# a) Calculate SS_{xx},\ SS_{yy}, and SS_{xy} (Lea

a) Calculate $$\displaystyle{S}{S}_{{\times}},\ {S}{S}_{{{y}{y}}}$$, and $$\displaystyle{S}{S}_{{{x}{y}}}$$ (Leave no cells blank - be certain to enter ''0'' wherever required. Round your answers to 2 decimal places.)
$\begin{array}{|c|c|}\hline \text{College} & \text{Student} & \text{Weekly} & \text{Earnings} & \text{Dollars}(n=5) \\ \hline \text{Hourse Worked (X)} & \text{Weekly Pay (Y)} & (x_{i}-\bar{x})^{2} & (y_{i}-\bar{y})^{2} & (x_{1}-\bar{x})(y_{i}-\bar{y}) \\ \hline 11 & 103 \\ \hline 16 & 187 \\ \hline 16 & 216 \\ \hline 16 & 157 \\ \hline 32 & 262 \\ \hline \\ \hline \bar{x} & \bar{y} & SS_{xx} & SS_{yy} & SS_{xy} \\ \hline \end{array}$
b) )Calculate the sample correlation coefficient. Check your work by using Excel's function =CORREL(array1,array2). (Round your answer to 4 decimal places.)

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Step 1
Given data of College student weekly earnings
$\begin{array}{|c|c|}\hline \text{Hours Worked (X)} & \text{Weekly Pay (Y)} \\ \hline 11 & 103 \\ \hline 16 & 187 \\ \hline 16 & 216 \\ \hline 16 & 157 \\ \hline 32 & 262 \\ \hline \end{array}$
The average hours worked and average weekly pay are calculated as shown below
$$\displaystyle\overline{{{X}}}={\frac{{\sum{X}}}{{{n}}}}$$
$$\displaystyle\overline{{{X}}}={\frac{{{11}+{16}+{16}+{16}+{32}}}{{{5}}}}={18.2}$$
$$\displaystyle\overline{{{Y}}}={\frac{{\sum{Y}}}{{{n}}}}$$
$$\displaystyle\overline{{{Y}}}={\frac{{{103}+{187}+{216}+{157}+{262}}}{{{5}}}}$$
Using the average values the missing values in table are calculated as shown below
$\begin{array}{|c|c|}\hline \text{College} & \text{Student} & \text{Weekly} & \text{Earnings} & \text{Dollars}(n=5) \\ \hline \text{Hourse Worked (X)} & \text{Weekly Pay (Y)} & (x_{i}-\bar{x})^{2} & (y_{i}-\bar{y})^{2} & (x_{1}-\bar{x})(y_{i}-\bar{y}) \\ \hline 11 & 103 & (11-18.2)^{2}=51.84 & (103-185)^{2}=6724 & (11-18.2)(103-185)=590.4 \\ \hline 16 & 187 & (16-18.2)^{2}=4.84 & (187-185)^{2}=4 & (16-18.2)(187-185)=-4.4 \\ \hline 16 & 216 & (16-18.2)^{2}=4.84 & (216-185)^{2}=961 & (16-18.2)(216-185)=-68.2 \\ \hline 16 & 157 & (16-18.2)^{2}=4.84 & (157-185)^{2}=784 & (16-18.2)(157-185)=61.6 \\ \hline 32 & 262 & (32-18.2)^{2} & (262-185)^{2}=5929 & (32-18.2)(262-185)=1062.6 \\ \hline \bar{x}=18.2 & \bar{y}=185 & S_{xx}=256.8 & S_{yy}=14402 & S_{xy}=1642 \\ \hline \bar{x} & \bar{y} & SS_{xx} & SS_{yy} & SS_{xy} \\ \hline \end{array}$
Step 2
b) The correlation coefficient is calculated using the below mentioned formula
$$\displaystyle{r}={\frac{{{S}_{{{x}{y}}}}}{{\sqrt{{{S}_{{\times}}{S}_{{{y}{y}}}}}}}}$$
$$\displaystyle{r}={\frac{{{1642}}}{{\sqrt{{{256.8}\times{14402}}}}}}={0.853816}$$
using the excel function CORREL(X data,Y data) we get the sample correlation coefficient as 0.853815903
Rounding to 4 decimal places we get same value of r in both methods. $$\displaystyle{r}={0.8538}$$
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Raymond Foley
Could you describe the solution to this problem in more detail? The answers were not found anywhere else. Thanks.