# A student wonders if tall women tend to date taller men than do short women. She measures herself, her dormitory roommate, and the women in the adjoining.

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A student wonders if tall women tend to date taller men than do short women. She measures herself, her dormitory roommate, and the women in the adjoining rooms. Then she measures the next man each woman date. Draw & discuss the scatterplot and calculate the correlation coefficient. Women (x) Men (y) 6672 6468 6670 6568 7071 65

Linear Regression

Guess the correlation coefficient http://istics.net/stat/Correlations/

Can we make a Line of Best Fit

Regression Line This is a line that describes how a response variable (y) changes as an explanatory variable (x) changes. It’s used to predict the value of (y) for a given value of (x). Unlike correlation, regression requires that we have an explanatory variable.

Let’s try some! http://illuminations.nctm.org/ActivityDetail.asp x?ID=146 http://illuminations.nctm.org/ActivityDetail.asp x?ID=146

Regression Line

The following data shows the number of miles driven and advertised price for 11 used Honda CR-Vs from the 2002-2006 model years (prices found at www.carmax.com). The scatterplot below shows a strong, negative linear association between number of miles and advertised cost. The correlation is -0.874. The line on the plot is the regression line for predicting advertised price based on number of miles. Thousand Miles Driven Cost (dollars) 2217998 2916450 3514998 3913998 4514599 4914988 5513599 5614599 6911998 7014450 8610998

The regression line is shown below…. Use it to answer the following. Slope: Y-intercept:

Predict the price for a Honda with 50,000 miles.

Extrapolation This refers to using a regression line for prediction far outside the interval of values of the explanatory variable x used to obtain the line. They are not usually very accurate predictions.

Slope: Y-int: Predict weight after 16 wk Predict weight at 2 years:

Residual

The equation of the least-squares regression line for the sprint time and long- jump distance data is predicted long-jump distance = 304.56 – 27.3 (sprint time). Find and interpret the residual for the student who had a sprint time of 8.09 seconds.

Regression Let’s see how a regression line is calculated.

Fat vs Calories in Burgers Fat (g)Calories 19410 31580 34590 35570 39640 39680 43660

Let’s standardize the variables FatCalz - x'sz - y's 19410-1.959-2 31580-0.42-0.1 34590-0.0360 355700.09-0.2 396400.60.56 396800.61 436601.120.78 The line must contain the point and pass through the origin.

Let’s clarify a little. (Just watch & listen) The equation for a line that passes through the origin can be written with just a slope & no intercept: y = mx. But, we’re using z-scores so our equation should reflect this and thus it’s Many lines with different slope pass through the origin. Which one fits our data the best? That is which slope determines the line that minimizes the sum of the squared residuals.

Line of Best Fit –Least Squares Regression Line It’s the line for which the sum of the squared residuals is smallest. We want to find the mean squared residual. Focus on the vertical deviations from the line. Residual = Observed - Predicted

Let’s find it. (just watch & soak it in) St. Dev of z scores is 1 so variance is 1 also. This is r!

Continue…… Since this is a parabola – it reaches it’s minimum at This gives us Hence – the slope of the best fit line for z-scores is the correlation coefficient → r.

Slope – rise over run A slope of r for z-scores means that for every increase of 1 standard deviation in, there is an increase of r standard deviations in. “Over 1 and up r” Translate back to x & y values – “over one standard deviation in x, up r standard deviations in y. Slope of the regression line is:

Why is correlation “r” Because it was calculated from the regression of y on x after standardizing the variables – just like we have just done – thus he used r to stand for (standardized) regression.

The number of miles (in thousands) for the 11 used Hondas have a mean of 50.5 and a standard deviation of 19.3. The asking prices had a mean of \$14,425 and a standard deviation of \$1,899. The correlation for these variables is r = -0.874. Find the equation of the least-squares regression line and explain what change in price we would expect for each additional 19.3 thousand miles.

So let’s write the equation! Fat (g)Calories 19410 31580 34590 35570 39640 39680 43660 Slope: Explain the slope:

Homework Page 191 (27-32, 35, 37, 39, 41)

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