the following is a set of data from a sample of n = 11 items. complete parts (a) through (c).\nx 18 19 9 7…

Explanation:

Step1: Calculate the means of X and Y

Let (x_i) be the values in the (X) - set and (y_i) be the values in the (Y) - set, (n = 11). The mean of (X), (\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}) (\sum_{i=1}^{11}x_i=18 + 19+9+7+15+12+17+1+3+16+20=137) (\bar{x}=\frac{137}{11}\approx12.455) The mean of (Y), (\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}) (\sum_{i = 1}^{11}y_i=54 + 57+27+21+45+36+51+3+9+48+60=361) (\bar{y}=\frac{361}{11}\approx32.818)

Step2: Calculate the numerator of the covariance formula

The formula for the sample covariance (S_{xy}=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{n - 1}) ((x_1-\bar{x})(y_1-\bar{y})=(18 - 12.455)(54-32.818)=5.545\times21.182 = 117.464) ((x_2-\bar{x})(y_2-\bar{y})=(19 - 12.455)(57-32.818)=6.545\times24.182 = 158.272) ((x_3-\bar{x})(y_3-\bar{y})=(9 - 12.455)(27-32.818)=- 3.455\times(-5.818)=20.191) ((x_4-\bar{x})(y_4-\bar{y})=(7 - 12.455)(21-32.818)=-5.455\times(-11.818)=64.467) ((x_5-\bar{x})(y_5-\bar{y})=(15 - 12.455)(45-32.818)=2.545\times12.182 = 30.903) ((x_6-\bar{x})(y_6-\bar{y})=(12 - 12.455)(36-32.818)=-0.455\times3.182=-1.448) ((x_7-\bar{x})(y_7-\bar{y})=(17 - 12.455)(51-32.818)=4.545\times18.182 = 82.648) ((x_8-\bar{x})(y_8-\bar{y})=(1 - 12.455)(3-32.818)=-11.455\times(-29.818)=341.655) ((x_9-\bar{x})(y_9-\bar{y})=(3 - 12.455)(9-32.818)=-9.455\times(-23.818)=225.299) ((x_{10}-\bar{x})(y_{10}-\bar{y})=(16 - 12.455)(48-32.818)=3.545\times15.182 = 53.821) ((x_{11}-\bar{x})(y_{11}-\bar{y})=(20 - 12.455)(60-32.818)=7.545\times27.182 = 205.188) (\sum_{i = 1}^{11}(x_i-\bar{x})(y_i - \bar{y})=117.464+158.272 + 20.191+64.467+30.903-1.448+82.648+341.655+225.299+53.821+205.188 = 1298.45)

Step3: Calculate the sample covariance

(S_{xy}=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{n - 1}=\frac{1298.45}{10}=129.845)

Answer:

(129.845)