3. 1/4 points details my notes aufqr1 5.4.005. an oceanographer measured the length, in meters, of a deep…
3. 1/4 points details my notes aufqr1 5.4.005. an oceanographer measured the length, in meters, of a deep - water wave and its speed, in meters per second. the results are shown in the following table.\nwave length (m) speed (m/s)\n100 14.6\n125 13.5\n130 15.9\n175 16.9\n210 21.9\n350 24.4\n400 24.8\n(a) find the equation of a linear regression line for the data where wave length is the independent variable, x, and speed is the dependent variable. (round your numerical values to two decimal places.)\n\\( \\hat{y}=\\) x
Answer
Explanation:
Step1: Calculate the means
Let $x$ be the wave - length and $y$ be the speed. $n = 7$ $\bar{x}=\frac{100 + 125+130+175+210+350+400}{7}=\frac{1490}{7}\approx212.86$ $\bar{y}=\frac{14.6 + 13.5+15.9+16.9+21.9+24.4+24.8}{7}=\frac{131}{7}\approx18.71$
Step2: Calculate the numerator and denominator for the slope $b_1$
$S_{xy}=\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})$ $S_{xx}=\sum_{i = 1}^{n}(x_i-\bar{x})^2$
| $x_i$ | $y_i$ | $x_i-\bar{x}$ | $y_i - \bar{y}$ | $(x_i-\bar{x})(y_i - \bar{y})$ | $(x_i-\bar{x})^2$ |
|---|---|---|---|---|---|
| 100 | 14.6 | $100 - 212.86=-112.86$ | $14.6-18.71=-4.11$ | $(-112.86)\times(-4.11) = 463.86$ | $(-112.86)^2=12736.38$ |
| 125 | 13.5 | $125 - 212.86=-87.86$ | $13.5 - 18.71=-5.21$ | $(-87.86)\times(-5.21)=457.75$ | $(-87.86)^2 = 7718.38$ |
| 130 | 15.9 | $130 - 212.86=-82.86$ | $15.9 - 18.71=-2.81$ | $(-82.86)\times(-2.81)=232.84$ | $(-82.86)^2=6865.78$ |
| 175 | 16.9 | $175 - 212.86=-37.86$ | $16.9 - 18.71=-1.81$ | $(-37.86)\times(-1.81)=68.53$ | $(-37.86)^2 = 1433.30$ |
| 210 | 21.9 | $210 - 212.86=-2.86$ | $21.9 - 18.71 = 3.19$ | $(-2.86)\times3.19=-9.12$ | $(-2.86)^2=8.18$ |
| 350 | 24.4 | $350 - 212.86 = 137.14$ | $24.4 - 18.71 = 5.69$ | $137.14\times5.69 = 779.33$ | $137.14^2=18817.38$ |
| 400 | 24.8 | $400 - 212.86 = 187.14$ | $24.8 - 18.71 = 6.09$ | $187.14\times6.09 = 1139.68$ | $187.14^2=34921.58$ |
$S_{xy}=463.86 + 457.75+232.84+68.53-9.12+779.33+1139.68=3132.87$ $S_{xx}=12736.38+7718.38+6865.78+1433.30+8.18+18817.38+34921.58=82491.98$
$b_1=\frac{S_{xy}}{S_{xx}}=\frac{3132.87}{82491.98}\approx0.04$
Step3: Calculate the intercept $b_0$
$b_0=\bar{y}-b_1\bar{x}$ $b_0 = 18.71-0.04\times212.86$ $b_0=18.71 - 8.51=10.20$
The equation of the linear regression line is $\hat{y}=b_0 + b_1x$.
Answer:
$\hat{y}=10.20+0.04x$