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Chapter 16 Example.pdf

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Department
Administrative Studies
Course
ADMS 3330
Professor
Michael Rochon
Semester
Winter

Description
© M.Rochon – All Rights Reserved. 2013. Consider the following data values for variables x and y: x 2 4 6 81013 y 7 11 17 21 27 36 a) Determine the least squares regression line. b) Find the predicted value of y for x = 9. c) What does the value of the slope of the regression line tell you? d) Calculate the coefficient of determination, and describe what this statistic tells you about the relationship between the two variables. e) Calculate Se and describe that value that is found. f) Can we conclude at the 5% significance level that higher cigarette consumption (x = number of cigarettes smoked) causes higher consumption of coffee (y = number of cups of coffee drank per week). g) Calculate the Coefficient of Correlation h) Predict with 95% confidence the number of coffees drank by a person if that person smokes 10 cigarettes. i) Find the Confidence Interval Estimator with 95% confidence of the number of coffees drank by a person if that person smokes 10 cigarettes. © M.Rochon – All Rights Reserved. 2013. A) I suggest putting the information that you require into tabular form (meaning simply organizing the data into a table reasolved). It allows you to organize the information, produce the necessary statistics you need to put into the formulas and allows you a quick way to check the work over. Th e following columns will be needed (shown in grey box below): 2 2 n xi yi x i y i xiyi 1 2 7 4 4914 2 4 1116 12144 3 6 1736 28102 4 8 2164 44168 5 10 27100 729270 6 13 36 169 1296 468 n=6 = 43 =119 = 389 = 2925 =1066 ∑ ∑ ∑ ∑ ∑ 43 119 x = 6 y = 6 =7.1667 = 19.8333 The SUMS of each of the columns will be used in the formulas that follow to get to the Least Squares Regression Line. Remember that this is a process and that the calculations you did above will feed into the rest of the question. © M.Rochon – All Rights Reserved. 2013. First we will calculate the Covariance Covariance ⎡ n n ⎤ ⎢ ∑ ∑xi yi⎥ 1 ⎢ n =1 i=1 ⎥ Sxy or cov( , ) = ∑ xi i − n 1 ⎢i=1 n ⎥ ⎢ ⎥ ⎣ ⎦ (43)(119) = 1 ⎡1066 − ⎤ 6 −1 ⎣ 6 ⎦ 1 = 1066 − 852.8333 5 = .2 [213.1667] = 42.6333 Next we will need the variance of x: Variance ⎡ ⎛ n ⎞2⎤ ⎢ ⎜∑ xi ⎟ ⎥ 2 1 ⎢ n 2 ⎝i=1 ⎠ ⎥ sx = ⎢∑ xi − ⎥ n −1 ⎢i=1 n ⎥ ⎢ ⎥ ⎣ ⎦ 1 ⎡ (43 2⎤ = ⎢389 − ⎥ 5 ⎣ 6 ⎦ =.20 389 − 308.1667 = 16.1667 © M.Rochon – All Rights Reserved. 2013. Now we are ready to calculate beta 1: Slope Coefficient S b = xy 1 s2 x 42.6333 = 16.1667 = 2.6371 It is now time to calculate the y-intercept: Y-Intercept b = y − b x 0 1 =19.8333− 2.6371(7.1667) =.934 Put the prior two calculations together and you get: Least Squares Line is y =.934 + 2.6371x © M.Rochon – All Rights Reserved. 2013. B) When x=9, the predictive value of y is: This is called a point predication. y= .934 + 2.6371(9) We simply use the least squares regression line to predict the value = 24.6679 of Y, given a value for X. C) The slope of the regression line tells us that xfincreases by one unit, yon average will increase by 2.6371. The above interpretation is a standard interpretation for the slope. If the slope was a negative number (-2.6371) then, as x increases by 1 unit, then y on average will decrease by 2.6371 D) Coefficient of Determination (**remember from the slides, this is one of our 3 assessment techniques) S 2 R = xy s s 2 x y First must calculate the variance of y: 2 ⎡ ⎛ n ⎞ ⎤ ⎢ ⎜ ∑ yi⎟ ⎥ 2 1 ⎢ n 2 ⎝ =1 ⎠ ⎥ sy = n − 1 ⎢∑ yi − n ⎥ ⎢=1 ⎥ ⎢ ⎥ ⎣ ⎦ = .2 [2925 – 2360.1667] = 112.9667 © M.Rochon – All Rights Reserved. 2013. Now we have all the rest of the information to answer the question: 2 2 S xy R = 2 2 sx y This is a standard conclusion for the (42.6333) 2 Coefficient of R = Determination and (16.1667)(112.9667) should be followed when you are = 1817.5983 interpreting this 1826.2987
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