25 multiple-choice questionsBring pencil, calculator, and note sheetWill have 1hr 50mins to finishSimilar questions will be used from the midterm reviewEstimate the median: middle value; 50% above/below; n=224 median would be 112-113 value; frequency is given; median is located in 1st bin; look at axis; median = 15; mean is larger than the median; mean would be between 20-40Approximate 1st and 3rd quartiles (Q1 and Q3): Q1=25% below and 75% above, Q3=75% below and 25% above; Q1=10, Q3=3rd or 4th bin = 60Calculate the IQR: Q3-Q1=IQR; 60-10=50Determine any outliers: compute upper and lower fence; Q3+1.5(IQR)=upper fence; Q1-1.5(IQR)=lower fence; 60+1.5(50)=135; 10-1.5(50)=-65; values higher than 135 are outliersFind equation of the least squares regression line: beers=x, score=y; find slope, take square root of R2, slope=r(Sy/Sx); r=-0.774, slope= -0.774(13.3/6.3)= -1.57; find the intercept; y-slopex, 72.8-(-1.57)(11.3) = 90.54; score=90.54+(-1.57)beersInterpret the slope: context: for every additional beer drank, we expect on average final exam score to decrease by 1.57 points/percent. Residual: plug in the value into the equation; score=90.54+(-1.57)(8)= 77.98; residual=observed-predicted; 87.1-77.98=On an exam, we want a positive residualTo determine association, compare percent of the 2 groups; diff=yes, same=no; doesn’t mean causationEmpirical rule: given mean and SD with unimodal symmetric distribution; #4=132Correlation: using regression equation; plot the points, make a table, r=1, slope=.95Box Plots: IQR is a measure of variability, IQR is the box; Q1=mean is right skewed; if Q2-Q1<Q3-Q2= right skewed; #7 is true; typical value would be mean or median, #8 is falseDescribing the z-score: right skewed; z=(obs-mean)/SD; mean is larger; (median-mean)/SD; z-score would be negative if mean is larger than the medianBox plots are best for comparing 2 numerical variablesDistributions must be able to fit the empirical ruleZip code is categorical, bar chart would be better; zip code tells
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