Answer to Question #346245 in Statistics and Probability for wiwili

Question

A study was conducted to investigate the relationship between roadway surface temperature (x) and pavement deflection (y). The data are as follows:

x

65.3

66.4

70.8

70.7

69.8

71.6

72.1

72.4

73.5

y

0.6144

0.6273

0.6668

0.6781

0.6497

0.6678

0.6783

0.6755

0.6873

Calculate the Pearson correlation coefficient. Round your answers to the nearest hundredths.

Answer 1

The following table needs to be used:

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} X & Y & XY & X^2 & Y^2 \\ \hline 65.3 & 0.6144 & 40.12032 & 44264.09 & 0.37748736 \\ \hdashline 66.4 & 0.6273 & 41.65272 & 4408.96 & 0.39350529 \\ \hdashline 70.8 & 0.6668 & 47.20944 & 5012.64 & 0.44462224 \\ \hdashline 70.7 & 0.6781 & 47.94167 & 4998.49 & 0.45981961 \\ \hdashline 69.8 & 0.6497 & 45.34906 & 4872.04 & 0.42211009 \\ \hdashline 71.6 & 0.6678 & 47.81448 & 5126.56 & 0.44595684 \\ \hdashline 72.1 & 0.6783 & 48.90543 & 5198.41 & 0.46009089 \\ \hdashline 72.4 & 0.6755 & 48.9062 & 5241.76 & 0.45630025 \\ \hdashline 73.5 & 0.6873 & 50.51655 & 5402.25 & 0.47238129 \\ \hdashline Sum =\\ 632.6 & 5.9452 & 418.41587 & 44525.2 & 3.93227386\\ \hdashline \end{array}$

$\bar{X}=\dfrac{1}{n}\sum _{i}X_i=\dfrac{632.6}{9}$ $=70.288889$ $\bar{Y}=\dfrac{1}{n}\sum _{i}Y_i=\dfrac{5.9452}{9}$ $=0.660578$ $SS_{XX}=\sum_iX_i^2-\dfrac{1}{n}(\sum _{i}X_i)^2$ $=44525.2-\dfrac{632.6^2}{9}=60.448889$ $SS_{YY}=\sum_iY_i^2-\dfrac{1}{n}(\sum _{i}Y_i)^2$ $=3.93227386-\dfrac{5.9452^2}{9}=0.005007$

$SS_{XY}=\sum_iX_iY_i-\dfrac{1}{n}(\sum _{i}X_i)(\sum _{i}Y_i)$ $=418.41587-\dfrac{632.6(5.9452)}{9}=0.534368$ $r=\dfrac{SS_{XY}}{\sqrt{SS_{XX}SS_{YY}}}=\dfrac{0.534368}{\sqrt{60.448889(0.005007)}}$

$=0.97132277$