Statistics Formula Sheet Part 1

Important Concepts not on the Statistics Formula Sheet Part I: IQR = Q3 – Q1 Test for an outlier: 1.5(IQR) above Q3 or below Q1 The calculator will run the test for you as long as you choose the boxplot with the oulier on it in STATPLOT Linear transformation: Addition: affects center NOT spread adds to , M, Q1 , Q3, IQR When describing data: describe center, spread, and shape. not σ Give a 5 number summary or mean and standard deviation when necessary. Multiplication: affects both center and spread multiplies , M, Q1 , Q3, IQR, σ skewed right Skewed left Stem and leaf Normal Probability Plot The 80th percentile means that 80% of the data is below that observation. residual = residual = observed – predicted y = a+bx Slope of LSRL(b): rate of change in y for every unit x y-intercept of LSRL(a): y when x = 0 Confounding: two variables are confounded when the effects of an RV cannot be distinguished. Exponential Model: y = abx take log of y Power Model: y = axb take log of x and y Ogive (cumulative frequency) Histogram: fairly symmetrical unimodal Boxplot (with an outlier) r: correlation coefficient, The strength of the linear relationship of data. Close to 1 or -1 is very close to linear HOW MANY STANDARD DEVIATIONS AN OBSERVATION IS FROM THE MEAN 68-95-99.7 Rule for Normality N(µ,σ) N(0,1) Standard Normal Explanatory variables explain changes in response variables. EV: x, independent RV: y, dependent r2: coefficient of determination. How well the model fits the data. Close to 1 is a good fit. “Percent of variation in y described by the LSRL on x” Lurking Variable: A variable that may influence the relationship bewteen two variables. LV is not among the EV’s Regression in a Nutshell Given a Set of Data: Enter Data into L1 and L2 and run 8:Linreg(a+bx) The regression equation is: predicted fat gain  3.5051  0.00344( NEA) y-intercept: Predicted fat gain is 3.5051 kilograms when NEA is zero. slope: Predicted fat gain decreases by .00344 for every unit increase in NEA. r: correlation coefficient r = - 0.778 Moderate, negative correlation between NEA and fat gain. r2: coefficient of determination r2 = 0.606 60.6% of the variation in fat gained is explained by the Least Squares Regression line on NEA. The linear model is a moderate/reasonable fit to the data. It is not strong. The residual plot shows that the model is a reasonable fit; there is not a bend or curve, There is approximately the same amount of points above and below the line. There is No fan shape to the plot. Predict the fat gain that corresponds to a NEA of 600. predicted fat gain  3.5051 0.00344(600) predicted fat gain  1.4411 Would you be willing to predict the fat gain of a person with NEA of 1000? No, this is extrapolation, it is outside the range of our data set. Residual: observed y - predicted y Find the residual for an NEA of 473 First find the predicted value of 473: predicted fat gain  3.5051  0.00344(473) predicted fat gain  1.87798 observed – predicted = 1.7 - 1.87798 = -0.17798 Transforming Exponential Data y = abx Transforming Power Data y = axb Take the log or ln of y. Take the log or ln of x and y. The new regression equation is: log(y) = a + bx The new regression equation is: log(y) = a + b log(x) Residual Plot examples: Linear mode is a Good Fit Curved Model would be a good fit Fan shape loses accuracy as x increases Inference with Regression Output: Construct a 95% Confidence interval for the slope of the LSRL of IQ on cry count for the 20 babies in the study. Formula: df = n – 2 = 20 – 2 = 18 b  t *SE b 1.4929  (2.101)(0.4870) 1.4929 1.0232 (0.4697, 2.5161) Find the t-test statistic and p-value for the effect cry count has on IQ. From the regression analysis t = 3.07 and p = 0.004 Or t 1.4929 b   3.07 SEb 0.4870 s = 17.50 This is the standard deviation of the residuals and is a measure of the average spread of the deviations from the LSRL.