When water flows across farmland, some soil is washed away, resulting in erosion. An experiment was conducted to investigate the effect of the rate of water flow (liters per second) on the amount of soil (kilograms) washed away. The data are given in the following table: Flow rate 0.31 0.85 1.26 2.47 3.75 Eroded soil 0.82 1.95 2.18 3.01 6.07 Let xx represent the flow rate variable and yy represent the variable for soil eroded. Then \bar x = 1.73, s_x = 1.38, \bar y = 2.81, s_y = 1.99xˉ=1.73,sx=1.38,yˉ=2.81,sy=1.99 Use this to complete the following calculation of the correlation coefficient for these data. r = \frac{1}{n-1}\left[\left(\frac{x_1 - \bar x}{s_x}\right)\left(\frac{y_1 - \bar y}{s_y}\right) + \left(\frac{x_2 - \bar x}{s_x}\right)\left(\frac{y_2 - \bar y}{s_y}\right) + \cdots + \left(\frac{x_n - \bar x}{s_x}\right)\left(\frac{y_n - \bar y}{s_y}\right)\right]r=n−11[(sxx1−xˉ)(syy1−yˉ)+(sxx2−xˉ)(syy2−yˉ)+⋯+(sxxn−xˉ)(syyn−yˉ)] \; \; \; = \frac{1}{5-1}\left[\left(\frac{x_1 - 1.73}{1.38}\right)\left(\frac{y_1 - 2.81}{1.99}\right) + \left(\frac{x_2 - 1.73}{1.38}\right)\left(\frac{y_2 - 2.81}{1.99}\right) + \cdots + \left(\frac{x_5 - \bar x}{s_x}\right)\left(\frac{y_5 - \bar y}{s_y}\right)\right] =5−11[(1.38x1−1.73)(1.99y1−2.81)+(1.38x2−1.73)(1.99y2−2.81)+⋯+(sxx5−xˉ)(syy5−yˉ)] \; \; \; = \frac{1}{4}\left[\left(\frac{0.31 - 1.73}{1.38}\right)\left(\frac{0.82 - 2.81}{1.99}\right) + \left(\frac{0.85 - 1.73}{1.38}\right)\left(\frac{1.95 - 2.81}{1.99}\right) + \cdots + \left(\frac{x_5 - \bar x}{s_x}\right)\left(\frac{y_5 - \bar y}{s_y}\right)\right] =41[(1.380.31−1.73)(1.990.82−2.81)+(1.380.85−1.73)(1.991.95−2.81)+⋯+(sxx5−xˉ)(syy5−yˉ)] \; \; \; = \frac{1}{4}\left[1.03 + 0.27 + 0.11 + 0.06 + \left(\frac{x_5 - \bar x}{s_x}\right)\left(\frac{y_5 - \bar y}{s_y}\right)\right] =41[1.03+0.27+0.11+0.06+(sxx5−xˉ)(syy5−yˉ)] \; \; \; = \frac{1}{4}\left[1.47 + \left(\frac{x_5 - \bar x}{s_x}\right)\left(\frac{y_5 - \bar y}{s_y}\right)\right] =41[1.47+(sxx5−xˉ)(syy5−yˉ)] What is the value of rr? Give your answer to two decimal places
When water flows across farmland, some soil is washed away, resulting in erosion. An experiment was conducted to investigate the effect of the rate of water flow (liters per second) on the amount of soil (kilograms) washed away. The data are given in the following table:
| Flow rate | 0.31 | 0.85 | 1.26 | 2.47 | 3.75 |
| Eroded soil | 0.82 | 1.95 | 2.18 | 3.01 | 6.07 |
Let xx represent the flow rate variable and yy represent the variable for soil eroded. Then
\bar x = 1.73, s_x = 1.38, \bar y = 2.81, s_y = 1.99xˉ=1.73,sx=1.38,yˉ=2.81,sy=1.99
Use this to complete the following calculation of the
r = \frac{1}{n-1}\left[\left(\frac{x_1 - \bar x}{s_x}\right)\left(\frac{y_1 - \bar y}{s_y}\right) + \left(\frac{x_2 - \bar x}{s_x}\right)\left(\frac{y_2 - \bar y}{s_y}\right) + \cdots + \left(\frac{x_n - \bar x}{s_x}\right)\left(\frac{y_n - \bar y}{s_y}\right)\right]r=n−11[(sxx1−xˉ)(syy1−yˉ)+(sxx2−xˉ)(syy2−yˉ)+⋯+(sxxn−xˉ)(syyn−yˉ)]
\; \; \; = \frac{1}{5-1}\left[\left(\frac{x_1 - 1.73}{1.38}\right)\left(\frac{y_1 - 2.81}{1.99}\right) + \left(\frac{x_2 - 1.73}{1.38}\right)\left(\frac{y_2 - 2.81}{1.99}\right) + \cdots + \left(\frac{x_5 - \bar x}{s_x}\right)\left(\frac{y_5 - \bar y}{s_y}\right)\right] =5−11[(1.38x1−1.73)(1.99y1−2.81)+(1.38x2−1.73)(1.99y2−2.81)+⋯+(sxx5−xˉ)(syy5−yˉ)]
\; \; \; = \frac{1}{4}\left[\left(\frac{0.31 - 1.73}{1.38}\right)\left(\frac{0.82 - 2.81}{1.99}\right) + \left(\frac{0.85 - 1.73}{1.38}\right)\left(\frac{1.95 - 2.81}{1.99}\right) + \cdots + \left(\frac{x_5 - \bar x}{s_x}\right)\left(\frac{y_5 - \bar y}{s_y}\right)\right] =41[(1.380.31−1.73)(1.990.82−2.81)+(1.380.85−1.73)(1.991.95−2.81)+⋯+(sxx5−xˉ)(syy5−yˉ)]
\; \; \; = \frac{1}{4}\left[1.03 + 0.27 + 0.11 + 0.06 + \left(\frac{x_5 - \bar x}{s_x}\right)\left(\frac{y_5 - \bar y}{s_y}\right)\right] =41[1.03+0.27+0.11+0.06+(sxx5−xˉ)(syy5−yˉ)]
\; \; \; = \frac{1}{4}\left[1.47 + \left(\frac{x_5 - \bar x}{s_x}\right)\left(\frac{y_5 - \bar y}{s_y}\right)\right] =41[1.47+(sxx5−xˉ)(syy5−yˉ)]
What is the value of rr? Give your answer to two decimal places
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