Page 308 - Statistics for Dummies

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Part V: Statistical Studies and the Hunt for a Meaningful Relationship
The slope, m, for the best-fitting line for the subset of cricket chirps versus
. So as the number of chirps
temperature data is
increases by 1 chirp per 15 seconds, the temperature is expected to increase
by 0.90 degrees Fahrenheit on average. To get a more meaningful interpre-
tation, you can multiply the top and bottom of the slope by 10 and say as
chirps increase by 10 (per 15 seconds) temperature increases 9 degrees
Fahrenheit.
Now, to find the y-intercept, b, you take
the best-fitting line for predicting temperature from cricket chirps based on
the data is y = 0.90x + 43.15, or temperature (in degrees Fahrenheit) = 0.90 ∗
(number of chirps in 15 seconds) + 43.2. Now can you use the y-intercept to pre-
dict temperature when no chirping is going on at all? Because no data was col-
lected at or near this point, you cannot make predictions for temperature in this
area. You can’t predict temperature using crickets if the crickets are silent.
Making Proper Predictions , or 67 – (0.90)(26.5) = 43.15. So
After you have determined a strong linear relationship and you find the equa-
tion of the best fitting line using y = mx + b, you use that line to predict (the
average) y for a given x-value. To make predictions, you plug the x-value into
the equation and solve for y. For example, if your equation is y = 2x + 1 and
you want to predict y for x = 1, then plug 1 into the equation for x to get
y = 2(1) + 1 = 3.
Keep in mind that you choose the values of X (the explanatory variable)
that you plug in; what you predict is Y, the response variable, which totally
depends on X. By doing this, you are using one variable that you can easily
collect data on to predict a Y variable that is difficult or not possible to mea-
sure. This process works well as long as X and Y are correlated. This concept
is the big idea of regression.
Using the examples from the previous section, the best-fitting line for the
crickets is y = 0.90x + 43.2. Say you’re camping outside, listening to the crick-
ets, and remember you can predict temperature by counting cricket chirps.
You count 35 chirps in 15 seconds, put in 35 for x, and find that y = 0.9(35) +
43.2 = 74.7. (Yeah, you memorized the formula before you went camping just
in case you needed it.) So because the crickets chirped 35 times in 15 sec-
onds, you figure the temperature is probably about 75 degrees Fahrenheit.
Just because you have a regression line doesn’t mean you can plug in any value
for X and do a good job of predicting Y. Making predictions using x-values that
fall outside the range of your data is a no-no. Statisticians call this extrapolation;
watch for researchers who try to make claims beyond the range of their data.

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