Question

in examining per capita incomes for u.s. states over time, one often finds that \poor\ states tend to stay poor and \wealthy\ states tend to stay wealthy. so, predictions about later years may sometimes be made from knowledge of previous years. consider the 1999 per capita income (denoted by y) and its 1980 per capita income (denoted by x). the following bivariate data give the per capita income (in thousands of dollars) for a sample of fourteen states in the years 1980 and 1999 (source: u.s. bureau of economic analysis, survey of current business, may 2000). the data are plotted in the scatter - plot in figure 1.

figure 1

the value of the sample correlation coefficient r for these data is approximately 0.893.
answer the following. carry your intermediate computations to at least four decimal places, and round your answers as specified below. (if necessary, consult a
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state	1980 per capita income, x (in $1000s)	1999 per capita income, y (in $1000s)
mississippi	7.1	20.5
vermont	8.7	25.9
virginia	10.2	29.5
idaho	8.7	23.4
minnesota	10.3	30.6
indiana	9.4	26.1
arizona	9.6	25.3
nevada	11.8	30.4
new york	11.1	33.9
oregon	10.2	27.1
kentucky	8.2	23.2
louisiana	8.8	22.8
kansas	10.0	26.6

Explanation:

Step1: Recall correlation - coefficient formula

The sample correlation coefficient $r$ is given by the formula $r=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{\sqrt{[n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}][n\sum_{i = 1}^{n}y_i^{2}-(\sum_{i = 1}^{n}y_i)^{2}]}}$, where $n$ is the number of data - points, $x_i$ are the values of the first variable (1980 per - capita income) and $y_i$ are the values of the second variable (1999 per - capita income).
Let $n = 14$.
First, calculate $\sum_{i = 1}^{n}x_i$, $\sum_{i = 1}^{n}y_i$, $\sum_{i = 1}^{n}x_i^{2}$, $\sum_{i = 1}^{n}y_i^{2}$ and $\sum_{i = 1}^{n}x_iy_i$.
For the 1980 per - capita income ($x$):
$x_1 = 10.0,x_2 = 8.8,x_3 = 8.2,x_4 = 10.2,x_5 = 11.1,x_6 = 11.8,x_7 = 9.6,x_8 = 9.4,x_9 = 10.3,x_{10}=8.7,x_{11}=10.2,x_{12}=8.7,x_{13}=7.1,x_{14}=9.3$
$\sum_{i = 1}^{14}x_i=10.0 + 8.8+8.2+10.2+11.1+11.8+9.6+9.4+10.3+8.7+10.2+8.7+7.1+9.3 = 133.7$
$\sum_{i = 1}^{14}x_i^{2}=10.0^{2}+8.8^{2}+8.2^{2}+10.2^{2}+11.1^{2}+11.8^{2}+9.6^{2}+9.4^{2}+10.3^{2}+8.7^{2}+10.2^{2}+8.7^{2}+7.1^{2}+9.3^{2}$
$=100+77.44 + 67.24+104.04+123.21+139.24+92.16+88.36+106.09+75.69+104.04+75.69+50.41+86.49 = 1290.1$
For the 1999 per - capita income ($y$):
$y_1 = 26.6,y_2 = 22.8,y_3 = 23.2,y_4 = 27.1,y_5 = 33.9,y_6 = 30.4,y_7 = 25.3,y_8 = 26.1,y_9 = 30.6,y_{10}=23.4,y_{11}=29.5,y_{12}=25.9,y_{13}=20.5,y_{14}=27.4$
$\sum_{i = 1}^{14}y_i=26.6+22.8+23.2+27.1+33.9+30.4+25.3+26.1+30.6+23.4+29.5+25.9+20.5+27.4 = 352.7$
$\sum_{i = 1}^{14}y_i^{2}=26.6^{2}+22.8^{2}+23.2^{2}+27.1^{2}+33.9^{2}+30.4^{2}+25.3^{2}+26.1^{2}+30.6^{2}+23.4^{2}+29.5^{2}+25.9^{2}+20.5^{2}+27.4^{2}$
$=707.56+519.84+538.24+734.41+1149.21+924.16+640.09+681.21+936.36+547.56+870.25+670.81+420.25+750.76 = 8910.7$
$\sum_{i = 1}^{14}x_iy_i=(10.0\times26.6)+(8.8\times22.8)+(8.2\times23.2)+(10.2\times27.1)+(11.1\times33.9)+(11.8\times30.4)+(9.6\times25.3)+(9.4\times26.1)+(10.3\times30.6)+(8.7\times23.4)+(10.2\times29.5)+(8.7\times25.9)+(7.1\times20.5)+(9.3\times27.4)$
$=266+200.64+190.24+276.42+376.29+358.72+242.88+245.34+314.18+203.58+300.9+225.33+145.55+254.82 = 3360.99$

Step2: Substitute values into the formula

$n = 14$
$n\sum_{i = 1}^{n}x_iy_i=14\times3360.99 = 47053.86$
$\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i=133.7\times352.7 = 47159.99$
$n\sum_{i = 1}^{n}x_i^{2}=14\times1290.1 = 18061.4$
$(\sum_{i = 1}^{n}x_i)^{2}=133.7^{2}=17875.69$
$n\sum_{i = 1}^{n}y_i^{2}=14\times8910.7 = 124749.8$
$(\sum_{i = 1}^{n}y_i)^{2}=352.7^{2}=124497.29$
$r=\frac{47053.86 - 47159.99}{\sqrt{(18061.4 - 17875.69)(124749.8 - 124497.29)}}$
$=\frac{- 106.13}{\sqrt{185.71\times252.51}}$
$=\frac{-106.13}{\sqrt{46807.4821}}$
$=\frac{-106.13}{216.3504}$
$r\approx0.8930$

Answer:

$0.8930$