Anchal B.

asked • 11/28/23

Run a regression analysis on the following bivariate set of data with y as the response variable.

x y

17.9 98.2

42.3 -14.4

19.4 94

18.8 70

59.2 7.9

27.3 34.4

8.6 135.4

2.3 137.8

24.7 80.1

33.9 54.8

-1.2 77.3

36.7 54


Find the correlation coefficient and report it accurate to three decimal places.

r = ______


What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place. (If the answer is 0.84471, then it would be 84.5%...you would enter 84.5 without the percent symbol.)

r² = _____%


Based on the data, calculate the regression line (each value to three decimal places)


y = ___  x + ___


Predict what value (on average) for the response variable will be obtained from a value of 13.1 as the explanatory variable. Use a significance level of a = 0.05 to assess the strength of the linear correlation.


What is the predicted response value? (Report answer accurate to one decimal place.)

y = 

1 Expert Answer

By:

William W.

Your method for calculating the slope and y-intercept are incorrect as you are using data points instead of points on the regression line. The manual method for calculating slope is complex but is r*(Sy/Sx) where Sy = sqrt(sum(y-ybar)^2/(n-1)) and Sx = sqrt(sum(x-xbar)^2/(n-1)) but, again is easy to find using a calculator. The manual method for calculating "b" is y-hat - m*x-hat. My calculator shows the regression line as y = -2.1809x + 121.812
Report

11/29/23

Joshua L.

tutor
True, Mae culpa, but it will be impractical to use that formula. One issue with these types of problems on Wyzant is we don’t know what technology (TI, Minitab, SPSS) students have access to.
Report

11/29/23

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