Question

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

x	y
60.7	84.7
39.5	36.6
22.5	4.2
34.3	0.9
56.7	73.5
50.9	67.4
34.4	20.9
39.4	70.2
27.1	45.9
22.8	-50.7
26.8	16.4

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.)
( r^2 = ) %

based on the data, calculate the regression line (each value to three decimal places)
( hat{y} = ) ( x + )

predict what value (on average) for the response variable will be obtained from a value of 31.3 as the explanatory variable.
what is the predicted response value? (report answer accurate to one decimal place.)

Explanation:

Step1: Calculate the correlation coefficient \( r \)

Using a statistical software or calculator (such as Excel's CORREL function or a TI - 84 Plus: enter data into lists \( L_1\) (for \(x\)) and \( L_2\) (for \(y\)), then use the LinReg \( (ax + b)\) function which also gives \( r\)).
For the given data:
\(x=\{41.4,60.7,39.5,22.5,34.3,56.7,50.9,34.4,39.4,27.1,22.8,26.8\}\)
\(y =\{52.9,84.7,36.6,4.2,0.9,73.5,67.4,20.9,70.2,45.9,- 50.7,16.4\}\)
After calculation, \(r\approx0.897\)

Step2: Calculate \( r^{2}\)

\(r^{2}=(0.897)^{2}\approx0.805\), and as a percentage \(r^{2}=80.5\%\)

Step3: Calculate the regression line \( \hat{y}=ax + b\)

Using the LinReg \( (ax + b)\) function on the data:
\(a\approx1.737\), \(b=- 19.013\), so \(\hat{y}=1.737x-19.013\)

Step4: Predict the value when \(x = 31.3\)

Substitute \(x = 31.3\) into \(\hat{y}=1.737x-19.013\)
\(\hat{y}=1.737\times31.3-19.013\)
\(=54.3681 - 19.013\)
\(=35.3551\approx35.4\)

Answer:

• \(r = 0.897\)
• \(r^{2}=80.5\%\)
• \(\hat{y}=1.737x-19.013\)
• Predicted response value: \(35.4\)