our computed statistic H 0 is rejected. 3 Sample questions The decimal point is at the Exercise 1 From the above stem and leaf plot give 1. The mode or modal class. 2. The median. 3. The minimum value. 4. The maximum value. Exercise 2 You are doing a market analysis for watches. 1. You would like to sell a large number of relatively inexpensive watches. Is the most relavent statistic the mean, median, or modal value of watches sold? 2. Which do you expect to be larger, the mean or median value of watches sold? Why? 4

5 Exercise 3 Write a statistical series which would correspond to the above boxplot. Exercise 4 We make the following R commands: >tb1=table(cdc$gender,cdc$smoke100) >tb1 0 1 m f > chisq.test(tb1) Pearson s Chi-squared test with Yates continuity correction data: tb1 X-squared = 204.6, df = 1, p-value < 2.2e How many males smoke? 2. How many females are there? 3. What is the frequency of females who smoke? 4. What H 0 is the χ 2 test testing? 5

6 5. Is H 0 accepted at the 5% significance level? 6. What does the p-value signify? Exercise 5 Consider the following R commands and their returns: > m1=lm(cdc$height ~ cdc$weight) > summary(m1) Call: lm(formula = cdc$height ~ cdc$weight) Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error t value Pr(> t ) (Intercept) 5.748e e <2e-16 *** cdc$weight 5.717e e <2e-16 *** --- Signif. codes: 0 *** ** 0.01 * Residual standard error: on degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: 8917 on 1 and DF, p-value: < 2.2e What model of the variables cdc$height and cdc$weight is the command lm evaluating? 2. What are the values of the coefficients that R has computed? 3. How much of the variance of height does the model explain? 4. What H 0 are the t values testing? Is H 0 accepted at the 1% significance level? 6