Combining survey and voter turnout data, a regression plot is created with %turnout as the dependent variable, and the percentage of previous survey respondents who stated they had a high likelihood of voting as the independent variable (also as a percentage). The units of analysis are countries within the EU before a parliamentary vote.
Our best-fitting regression equation is as follows:
y=% predicted actual country voter turnout
x = % expected turnout from survey
y = -50.387% + 1.3789x
If the % expected turnout in a country is 70%, what would be the predicted actual turnout?
Next, the intercept is -50.387%. Although not seeming reasonable, interpret this intercept.
Then, y=% predicted actual country voter turnout
x = % expected turnout from survey
y = -50.387% + 1.3789x
Two countries differ in their survey expectation by 20%. Using this regression estimate, by what percentage would they be predicted to differ in their actual turnout and then find the value R-squared value for our equation is 0.4068. What does that value tell us about the relationship between expected and actual turnout?
Don't just write--"it's strong" or "moderate" or "weak." Make sure '40.68' or '.4068' appears somewhere in your answer.
Solution1:
given y=-50.387%+1.3789x
where
x=% expecetd turnout from survey
y=% predicted actual country voter turn out
Given x=70%
predicted actual turn out=
-50.387+1.3789*70
=46.136%
predicted ctual turn out=46.136
Solution2:
given y=-50.387%+1.3789x
where
x=% expecetd turnout from survey
y=% predicted actual country voter turn out
for y intercept put x=0
y=-50.387%+1.3789*0
y=-50.387%
that is when there are no % expected turnout from survey,
predicted actual country voter turn out decreases by 50.387%
y intercept ahs no meaning
Solution3:
r sq=0.4068
40.68% variance in predicted actual country voter turn out is explained by expected turnout from survey
expalined variance=40.68%
unexplained variance=100-40.68=59.32%
rq=0.4068
r=sqrt(0.4068)=0.6378
there exists a strong positive relationship between y and x.
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