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Review. A quick review of each chapter. Chapter 1. One-Variable Statistics. How old is our teacher?. N=82. Describe the distribution. CUSS: C enter(mean and median), U nusual features(are there outliers?), S hape(skewed left,symmetric , skewed right), S pread(max to min).

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## Review

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**Review**A quick review of each chapter**Chapter 1**One-Variable Statistics**How old is our teacher?**N=82 Describe the distribution. CUSS: Center(mean and median), Unusual features(are there outliers?), Shape(skewed left,symmetric, skewed right), Spread(max to min)**How old is our teacher?**The distribution of age guesses are slightly skewed to the left, with a mean age guess of 36.4 which slightly larger than the median guess of 36. Guesses of Mr. Pines’ age ranged from 19 to 48 years old.**5# Summary**Definition: The five-number summary of a distribution consists of the smallest observation, the first quartile, the median, the third quartile, and the largest observation, written in order from smallest to largest. Minimum Q1 M Q3 Maximum**Box Plots**25% 25% 25% 25% Min Q1 Q2 Q3 MAX**Outliers in a boxplot**1,4,7,12,13,13,21,36 Possible outlier Checking for outliers we use the formulas: Q1 – 1.5(IQR)…..any number LOWER is an outlier Q3 + 1.5(IQR)…..any number LARGER is an outlier IQR = Q3-Q1**5# Summary of Histogram**10 8 6 4 3 1 1 n=33**Half of 33 is 16.5 so the median falls in the “6” column**Half of 16.5 is 8.25 which makes Q1 in the “4” column On the other half Q3 falls in the “7” column 10 8 6 4 3 1 1 n=33**Chapter 2**Normal Distribution**The Empirical Rule**68,95,99.7 68% of the data is within 1 StDev of the mean 95% of the data is within 2 StDev of the mean 99.7% of the data is within 3 StDev of the mean**Standard Normal Curve**N(0,1)**Outliers on a Normal Curve**You need to remember the outlier formulas: Q1 – 1.5(IQR) Q3 +1.5(IQR)**Outliers on a Normal Curve**Q1 is the 25th percentile Q3 is the 75th percentile Why?**Outliers on a Normal Curve**invNorm(.25) = -.67 invNorm(.75) = .67 IQR = Q3 – Q1 IQR = .67 – (-.67) = 1.34 Q1 – 1.5(IQR) Q3 +1.5(IQR) -.67 – 1.5(1.34) =-2.68 .67 + 1.5(1.34) = 2.68**Outliers on a Normal Curve**z= 2.68 z= -2.68**Outliers on a Normal Curve**z= -.67 z= 2.68 z= -2.68 z= .67 Q1 Q3**Outliers on a Normal Curve**IQR = 1.34 Q1 Q3**Chapter 3**Scatterplots and Linear Regression**Study Time and GPA**Residual Plot A randomly scattered residual plot shows that a linear model is appropriate.**Study Time and GPA**Write the linear equation: GPA = 1.8069326 + .4247748(Study Time)**Study Time and GPA**Interpret the Slope(b): For every hour of study our model predicts an avg increase of .4247748319 in GPA.**Study Time and GPA**Interpret the y-intercept(a): At 0 hours of study our model predicts a GPA of 1.8069326.**Study Time and GPA**Interpret the correlation(r): There is a strong positive linear association between hours of study and GPA.**Study Time and GPA**Interpret the Coefficient of Determination(r2): 66.6% of the variation in GPA can be explained by the approximate linear relationship with hours of study.**Tootsie Pop Grab**If this point was removed, the slope of the line would increase and the correlation would become stronger. Are there any outliers or influential points?**Scatterplot vs Residual Plot**The residual plot uses the same x-axis but the y-axis is the residuals. The residual plot shows the actual points. It shows whether they were above or below the prediction line.**Scatterplot vs Residual Plot**Prediction line**Tootsie Pop Grab**What was the predicted # of pops for a handspanof 24? Predicted # of Pops = -12.9362 + 1.57199(Handspan)**Tootsie Pop Grab**What was the predicted # of pops for a handspan of 24? Predicted # of Pops = -12.9362 + 1.57199(24) 24.79**Tootsie Pop Grab**Check the residual plot for this. What was the ACTUAL # of pops for a handspan of 24? It’s predicted +/- residual. 24.79 + 4 = 28.79**Correlation**r = ± .70 to ± .99 Strong Correlation r = ± .40 to ±.69 Moderate Correlation r = ± .01 to ± .39 Weak Correlation**Scatterplot & Residual Plot**Sometimes you can spot a curved residual plot in the scatterplot**Chapter 12.2**Transformations**Transformations**If your Linear Model x,y is not Appropriate….. There are a few options to try….. Exponential Model x ,Log(y) Power Model Log(x), Log(y) Try the above options in that order, check r2 and the residual plot,……if r2 is high and the residual plot looks good then you have found a suitable model CAUTION…Real data may not have a perfect model….sometimes you have to settle on “good enough”**Baseball Salaries**Ballplayers have been signing very large contracts. The highest salaries (in millions of dollars per season) for some notable players are given in the following table.**Year VS SALARY**R2 is high, however the scatterplot appears to have a curved pattern. A linear model may not be appropriate.**Year vs Log(salary)**This is an exponential model. R2 is very high and the scatterplot shows no curvature. This appears to be a good fit for this data. Make sure to check the residual plot to make sure.**Residual Plot**This residual plot shows no curved pattern and the residuals are randomly scattered above and below the axis…this shows that your model is a good fit.**Exponential model**Use the data from your new model to write the equation. Make a prediction for the salary in the year 2006 Make a prediction for the salary in the year 2015**Exponential model**Log(salary) = -109.133 + 0.05516YEAR Make a prediction using your model for a salary in 2006. About 33 million a year

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