Line of Best Fit Our objective is to fit a line in the scatterplot that fits the data the best Line of best fit looks like:

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1 Line of Best Fit Our ojective is to fit a line in the scatterplot that fits the data the est Line of est fit looks like: ŷ = + x Least Square Regression Line That s a hat on the y, meaning that it is a prediction not the actual y values. VERY IMPORTANT!!! The line of est fit we calculate is called the LEAST SQUARES REGRESSION LINE (LSRL) Called this ecause it minimizes the sum of the squared residuals y ( yˆ ) 2 Slope in the z s If we look at the scatterplot of the z-scores we find that the line of est fit must go through (,) The slope of the line that minimizes the sum of squares in the z-scores will always e r. This tells you that on average for each increase of standard deviation in x there is a change of r standard deviations in y. Example: Square Foot vs. Selling Price for Houses in Boulder, CO (MM Tale 2.3) House Sales r = Square_feet

2 Here is the scatterplot of the z-scores with the line that minimizes the sum of squares. House Sales 3 Slope in the z scores Ex. zˆ y =.677zx If a house is standard deviation aove the mean for square feet in space we would expect the price to e.677 standard deviations aove the mean price z_sf z_sp =.67z_sf zˆ =.677z y x If a house is.4 standard deviations elow the mean for square feet in space we would expect the price to e.677 x -.4 = standard deviations elow the mean price Switching Variales If the explanatory and response variale are switched the correlation remains the same Therefore the slope remains the same if they are switched as well. zˆ y =.677zx zˆ x =.677zy Ex. If a house is.2 standard deviations elow the mean for price we would expect the space to e.82 st. dev. elow the mean square feet. Slope in the actual scatterplot Since the slope of the line in the z-scores compares the standard deviations we include these ack to get the slope of the line in the scatterplot of the data. Thus the slope of the line in the regular scatterplot ecomes sy = r s x

3 House sales s price = s square _ feet = 64.5 r =.677 = s y r s =.677 x 64.5 = Interpretation of Slope On average for every increase of (x unit) in the (x variale) there tends to e an increase/decrease of (Slope) (y units) in the (y variale). House sales = $ per Sq. Foot Interpretation: On average for every increase of sq. ft. in the size there tends to e an increase of $47.73 in the selling price. Finding intercept, Now that we have the slope, we only need a point that the line runs through to get the intercept. We have one: ( x, y) So the equation for intercept ecomes: = y x Interpretation of the intercept is generally meaningless. So e careful!

4 House Sales x = y = 7733 = = y x = ( ) = LSRL : yˆ = x House Sales price = ( square feet) Square_feet Selling_Price = ( Square_feet) Making Predictions We use the model to make predictions on what the price would e for a certain size of house. Ex. What price would we expect from a house that was 2 sq. ft? price ˆ = (2) price ˆ = $ Residuals Residuals are difference etween the actual data and predictions in the data. Or oserved minus expected. e= y yˆ Positive residuals represent underestimations Negative residuals represent overestimations The sum of the residuals should e zero. Ex. From the data the cost of one of the houses that was 2 sq. ft. was $75,. price = $ e = e = $27,69.27

5 A survey was conducted in the United States and countries of Western Europe to determine the percentage of teenagers who had used marijuana and other drugs. Ch7_Drug_ause Marijuana_ Ch7_Drug_ause Other_Drugs_ Marijuana_.934. Descrie the scatterplot Positive, Linear, Strong (r =.934) Possile outlier at (55,32) ut it s with in the linear pattern 2. Does the linear model seem appropriate? Yes. There is no ovious curve. Ch7_Drug_ause Other_Drugs_ Marijuana_.934 S = correlation ( ) 3. If you knew a country had a percent of teenagers that used marijuana that was.5 standard deviations aove the mean, what would you predict aout the percent of teenagers that used other drugs? I would expect it to e.4 standard deviations aove the mean percent of other drug usage. 4. If you knew a country s percent of teenagers that used other drugs was.8 standard deviations elow the mean, what would you predict aout the percent of teenagers that used marijuana for that country? I would expect it to e.68 standard deviations elow the mean percent of marijuana usage. S = correlation ( ) = 5. Calculate the equation for the LSRL =.65 yˆ = x = (23.99) = 3.68 x - % of teenagers that use marijuana y - % of teenagers that use other drugs 6. Interpret the slope in the context of the prolem. On average for every increase of percent in marijuana usage there tends to e an increase of.65 percent in other drug usage. 7. If a country had a percent of teenagers that used marijuana at 9%, what would you predict that the percent of teenagers that used other drugs to e for that country? yˆ = (9) yˆ = 8.67% 8. If the actual percent of teenagers that used other drugs for the country in #7 is 8%, what is the residual? Did the model overestimate or underestimate the actual value? yˆ = 8.67%; y = 8% e = e =.67% Overestimation

6 9. If a country had a percent of teenagers that used marijuana of 68%, what would your model predict for the percent of teenagers that used other drugs for that country? How confident are you on that prediction? yˆ = (68) yˆ = % I would not e confident in this prediction ecause the value 68% is far eyond the range of the data. Any prediction made would e an extrapolation.