Question

The well-known psychologist Dr. Elbod has established what he calls his Generalized Anxiety Scale (GAS). The GAS, which is a scale from 0 to 10, measures the "general anxiety of an individual, with higher GAS scores corresponding to more anxiety. (Dr. Elbod's assessment of anxiety is based on a variety of measurements, both physiological and psychological.) We're interested in making predictions about individuals' sleep behavior based on their GAS scores. The bivariate data below give the GAS score (denoted by x) and the number of hours of sleep last night (denoted by y) for each of the fifteen adults participating in a study. A scatter plot of the data is shown in Figure 1. Also given are the products of the GAS scores and sleep times for each of the fifteen adults. (These products, written in the column labelled "xy," may aid in calculations.) GAS score, x Sleep time, y (in hours) xy 10 8.0 1.4 6.6 3.6 1.1 9.1 3.1 5.4 8.6 5.9 7.5 7.4 5.7 7.1 6.3 8.5 43.2 12.04 38.94 27 8.14 51.87 22.01 44.73 34 28.4 46.4 16.4 28.32 56.16 56.7 7.1 4.0 4.0 5.8 2.0 4.8 7.8 9.0 7.1 8.0 8.2 5.9 7.2 6.3 Figure 1 Send data be Excel Answer the following. Carry your intermediate computations to at least four decimal places, and round your answer as specified below. What is the value of the sample correlation coefficient for these data? Round your answer to at least three decimal places. 10 х 6 ?

Answer 1

The correlation coefficient between two series x and y is given by

$r=\frac{\sum(x_i-\overline x )(y_i- \overline y)} {\sqrt(\sum(x_i-\overline x )^2\sum(y_i-\overline y )^2)}$

    Equation II

Σ(rigi – ηZ 9) Σ? – ηπ?)(Συ? - my2)

The table with calculation is given below

А - х y ху х^2 у*2 1 8.0 5.4 43.20 64.00 29.16 2 1.4 8.6 12.04 1.96 73.96 3 6.6 5.9 38.94 43.56 34.81 4 3.6 7.5 27.00 12.96 56.25 5 1.1 7.4 8.14 1.21 54.76 6 9.1 5.7 51.87 82.81 32.49 7 3.1 7.1 22.01 9.61 50.41 8 7.1 6.3 44.73 50.41 39.69 9 4.0 8.5 34.00 16.00 72.25 10 4.0 7.1 28.40 16.00 50.41 11 5.8 8.0 46.40 33.64 64.00 12 2.0 8.2 16.40 4.00 67.24 13 4.8 5.9 28.32 23.04 34.81 14 7.8 7.2 56.16 60.84 51.84 15 9.0 6.3 56.70 81.00 39.69

From above table we can get following results

$\overline x=$5.16

$\overline y=$7.006667

$\sum x_iy_i=$ 514.31

$\sum x_i^2=$501.04

$\sum y_i^2=$751.77

n=5

Using these value values in equation II , we have,

r   $=\frac{\sum(x_iy_i- n\overline x~\overline y)} {\sqrt(\sum(x_i^2-n\overline x ^2)\sum(y_i^2-n\overline y ^2))}$

=$=\frac{514.31-15*5.16*7.006667} {\sqrt(501.04-15*(5.16)^2)(751.77-15*(7.006667)^2)}=~~-0.70853$   

$r\approx -0.709$ ( rounded off 3 decimal place)