Question

distance from the beach, x (in miles) selling price, y (in thousands of dollars) xy
5.4 268.3 1448.82
12.0 280.2 3362.4
6.3 248.4 1564.92
10.1 301.2 3042.12
11.9 186.1 2214.59
11.1 193.7 2150.07
12.3 206.4 2538.72
11.4 269.5 3072.3
11.9 218.7 2602.53
18.3 224.6 4110.18
13.3 268.8 3575.04
7.0 304.9 2134.3
7.6 226.0 1717.6
7.5 220.7 1655.25
4.0 273.2 1092.8
2.9 320.7 930.03
send data to calculator send data to excel
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 \(n = 16\) (number of data points).
\(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\)
\(\sum_{i=1}^{16}x_{i}=5.4 + 12.0+6.3 + 10.1+11.9+11.1+12.3+11.4+11.9+18.3+13.3+7.0+7.6+7.5+4.0+2.9=154.2\)
\(\bar{x}=\frac{154.2}{16}=9.6375\)

\(\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}\)
\(\sum_{i = 1}^{16}y_{i}=268.3+280.2 + 248.4+301.2+186.1+193.7+206.4+269.5+218.7+224.6+268.8+304.9+226.0+220.7+273.2+320.7 = 4015.1\)
\(\bar{y}=\frac{4015.1}{16}=250.94375\)

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

\(S_{xx}=\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=\sum_{i = 1}^{n}x_{i}^{2}-n\bar{x}^{2}\)
\(\sum_{i = 1}^{n}x_{i}^{2}=5.4^{2}+12.0^{2}+6.3^{2}+10.1^{2}+11.9^{2}+11.1^{2}+12.3^{2}+11.4^{2}+11.9^{2}+18.3^{2}+13.3^{2}+7.0^{2}+7.6^{2}+7.5^{2}+4.0^{2}+2.9^{2}=1877.94\)
\(S_{xx}=1877.94-16\times(9.6375)^{2}=1877.94 - 16\times92.88265625=1877.94 - 1486.1225=391.8175\)

\(S_{yy}=\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}=\sum_{i = 1}^{n}y_{i}^{2}-n\bar{y}^{2}\)
\(\sum_{i = 1}^{n}y_{i}^{2}=268.3^{2}+280.2^{2}+248.4^{2}+301.2^{2}+186.1^{2}+193.7^{2}+206.4^{2}+269.5^{2}+218.7^{2}+224.6^{2}+268.8^{2}+304.9^{2}+226.0^{2}+220.7^{2}+273.2^{2}+320.7^{2}=1031997.31\)
\(S_{yy}=1031997.31-16\times(250.94375)^{2}=1031997.31-16\times62972.779297=1031997.31 - 1007564.46875 = 24432.84125\)

\(S_{xy}=\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})=\sum_{i = 1}^{n}x_{i}y_{i}-n\bar{x}\bar{y}\)
\(\sum_{i = 1}^{n}x_{i}y_{i}=1448.82+3362.4+1564.92+3042.12+2214.59+2150.07+2538.72+3072.3+2602.53+4110.18+3575.04+2134.3+1717.6+1655.25+1092.8+930.03 = 37213.76\)
\(S_{xy}=37213.76-16\times9.6375\times250.94375=37213.76 - 16\times2419.3359375=37213.76 - 38709.375=-1495.615\)

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

\(r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}\)
\(r=\frac{- 1495.615}{\sqrt{391.8175\times24432.84125}}\)
\(\sqrt{391.8175\times24432.84125}=\sqrt{9575197.777}\approx3094.3817\)
\(r=\frac{-1495.615}{3094.3817}\approx - 0.483\)

Answer:

\(-0.483\)