The following data gave X = the water content of snow on April 1 and Y = the yield from April to July (in inches) on the Snake River watershed in Wyoming for 1919 to 1935. (The data were taken from an article in Research Notes,Vol. 61, 1950, Pacific Northwest Forest Range Experiment Station, Oregon). x 23.1 32.8 31.8 32.0 30.4 24.0 39.5 24.2 52.5 37.9 30.5 25.1 12.4 35.1 31.5 21.1 27.6 y 10.5 16.7 18.2 17.0 16.3 10.5 23.1 12.4 24.9 22.8 14.1 12.9 8.8 17.4 14.9 10.5 16.1 (a) Estimate the correlation between Y and X. Round your answer to 3 decimal places

Answer:r=0.933Step-by-step explanation:X = the water content of snow on April 1 Y = the yield from April to July (in inches) on the Snake River watershed in Wyoming for 1919 to 1935.x: 23.1 32.8 31.8 32.0 30.4 24.0 39.5 24.2 52.5 37.9 30.5 25.1 12.4 35.1 31.5 21.1 27.6y: 10.5 16.7 18.2 17.0 16.3 10.5 23.1 12.4 24.9 22.8 14.1 12.9 8.8 17.4 14.9 10.5 16.1We can construct the following tablen  [tex]\sum x[/tex]     [tex]\sum y[/tex]     [tex]\sum x^2[/tex]       [tex]\sum y^2[/tex]        [tex]\sum xy[/tex]______________________________________________1    23.1    10.5   533.61     110.25     242.552   32.8   16.7    1075.84   278.89   547.763   31.8    18.2    1011.24     331.24    578.764   32.0   17.0    1024         289        5445   30.4   16.3    924.16      265.69   495.526   24.0   10.5    600.25     110.25     2527   39.5    23.1    1560.25    533.61    912.458   24.2    12.4    585.64      153.76    300.489   52.5    24.9   2756.25    620.01   1307.2510  37.9    22.8   1436.41      519.84    864.1211   30.5    14.1     930.25      198.81     430.0512   25.1    12.9    630.01       166.41     323.7913   12.4     8.8     153.76       77.44      109.1214   35.1     17.4    1232.01      302.76   610.7415   31.5     14.9    992.25      222.01    469.3516   21.1      10.5    445.21       110.25     221.5517    27.6    16.1     761.76       259.21    444.36____________________________________________On this case n=17, [tex]\sum x = 511.5[/tex]   [tex]\sum y=267.1[/tex]   [tex]\sum x^2 =16628.65[/tex]  [tex]\sum y^2 =4549.43[/tex]   [tex]\sum xy =8653.45[/tex]And we can use the following formula to calculate the correlation coefficient:[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2-(\sum x)^2][n\sum y^2-(\sum y)^2]}}[/tex]And replacing we have:[tex]r=\frac{17(8653.45)-(511.5)(267.1)}{\sqrt{[17(16628.65)-(511.5)^2][17(4549.43)-(267.1)^2]}}=0.933[/tex]