Question

a pediatrician wants to determine the relation that may exist between a childs height and head circumference. she randomly selects 8 children, measures their height and head circumference, and obtains the data shown in the table. the pediatrician wants to use height to predict head circumference. compute the linear correlation coefficient between the height and head circumference of a child. r = (round to three decimal places as needed.)

Explanation:

Step1: Calculate the means of \(x\) (height) and \(y\) (head circumference)

Let \(x\) be the height values and \(y\) be the head - circumference values.
\(\bar{x}=\frac{27 + 25.25+26.5+25.5+27.25+26.5+26.25+27.25}{8}=\frac{211.5}{8}=26.4375\)
\(\bar{y}=\frac{17.4 + 17.2+17.3+16.9+17.4+17.3+17.2+17.4}{8}=\frac{138.1}{8}=17.2625\)

Step2: Calculate \(\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})\), \(\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\) and \(\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}\)

\((x_1-\bar{x})(y_1 - \bar{y})=(27 - 26.4375)(17.4-17.2625)=0.5625\times0.1375 = 0.0771484375\)
\((x_2-\bar{x})(y_2 - \bar{y})=(25.25 - 26.4375)(17.2-17.2625)=(- 1.1875)\times(-0.0625)=0.07421875\)
\((x_3-\bar{x})(y_3 - \bar{y})=(26.5 - 26.4375)(17.3-17.2625)=0.0625\times0.0375 = 0.00234375\)
\((x_4-\bar{x})(y_4 - \bar{y})=(25.5 - 26.4375)(16.9-17.2625)=(-0.9375)\times(-0.3625)=0.34033203125\)
\((x_5-\bar{x})(y_5 - \bar{y})=(27.25 - 26.4375)(17.4-17.2625)=0.8125\times0.1375 = 0.11181640625\)
\((x_6-\bar{x})(y_6 - \bar{y})=(26.5 - 26.4375)(17.3-17.2625)=0.0625\times0.0375=0.00234375\)
\((x_7-\bar{x})(y_7 - \bar{y})=(26.25 - 26.4375)(17.2-17.2625)=(-0.1875)\times(-0.0625)=0.01171875\)
\((x_8-\bar{x})(y_8 - \bar{y})=(27.25 - 26.4375)(17.4-17.2625)=0.8125\times0.1375 = 0.11181640625\)
\(\sum_{i = 1}^{8}(x_{i}-\bar{x})(y_{i}-\bar{y})=0.0771484375+0.07421875+0.00234375+0.34033203125+0.11181640625+0.00234375+0.01171875+0.11181640625 = 0.73193828125\)

\((x_1-\bar{x})^{2}=(27 - 26.4375)^{2}=0.31640625\)
\((x_2-\bar{x})^{2}=(25.25 - 26.4375)^{2}=1.41015625\)
\((x_3-\bar{x})^{2}=(26.5 - 26.4375)^{2}=0.00390625\)
\((x_4-\bar{x})^{2}=(25.5 - 26.4375)^{2}=0.87890625\)
\((x_5-\bar{x})^{2}=(27.25 - 26.4375)^{2}=0.66015625\)
\((x_6-\bar{x})^{2}=(26.5 - 26.4375)^{2}=0.00390625\)
\((x_7-\bar{x})^{2}=(26.25 - 26.4375)^{2}=0.03515625\)
\((x_8-\bar{x})^{2}=(27.25 - 26.4375)^{2}=0.66015625\)
\(\sum_{i = 1}^{8}(x_{i}-\bar{x})^{2}=0.31640625 + 1.41015625+0.00390625+0.87890625+0.66015625+0.00390625+0.03515625+0.66015625=3.96875\)

\((y_1-\bar{y})^{2}=(17.4 - 17.2625)^{2}=0.01890625\)
\((y_2-\bar{y})^{2}=(17.2 - 17.2625)^{2}=0.00390625\)
\((y_3-\bar{y})^{2}=(17.3 - 17.2625)^{2}=0.00140625\)
\((y_4-\bar{y})^{2}=(16.9 - 17.2625)^{2}=0.13140625\)
\((y_5-\bar{y})^{2}=(17.4 - 17.2625)^{2}=0.01890625\)
\((y_6-\bar{y})^{2}=(17.3 - 17.2625)^{2}=0.00140625\)
\((y_7-\bar{y})^{2}=(17.2 - 17.2625)^{2}=0.00390625\)
\((y_8-\bar{y})^{2}=(17.4 - 17.2625)^{2}=0.01890625\)
\(\sum_{i = 1}^{8}(y_{i}-\bar{y})^{2}=0.01890625+0.00390625 + 0.00140625+0.13140625+0.01890625+0.00140625+0.00390625+0.01890625=0.20171875\)

Step3: Calculate the correlation coefficient \(r\)

The formula for the linear correlation coefficient is \(r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}\)
Substitute the values we found:
\(r=\frac{0.73193828125}{\sqrt{3.96875\times0.20171875}}=\frac{0.73193828125}{\sqrt{0.8000000000000001}}\approx\frac{0.73193828125}{0.894427191}=\approx0.818\)

Answer:

\(r = 0.818\)