Question

standardized test score, x grade point average, y xy
950 2.19 2080.5
1200 2.97 3564
1060 2.96 3137.6
1100 2.39 2629
1000 2.94 2940
1500 3.19 4785
1290 2.99 3857.1
1500 3.33 4995
1340 3.52 4716.8
790 2.47 1951.3
1400 3.07 4298
1000 2.32 2320
840 2.26 1898.4
900 2.57 2313
1240 3.45 4278
what is the sample correlation coefficient for these data? carry your intermediate computations to at least four places and round your answer to at least three decimal places. (if necessary, consult a list of formulas.)

Explanation:

Step1: Calculate the means of \(x\) and \(y\)

Let \(n = 15\) (the number of data points).
\(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{950 + 1200+1060+1100+1000+1500+1290+1500+1340+790+1400+1000+840+900+1240}{15}=\frac{17950}{15}\approx1196.67\)
\(\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}=\frac{2.19+2.97+2.96+2.39+2.94+3.19+2.99+3.33+3.52+2.47+3.07+2.32+2.26+2.57+3.45}{15}=\frac{41.6}{15}\approx2.77\)

Step2: Calculate \(S_{xx}\), \(S_{yy}\), and \(S_{xy}\)

\(S_{xx}=\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\)
\(S_{yy}=\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}\)
\(S_{xy}=\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})\)

First, calculate \(S_{xx}\):
\(\sum_{i = 1}^{n}x_{i}^{2}=950^{2}+1200^{2}+1060^{2}+1100^{2}+1000^{2}+1500^{2}+1290^{2}+1500^{2}+1340^{2}+790^{2}+1400^{2}+1000^{2}+840^{2}+900^{2}+1240^{2}\)
\(=902500 + 1440000+1123600+1210000+1000000+2250000+1664100+2250000+1795600+624100+1960000+1000000+705600+810000+1537600\)
\(=22077700\)
\(S_{xx}=\sum_{i = 1}^{n}x_{i}^{2}-n\bar{x}^{2}=22077700-15\times(1196.67)^{2}\)
\(=22077700 - 15\times1432026.69\)
\(=22077700-21480399.95 = 597300.05\)

Next, calculate \(S_{yy}\):
\(\sum_{i = 1}^{n}y_{i}^{2}=2.19^{2}+2.97^{2}+2.96^{2}+2.39^{2}+2.94^{2}+3.19^{2}+2.99^{2}+3.33^{2}+3.52^{2}+2.47^{2}+3.07^{2}+2.32^{2}+2.26^{2}+2.57^{2}+3.45^{2}\)
\(=4.7961+8.8209+8.7616+5.7121+8.6436+10.1761+8.9401+11.0889+12.3904+6.1009+9.4249+5.3824+5.1076+6.6049+11.9025\)
\(=122.853\)
\(S_{yy}=\sum_{i = 1}^{n}y_{i}^{2}-n\bar{y}^{2}=122.853-15\times(2.77)^{2}\)
\(=122.853-15\times7.6729\)
\(=122.853 - 115.0935=7.7595\)

Then, calculate \(S_{xy}\):
\(\sum_{i = 1}^{n}x_{i}y_{i}=2080.5 + 3564+3137.6+2629+2940+4785+3857.1+4995+4716.8+1951.3+4298+2320+1898.4+2313+4278\)
\(=50763.7\)
\(S_{xy}=\sum_{i = 1}^{n}x_{i}y_{i}-n\bar{x}\bar{y}=50763.7-15\times1196.67\times2.77\)
\(=50763.7-15\times3314.7759\)
\(=50763.7 - 49721.6385=1042.0615\)

Step3: Calculate the sample correlation coefficient \(r\)

The formula for the sample correlation coefficient is \(r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}\)
\(r=\frac{1042.0615}{\sqrt{597300.05\times7.7595}}\)
\(=\frac{1042.0615}{\sqrt{4630319.9}}\)
\(=\frac{1042.0615}{2151.82}\approx0.484\)

Answer:

\(0.484\)